Frank Brothers Solutions for ICSE Class 10 Physics Chapter 4 Current Electricity

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Question 1. What is an electric cell? State the energy transformation that occurs in a cell when it is in use.
Answer: An electric cell is a system that maintains a steady potential difference between its two conductor terminals through a chemical process. Consequently, the cell functions as a source of electric current or electrons. During its operation, it converts stored chemical energy into electrical energy.
In simple words: An electric cell is a device that uses chemical reactions to keep electricity flowing. It works by turning chemical energy into electrical energy.

Exam Tip: Remember that a cell maintains potential difference through a chemical reaction. Mention 'conversion of chemical energy to electrical energy' for full marks.

Question 2. What is a battery? Explain how cells are connected to form a battery, and represent an electric cell and a battery of two cells with their symbols.
Answer: An electric cell generates electricity from stored chemical substances. When these internal chemicals are fully consumed, the cell ceases to produce current. It contains two terminals, namely positive and negative.
Connecting multiple cells together, such that the positive end of one cell links to the negative end of the succeeding cell, forms a battery. Thus, a battery represents a linked group of two or more individual cells.
The symbols are represented as:
In simple words: A battery is a group of two or more cells joined together. The positive side of one cell is connected to the negative side of the next to make the electricity stronger.

Exam Tip: Clearly show the difference in line lengths for positive and negative terminals in the cell symbol. The longer line is always positive.

Question 3. Define an electric circuit. How is it usually represented?
Answer: An electrical circuit is a continuous, closed path made of conducting wires and electrical resistances connected between the terminals of a battery, facilitating the flow of current. It is conventionally illustrated using a circuit diagram.
In simple words: An electric circuit is an unbroken loop of wire and other parts that lets electricity travel from one end of a battery to the other.

Exam Tip: Always define an electric circuit as a 'continuous and closed path'. Leaving out 'closed' is a very common trap.

Question 4. List four essential components of an electrical circuit and show their standard circuit symbols.
Answer: The four primary parts of an electrical loop include:
(i) **Cell**: Serves as the voltage source.
(ii) **A load (bulb)**: Consumes the electrical energy and indicates current flow.
(iii) **Key**: Controls the opening or closing of the circuit.
(iv) **Connecting wire**: Provides the path for current to flow.
The symbols are represented below:
(i) Cell
(ii) A load (bulb)
(iii) Key
(iv) Connecting wire
In simple words: An electrical circuit is made of a power source (cell), a user of electricity (bulb), a control switch (key), and metal tracks (wires) to join them together.

Exam Tip: In diagrams, draw key symbols carefully. A closed key must have a central dot, whereas an open key must not.

Question 5. What is a circuit diagram?
Answer: A circuit diagram is a graphical representation that shows how different electrical components are connected to one another using their standard symbolic representations.
In simple words: A circuit diagram is a drawing that uses simple symbols to show how wires, batteries, and bulbs are connected in a circuit.

Exam Tip: A circuit diagram must always use standard symbols. Never draw actual physical sketches of bulbs or batteries.

Question 6. In a given electrical circuit, an ammeter is connected in parallel and a voltmeter is connected in series. Explain why this setup is incorrect and draw the correct circuit diagram.
Answer: In the described configuration, the ammeter is incorrectly wired in parallel while the voltmeter is incorrectly placed in series. Additionally, the polarity of both devices is improperly linked to the battery. For correct operation, an ammeter (which has low resistance) must be wired in series to measure current, and a voltmeter (which has high resistance) must be connected in parallel across the resistor to measure voltage drop.
The correct configuration is illustrated below:
In simple words: An ammeter measures current and must go in the main loop (series), while a voltmeter measures voltage drop across a device and must connect across it in parallel.

Exam Tip: Remember: Ammeters have low resistance and must be connected in series, while voltmeters have high resistance and must be connected in parallel.

Question 7. State Ohm's law. Is this law universally applicable to all electrical conductors? Explain with examples.
Answer: Ohm's law states that the current flowing through a conductor is directly proportional to the potential difference applied across its two ends, provided the physical conditions (such as temperature, tension, etc.) of the conductor remain constant.
No, this relationship is not always true. Certain devices, like diode valves, junction diodes, transistors, and thermistors, do not obey Ohm's law and are classified as non-ohmic conductors.
In simple words: Ohm's law says that current increases in direct proportion to voltage, as long as the temperature of the wire doesn't change. This isn't true for everything; devices like diodes do not follow this rule.

Exam Tip: Ohm's law holds true only if 'temperature and physical conditions remain constant'. This is a critical qualifying phrase.

Question 8. What is electrical resistance? Compare it with a mechanical phenomenon and state its SI unit.
Answer: Electrical resistance is the opposition or obstruction offered by a conducting wire to the flow of electric current. It is analogous to mechanical friction, arising from collisions between moving free electrons and the fixed atoms of the conductor. Its standard SI unit is the 'ohm' (\(\Omega\)).
In simple words: Resistance is how much a wire slows down the flow of electricity. It is like friction for electrons as they bump into atoms inside the wire. It is measured in ohms.

Exam Tip: Do not confuse resistance with resistivity. Resistance depends on length and area, whereas resistivity is a constant material property.

Question 9. Name the physical quantity whose SI unit is 'ohm'.
Answer: The physical quantity measured in ohms is 'resistance'.
In simple words: The unit 'ohm' is used to measure electrical resistance.

Exam Tip: Ensure you write the symbol Ohm (\(\Omega\)) correctly. Do not write the word 'ohm' if the unit symbol is requested.

Question 10. Define 1 ohm of resistance.
Answer: The resistance of a conductor is defined as 1 ohm if a current of 1 ampere passes through it when a potential difference of 1 volt is applied across its ends.
In simple words: A wire has a resistance of 1 ohm if 1 volt of electrical push makes 1 ampere of current flow through it.

Exam Tip: To define 1 ohm, use the ratio of 1 volt to 1 ampere. This makes the definition mathematically rigorous.

Question 11. Does Ohm's law hold good for semiconductors and electrolytic solutions?
Answer: No, Ohm's law does not apply to semiconductors or electrolytic solutions.
In simple words: No, semiconductors and chemical solutions do not follow Ohm's law.

Exam Tip: Always remember that semiconductors and electrolytes are non-ohmic conductors and do not obey Ohm's law.

Question 12. List the factors on which the electrical resistance of a conductor depends, explaining how each factor affects the resistance.
Answer: The resistance of a conducting medium depends on the following four factors:
(i) **Nature of the material**: Different substances have varying concentrations of free charge carriers (electrons), meaning resistance is highly material-dependent.
(ii) **Length of the conductor**: The resistance of a conductor is directly proportional to its length (\( R \propto l \)).
(iii) **Cross-sectional area**: The resistance is inversely proportional to the cross-sectional area of a uniform wire (\( R \propto \frac{1}{A} \)).
(iv) **Temperature**: For metallic conductors, resistance generally increases as the temperature rises.
In simple words: The resistance of a wire depends on: (i) what it is made of, (ii) how long it is (longer means more resistance), (iii) how thick it is (thinner means more resistance), and (iv) its temperature (hotter metals have more resistance).

Exam Tip: This is a frequent long-answer question. List all four factors clearly with their mathematical proportionalities.

Question 13. What is meant by an equivalent resistor?
Answer: An equivalent resistor is a single resistor that can replace a combination of multiple resistors in a circuit without changing the total current flowing through the circuit.
In simple words: An equivalent resistor is one single resistor that does the exact same job as a group of resistors connected together.

Exam Tip: The equivalent resistor must draw the exact same total current from the source as the combination it replaces.

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Question 14. Calculate the current flowing through a resistor of resistance \( 60\ \Omega \) when connected across a potential difference of \( 24\ \text{V} \).
Answer: Using Ohm's law, the relation for electric current is: \[ \text{Current } (I) = \frac{\text{Potential difference } (V)}{\text{Resistance } (R)} \] Given that:
- Potential difference, \( V = 24\ \text{V} \)
- Resistance, \( R = 60\ \Omega \)
Substituting these values: \[ I = \frac{24}{60} = 0.4\ \text{A} \] Therefore, the current passing through the resistor is \( 0.4\ \text{A} \).
In simple words: To find the current, divide the voltage (24 V) by the resistance (60 ohms). This gives 0.4 amperes.

Exam Tip: In Ohm's law calculations, always check your units and state the formula \( I = V/R \) before substituting values.

Question 15. What physical quantity is represented by the slope of a potential difference versus current (\(V\text{-}I\)) graph?
Answer: The slope of the potential difference versus current graph, represented as \( \frac{dV}{dI} \), indicates the electrical resistance of the conductor.
In simple words: The slope of a voltage-current graph tells us the resistance of the wire.

Exam Tip: Remember: Slope \( dV/dI \) represents resistance, whereas slope \( dI/dV \) represents conductance. Read the axes carefully.

Question 16. Define the term resistivity (or specific resistance) of a material.
Answer: The resistivity of a material is defined as the electrical resistance of a wire made of that material having a unit length and a unit cross-sectional area.
In simple words: Resistivity is the natural resistance of a standard piece of a material that is 1 meter long and has an area of 1 square meter.

Exam Tip: Resistivity is independent of the length and area of the wire; it only depends on the material type and temperature.

Question 17. State the SI unit of specific resistance (resistivity).
Answer: The SI unit used to measure specific resistance is the ohm-meter (\( \Omega \cdot \text{m} \)).
In simple words: Resistivity is measured in a unit called ohm-meter.

Exam Tip: The correct spelling and unit symbol is ohm-meter (\(\Omega \cdot \text{m}\)). Do not write it as ohm/meter.

Question 18. What is electrical conductance? State its SI unit.
Answer: Conductance represents the reciprocal of the electrical resistance of a conductor. Its SI unit is the 'mho' (\( \Omega^{-1} \) or siemens).
In simple words: Conductance is the opposite of resistance. It measures how easily electricity flows and its unit is called mho.

Exam Tip: Conductance is the mathematical reciprocal of resistance. Its unit 'mho' is the reverse spelling of 'ohm'.

Question 19. Give one example each of: (i) a metal, (ii) an alloy, and (iii) a semiconductor.
Answer:
(i) **Metal**: Copper (\(\text{Cu}\))
(ii) **Alloy**: Constantan
(iii) **Semiconductor**: Germanium (\(\text{Ge}\))
In simple words: Copper is a metal, constantan is an alloy, and germanium is a semiconductor.

Exam Tip: Keep standard examples like copper for metals and germanium for semiconductors memorized for quick marks.

Question 20. In a given circuit, two \( 2\ \Omega \) resistors are connected in series, and this combination is connected in parallel with another \( 2\ \Omega \) resistor. Find the equivalent resistance of the circuit and the current flowing through it when connected across a potential difference of \( 3\ \text{V} \).
Answer: In the specified circuit, two \( 2\ \Omega \) resistors connected in series are arranged in parallel with a third \( 2\ \Omega \) resistor.
First, the series combination resistance is: \[ R_{\text{series}} = 2\ \Omega + 2\ \Omega = 4\ \Omega \]
Next, the total equivalent resistance \( R \) of the parallel arrangement is: \[ R = \left[ \frac{1}{4} + \frac{1}{2} \right]^{-1} \] \[ R = \left[ \frac{3}{4} \right]^{-1} = \frac{4}{3}\ \Omega \]
Using Ohm's law with potential difference \( V = 3\ \text{V} \), the total current \( I \) is: \[ I = \frac{V}{R} = \frac{3}{\frac{4}{3}} = 2.25\ \text{A} \] Thus, the equivalent resistance is \( \frac{4}{3}\ \Omega \) and the current is \( 2.25\ \text{A} \).
In simple words: First, add the two series resistors (2 + 2 = 4 ohms). Then, find the parallel resistance of 4 ohms and 2 ohms, which is 1.33 ohms. Finally, divide 3 volts by 1.33 ohms to get 2.25 amperes of current.

Exam Tip: For combined networks, solve the series parts first, then calculate the parallel equivalent to simplify your working.

Question 21. State the relationship between current and potential difference as defined by Ohm's law.
Answer: Ohm's law defines the relation between the electric current in a conductor and the potential difference maintained across its terminals. It states that the current is directly proportional to this potential difference, assuming physical parameters and temperature remain constant.
In simple words: Ohm's law states that current is directly proportional to voltage as long as temperature remains unchanged.

Exam Tip: State Ohm's law clearly as a direct proportionality between current and voltage at a constant temperature.

Question 22. Explain why a thinner wire offers more electrical resistance than a thicker wire of the same material. Also, solve the following numerical problem: In a circuit, the heat dissipated is \( 60\ \text{J} \), charge is \( 20\ \text{C} \), and time is \( 5\ \text{s} \). Find (a) the potential difference across the resistor, (b) the resistance of the resistor, and (c) the power dissipated.
Answer: When a conductor is made thinner, the available cross-sectional area for electron flow decreases, which results in a greater number of collisions and thus a larger resistance.
Given:
- Heat/Work done, \( W = 60\ \text{J} \)
- Charge, \( Q = 20\ \text{C} \)
- Time, \( t = 5\ \text{s} \)

**(a) Potential difference across the resistor (\(V\)):** \[ V = \frac{W}{Q} = \frac{60}{20} = 3\ \text{V} \]
**(b) Resistance of the resistor (\(R\)):** Using the relation for electrical power: \[ P = \frac{W}{t} = \frac{V^2}{R} \] \[ R = V^2 \times \frac{t}{W} = (3)^2 \times \frac{5}{60} = 9 \times \frac{1}{12} = 0.75\ \Omega \]
**(c) Power dissipated (\(P\)):** \[ P = \frac{W}{t} = \frac{60}{5} = 12\ \text{W} \]
In simple words: Thinner wires have less room for electrons to pass, causing more resistance. (a) Voltage is energy divided by charge, which is 3 V. (b) Power is energy over time (12 W), which lets us calculate resistance as 0.75 ohms. (c) Power is 12 watts.

Exam Tip: For power calculations, keep the units consistent. Convert energy from Joules and time to seconds.

Question 23. If the current flowing through a resistor is plotted against the potential difference at a constant temperature, what will be the shape of the graph? Sketch this graph.
Answer: At a constant temperature, plotting the current against the potential difference yields a straight line passing through the origin, where the slope of the line relative to the axes represents the electrical conductance.
In simple words: The graph of current versus voltage is a straight line starting at zero, showing that current increases in step with voltage.

Exam Tip: A straight line starting from the origin in a V-I graph confirms that the conductor is ohmic.

Question 24. What instrument is used to measure electric current? How is it connected in a circuit and what is the nature of its resistance?
Answer: An ammeter is the instrument utilized to measure electric current. It is characterized by having a very low electrical resistance and is always connected in series with the circuit components.
In simple words: An ammeter measures current. It has low resistance and must always be connected in series.

Exam Tip: An ammeter must have low resistance to avoid altering the current in the circuit branch it measures.

Question 25. What instrument is used to measure potential difference? How is it connected in a circuit and what is the nature of its resistance?
Answer: A voltmeter is the instrument used to measure potential difference. It possesses a very high electrical resistance and is always connected in parallel across the component whose voltage is being measured.
In simple words: A voltmeter measures voltage. It has high resistance and must always be connected in parallel.

Exam Tip: A voltmeter must have high resistance so that it does not draw significant current from the parallel branch.

Question 26. State whether an ammeter has high or low resistance.
Answer: An ammeter possesses a very low electrical resistance.
In simple words: Ammeters have very low resistance.

Exam Tip: Ammeter resistance should ideally be zero, but practically it is kept as low as possible.

Question 27. State whether a voltmeter has high or low resistance.
Answer: A voltmeter possesses a very high electrical resistance.
In simple words: Voltmeters have very high resistance.

Exam Tip: Voltmeter resistance should ideally be infinite to prevent altering circuit current.

Question 28. What is the function of a key (or switch) in an electrical circuit?
Answer: A key functions as a switch in an electrical circuit, helping to complete (close) or break (open) the path of current flow as needed.
In simple words: A key works like a switch to turn a circuit on or off.

Exam Tip: A key acts as an electrical gate. When open, current is zero because the air gap has infinite resistance.

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Question 29. What is a rheostat? What is its other name?
Answer: A rheostat is an electrical component used to adjust or control the magnitude of current in a circuit without altering the voltage of the source. It is commonly referred to as a variable resistance.
In simple words: A rheostat lets you change the amount of current in a circuit by adjusting the resistance without changing the battery.

Exam Tip: A rheostat adjusts current by physically changing the length of the resistance wire in the circuit path.

Question 30. Calculate the current passing through a resistor of \( 60\ \Omega \) when connected to a potential difference of \( 30\ \text{V} \).
Answer: According to Ohm's law, we have: \[ I = \frac{V}{R} \] Given:
- Potential difference, \( V = 30\ \text{V} \)
- Resistance, \( R = 60\ \Omega \)
Substituting the values: \[ I = \frac{30}{60} = 0.5\ \text{A} \] Therefore, the electric current is \( 0.5\ \text{A} \).
In simple words: Divide the voltage (30 V) by the resistance (60 ohms) to find the current, which is 0.5 amperes.

Exam Tip: Always write out Ohm's law formula before plugging in values to get partial credit if your calculation goes wrong.

Question 31. Find the resistance of a conductor if a potential difference of \( 0.1\ \text{V} \) across its ends produces a current of \( 2\ \text{A} \).
Answer: Using Ohm's law: \[ R = \frac{V}{I} \] Given values:
- Potential difference, \( V = 0.1\ \text{V} \)
- Current, \( I = 2\ \text{A} \)
Substituting these: \[ R = \frac{0.1}{2} = 0.05\ \Omega \] Hence, the resistance of the conductor is \( 0.05\ \Omega \).
In simple words: Divide the voltage (0.1 V) by the current (2 A) to get a resistance of 0.05 ohms.

Exam Tip: A common mistake in decimals is placing the decimal point incorrectly. Double-check your division: \(0.1 / 2 = 0.05\).

Question 32. An electric iron connected to a \( 240\ \text{V} \) supply line draws a current of \( 6\ \text{A} \). Find its resistance.
Answer: By Ohm's law, we write: \[ R = \frac{V}{I} \] Here:
- Potential difference, \( V = 240\ \text{V} \)
- Current, \( I = 6\ \text{A} \)
Substituting the values: \[ R = \frac{240}{6} = 40\ \Omega \] Thus, the resistance of the electric iron is \( 40\ \Omega \).
In simple words: Divide the voltage (240 V) by the current (6 A) to find the resistance, which is 40 ohms.

Exam Tip: An electric iron draws a relatively high current, so its resistance must be lower compared to low-power appliances.

Question 33. Two \( 2\ \Omega \) resistors are connected in parallel, and this combination is connected in series with a \( 1\ \Omega \) resistor across a \( 2\ \text{V} \) battery. Calculate the equivalent resistance of the circuit and the current flowing through it.
Answer: Let \( R \) be the total equivalent resistance, and \( I \) be the current flowing through the circuit.
First, find the equivalent resistance of the parallel combination of the two \( 2\ \Omega \) resistors: \[ R_p = \left( \frac{1}{2} + \frac{1}{2} \right)^{-1} = 1\ \Omega \]
Now, adding the \( 1\ \Omega \) series resistor: \[ R = 1\ \Omega + 1\ \Omega = 2\ \Omega \]
Using Ohm's law with potential difference \( V = 2\ \text{V} \): \[ I = \frac{V}{R} = \frac{2}{2} = 1\ \text{A} \] Therefore, the equivalent resistance is \( 2\ \Omega \) and the current is \( 1\ \text{A} \).
In simple words: Two 2-ohm resistors in parallel equal 1 ohm. Adding the 1-ohm resistor in series gives 2 ohms in total. 2 volts divided by 2 ohms gives 1 ampere of current.

Exam Tip: For mixed parallel-series circuits, always simplify the parallel section first before adding the series resistance.

Question 34. What potential difference is needed to send a current of \( 1\ \text{A} \) through a resistor of \( 5\ \Omega \)?
Answer: Let \( V \) represent the required potential difference.
Given:
- Current, \( I = 1\ \text{A} \)
- Resistance, \( R = 5\ \Omega \)
According to Ohm's law: \[ V = I \times R = (1) \times (5) = 5\ \text{V} \] Thus, the potential difference is \( 5\ \text{V} \).
In simple words: Multiply the current (1 A) by the resistance (5 ohms) to get the potential difference, which is 5 volts.

Exam Tip: The formula \( V = IR \) is the most common tool in circuit analysis. Always verify the calculation using division.

Question 35. Find the current flowing through a resistor of \( 3\ \Omega \) connected across a potential difference of \( 6\ \text{V} \).
Answer: Let \( I \) be the current.
Given:
- Potential difference, \( V = 6\ \text{V} \)
- Resistance, \( R = 3\ \Omega \)
Using Ohm's law: \[ I = \frac{V}{R} = \frac{6}{3} = 2\ \text{A} \] Thus, the current is \( 2\ \text{A} \).
In simple words: Divide the voltage (6 V) by the resistance (3 ohms) to get 2 amperes of current.

Exam Tip: Current is directly proportional to voltage; doubling the voltage across a constant resistor will double the current.

Question 36. Calculate the resistance of a conductor if a potential difference of \( 20\ \text{V} \) is applied across it to produce a current of \( 2\ \text{A} \).
Answer: By Ohm's law, we have: \[ R = \frac{V}{I} \] Given:
- Potential difference, \( V = 20\ \text{V} \)
- Current, \( I = 2\ \text{A} \)
Substituting these values: \[ R = \frac{20}{2} = 10\ \Omega \] Hence, the resistance is \( 10\ \Omega \).
In simple words: Divide the voltage (20 V) by the current (2 A) to find the resistance, which is 10 ohms.

Exam Tip: Resistance is a physical constant of the conductor at a given temperature; changing voltage does not change the resistance.

Question 37. Between two points A and B, a series combination of two resistors of \( 2\ \Omega \) each is connected in parallel with a third \( 1\ \Omega \) resistor. Find the equivalent resistance between A and B.
Answer: The series combination of the two \( 2\ \Omega \) resistors yields: \[ R_s = 2 + 2 = 4\ \Omega \] Now, this \( 4\ \Omega \) equivalent resistance is in parallel with the \( 1\ \Omega \) resistor between points A and B: \[ R_p = \left( \frac{1}{4} + \frac{1}{1} \right)^{-1} = \left( \frac{5}{4} \right)^{-1} = \frac{4}{5}\ \Omega = 0.8\ \Omega \] Therefore, the equivalent resistance between A and B is \( \frac{4}{5}\ \Omega \) (or \( 0.8\ \Omega \)).
In simple words: First, add the two 2-ohm resistors in series to get 4 ohms. Then, find the parallel resistance of 4 ohms and 1 ohm, which is 0.8 ohms.

Exam Tip: When resistors are in series, their equivalent resistance is always greater than any individual resistor.

Question 38. Five resistors of \( 2\ \Omega \) each are connected (a) in series, and (b) in parallel. Find the equivalent resistance in each case.
Answer:
**(a) In series:** The equivalent resistance is the sum of the individual resistances: \[ R_s = 2 + 2 + 2 + 2 + 2 = 10\ \Omega \]
**(b) In parallel:** The equivalent resistance is: \[ R_p = \frac{\text{Resistance of one resistor}}{\text{Number of resistors}} = \frac{2}{5}\ \Omega = 0.4\ \Omega \]
In simple words: (a) In series, simply add the five resistors (2+2+2+2+2) to get 10 ohms. (b) In parallel, divide the value of one resistor (2 ohms) by the total number of resistors (5) to get 0.4 ohms.

Exam Tip: In parallel, the total resistance is always less than the smallest individual resistor in the combination.

Question 39. How should five resistors of \( 1\ \Omega \) each be connected to obtain an equivalent resistance of \( 0.2\ \Omega \)?
Answer: To achieve an equivalent resistance of \( 0.2\ \Omega \), all five \( 1\ \Omega \) resistors must be connected in a parallel configuration. \[ R_p = \frac{\text{Resistance of each resistor}}{\text{Number of resistors}} = \frac{1}{5}\ \Omega = 0.2\ \Omega \]
In simple words: Connecting the five 1-ohm resistors in parallel gives an equivalent resistance of 0.2 ohms.

Exam Tip: To get a very small resistance from several larger identical resistors, always connect them in parallel.

Question 40. State two practical uses of electrical conductors.
Answer: Two common applications of electrical conductors are:
(i) Copper is widely used to manufacture connecting wires due to its low resistivity.
(ii) Conducting solutions are utilized as electrolytes in batteries and chemical cells.
In simple words: Conductors are used to make electrical wires (like copper wires) and as the liquid inside battery cells.

Exam Tip: Copper is chosen for connecting wires because of its high conductivity, which minimizes energy losses due to heating.

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Question 41. State the SI units of the following physical quantities: (a) Electric current, (b) Potential difference, (c) Charge.
Answer: The standard SI units are as follows:
(a) The SI unit for electric current is the **Ampere**.
(b) The SI unit for potential difference is the **Volt**.
(c) The SI unit for electric charge is the **Coulomb**.
In simple words: (a) We measure current in amperes, (b) voltage in volts, and (c) electric charge in coulombs.

Exam Tip: Make sure to use the correct symbol abbreviations (\(\text{A}\), \(\text{V}\), \(\text{C}\)) in numerical solutions to avoid losing presentation marks.

Question 42. Define the term potential difference between two points.
Answer: The potential difference across any two points represents the amount of work required to transport a unit positive charge from the starting point to the destination point.
In simple words: Potential difference is the work needed to carry a single positive charge between two spots.

Exam Tip: Highlight the phrase 'unit positive charge' in your definition, as it is a crucial keyword examiners look for.

Question 43. Is electrical potential a scalar or a vector quantity?
Answer: Electrical potential is classified as a scalar physical quantity because it possesses only magnitude and lacks directional attributes.
In simple words: Electric potential is a scalar quantity, meaning it has a size but no specific direction.

Exam Tip: Remember that work and charge are scalars, which is why electrical potential (work/charge) is also a scalar quantity.

Question 44. Define electric intensity. State its SI unit and whether it is a scalar or a vector quantity?
Answer: Electric intensity at any given location is the electrostatic force exerted on a unit positive test charge placed at that specific spot. It is measured in the SI unit of newtons per coulomb (\(\text{N/C}\) or \(\text{N C}^{-1}\)). Since it has a specific direction, it is a vector quantity.
In simple words: Electric intensity is the force pushing on a unit positive charge at a certain spot. It is a vector quantity and its unit is newton per coulomb.

Exam Tip: Do not confuse electric potential (scalar) with electric intensity (vector). Electric intensity always has a defined direction parallel to the force on a positive charge.

Question 45. What is the other name of the unit joule per coulomb (J/C)?
Answer: The ratio of joules per coulomb (\(\text{J/C}\)) is commonly referred to as the **volt** (\(\text{V}\)).
In simple words: One Joule per Coulomb is called a volt.

Exam Tip: This relation (\(1\text{ V} = 1\text{ J/C}\)) is extremely helpful for cross-checking dimensions in electrostatic formulas.

Question 46. Calculate the work done in moving a charge of \(5\ \text{C}\) across a potential difference of \(1\ \text{V}\).
Answer: The work performed (\(W\)) during the displacement of a charge is given by: \[ W = \text{charge } (Q) \times \text{potential difference } (V) \] Given values:
- Charge, \(Q = 5\ \text{C}\)
- Potential difference, \(V = 1\ \text{V}\)
Substituting these: \[ W = 5 \times 1 = 5\ \text{J} \] Thus, the work done is \(5\ \text{J}\).
In simple words: To find the work done, multiply the charge of 5 coulombs by the 1-volt push, which gives 5 Joules of energy.

Exam Tip: Always specify the final unit in Joules (\(\text{J}\)) since work is a form of energy.

Question 47. Which of the following physical quantities has 'volt' as its SI unit? (a) Electric current, (b) Potential difference, (c) Resistance, (d) Electrical power.
Answer: (b) potential difference
In simple words: The unit of potential difference is the volt.

Exam Tip: Remember that both potential difference and electromotive force (emf) are measured in volts.

Question 48. Which instrument is used to measure potential difference? (a) Ammeter, (b) Galvanometer, (c) Rheostat, (d) Voltmeter.
Answer: (d) voltmeter
In simple words: We use a voltmeter to measure the voltage across a resistor.

Exam Tip: In multiple-choice questions, remember that a voltmeter is connected in parallel, while an ammeter is connected in series.

Question 49. Under what condition do free charges move in a definite direction in a conductor?
Answer: The mobile charge carriers inside a conducting material align and travel in a coordinated direction once an external potential difference is established across the conductor's boundaries.
In simple words: Charges flow in one set direction through a wire only when you connect it to a voltage source like a battery.

Exam Tip: Explain that without a potential difference, free electrons move randomly in all directions, resulting in zero net current.

Question 50. How can you maintain a constant potential difference across the ends of a conductor?
Answer: A steady potential difference can be successfully maintained at the boundaries of a conducting path by linking its opposite terminals directly to the positive and negative ends of an electric cell or battery.
In simple words: You can keep a steady voltage push across a wire by connecting its ends to a battery.

Exam Tip: Emphasize that chemical reactions inside the cell continuously replenish the charges to maintain this potential difference.

Question 51. Define electric current and state its mathematical expression.
Answer: The rate at which electrical charge migrates across a cross-section of a conducting wire is defined as electric current. It represents the flow rate of charge over time: \[ I = \frac{Q}{t} \] where \(I\) stands for current, \(Q\) represents charge, and \(t\) is time.
In simple words: Electric current is how much electric charge flows through a wire every second.

Exam Tip: Ensure you define current as the *rate* of charge flow. Simply writing 'flow of charges' is technically incomplete.

Question 52. Compare the direction of conventional current with the direction of electronic current.
Answer: By scientific convention, the path of conventional current is aligned with the trajectory of positive charge carriers. Conversely, the flow direction of negative electrons (electronic current) is directly opposite to this conventional path.
In simple words: Conventional current is imagined to go from positive to negative, which is the exact opposite of how real electrons actually travel.

Exam Tip: When labeling circuit diagrams, always draw the current arrow in the direction of conventional current (away from the positive terminal of the cell).

Question 53. What does electric current define in a conductor?
Answer: Electric current characterizes how rapidly electric charge carriers pass through a given cross-section of a conducting path.
In simple words: Current is the measure of how fast electric charges move through a wire.

Exam Tip: Remember that the unit of current is Ampere, which is equal to one Coulomb per second (\(1\text{ A} = 1\text{ C/s}\)).

Question 54. Name the instrument used to measure electric current.
Answer: An **ammeter** is the specialized device utilized to gauge the amount of electric current flowing in a circuit.
In simple words: We use an ammeter to measure electric current.

Exam Tip: State clearly that an ammeter has very low resistance so it does not alter the current it is measuring.

Question 55. If the charge on a single electron is \(1.6 \times 10^{-19}\ \text{C}\), calculate the number of electrons that constitute \(1\ \text{C}\) of charge.
Answer: Given that the charge of a single electron is \(1.6 \times 10^{-19}\ \text{C}\), the total count of electrons (\(n\)) required to make up a charge of \(1\ \text{C}\) is calculated using the quantization of charge formula: \[ Q = n \cdot e \implies n = \frac{Q}{e} \] Substituting \(Q = 1\ \text{C}\) and \(e = 1.6 \times 10^{-19}\ \text{C}\): \[ n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}\text{ electrons} \]
In simple words: It takes about \(6.25 \times 10^{18}\) electrons grouped together to make just 1 coulomb of electric charge.

Exam Tip: Show the calculation step clearly using division, as this is a very common two-mark numerical problem.

Question 56. A battery of emf \(15\ \text{V}\) and internal resistance \(3\ \Omega\) is connected in series with two resistors of resistances \(3\ \Omega\) and \(6\ \Omega\). Calculate: (i) the electric current in the circuit, and (ii) the terminal potential difference across the battery.
Answer: Given:
- Electromotive force of cell, \(e = 15\ \text{V}\)
- Internal resistance, \(r = 3\ \Omega\)
- External resistance consisting of two series-connected resistors: \[ R_s = 3\ \Omega + 6\ \Omega = 9\ \Omega \]
**(i) Calculation of Electric Current (\(I\)):** The total resistance of the entire circuit is the sum of the external and internal resistances: \[ R_{\text{total}} = R_s + r = 9\ \Omega + 3\ \Omega = 12\ \Omega \] Using Ohm's law, current is: \[ I = \frac{e}{R_{\text{total}}} = \frac{15}{12} = 1.25\ \text{A} \]
**(ii) Calculation of Terminal Potential Difference (\(V\)):** The internal potential drop across the cell's internal resistance is: \[ \text{Voltage drop} = I \times r = 1.25 \times 3 = 3.75\ \text{V} \] Thus, the terminal potential difference is: \[ V = e - (I \cdot r) = 15 - 3.75 = 11.25\ \text{V} \]
In simple words: (i) First, add up all resistances (3 + 6 + 3 = 12 ohms). Dividing the 15 V battery voltage by 12 ohms gives 1.25 amperes of current. (ii) The battery loses 3.75 volts inside itself, so the voltage left at the terminals is 11.25 volts.

Exam Tip: Remember that terminal potential difference is always less than the EMF when a current is being drawn from the cell.

Question 57. A wire of uniform thickness and resistance \(27\ \Omega\) is cut into three equal pieces. If these pieces are then connected in parallel, find their equivalent resistance.
Answer: When a uniform wire having an initial resistance of \(27\ \Omega\) is sliced into three identical pieces, the resistance of each segment drops proportionally: \[ R_{\text{piece}} = \frac{27}{3} = 9\ \Omega \] If these three \(9\ \Omega\) segments are connected in a parallel configuration, their total equivalent resistance (\(R_p\)) is calculated as: \[ \frac{1}{R_p} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} \] \[ R_p = \frac{9}{3} = 3\ \Omega \]
In simple words: Cutting the wire into three equal parts gives three 9-ohm wires. Connecting them in parallel divides the resistance of one piece by three, resulting in a total resistance of 3 ohms.

Exam Tip: For \(N\) identical resistors connected in parallel, the equivalent resistance is simply \(R/N\).

Question 58. If you are given a combination of resistors and want to obtain an equivalent resistance of less than \(2\ \Omega\), how should you connect them? Explain the reason.
Answer: To obtain a net resistance that is lower than \(2\ \Omega\), the components must be configured in a parallel arrangement. This is because a parallel connection always produces an equivalent resistance that is smaller than the resistance of the single smallest resistor included in the network.
In simple words: You should connect them in parallel. This is because a parallel setup always lowers the overall resistance below the value of the smallest resistor used.

Exam Tip: In comparison, series connections always yield an equivalent resistance larger than the largest individual resistor.

Question 59. In a given circuit network, a series combination of two resistors \(r_1\) and \(r_2\) is further connected in series with a parallel combination of two other resistors \(r_3\) and \(r_4\). Write down the expression for the total equivalent resistance of this network.
Answer: In this electrical network, we analyze the circuit in separate stages:
- First, the equivalent resistance of the series-connected pair \(r_1\) and \(r_2\) is: \[ R_s = r_1 + r_2 \]
- Second, the equivalent resistance of the parallel-connected pair \(r_3\) and \(r_4\) is: \[ R_p = \left( \frac{1}{r_3} + \frac{1}{r_4} \right)^{-1} = \frac{r_3 \cdot r_4}{r_3 + r_4} \]
- Finally, since these two groups are connected in series with each other, the total net resistance (\(R\)) of the entire circuit is the sum of these two parts: \[ R = R_s + R_p = r_1 + r_2 + \frac{r_3 \cdot r_4}{r_3 + r_4} \]
In simple words: First, add \(r_1\) and \(r_2\) together because they are in series. Then, work out the parallel combination of \(r_3\) and \(r_4\). Add these two parts together to get the total resistance.

Exam Tip: Break complex networks down into simpler series and parallel sub-circuits before attempting to find the overall resistance.

Question 60. Twelve identical resistors, each of resistance \(2\ \Omega\), are arranged in four parallel branches. Each branch consists of three of these resistors connected in series. Find the equivalent resistance of the entire combination.
Answer: Based on the circuit layout, the network consists of four parallel branches:
- Each of these four paths contains three \(2\ \Omega\) resistors connected in series. The resistance of any single branch is: \[ R_{\text{branch}} = 2\ \Omega + 2\ \Omega + 2\ \Omega = 6\ \Omega \]
- Since all four identical branches (each with a resistance of \(6\ \Omega\)) are in parallel, the total equivalent resistance (\(R_{XY}\)) across the network terminals is: \[ \frac{1}{R_{XY}} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} \] \[ R_{XY} = \frac{6}{4} = 1.5\ \Omega \] Thus, the total equivalent resistance of the network is \(1.5\ \Omega\).
In simple words: Each of the four parallel branches is made of three 2-ohm resistors in series, which equals 6 ohms per branch. When you connect four of these 6-ohm branches in parallel, the total equivalent resistance is 1.5 ohms.

Exam Tip: When identical resistors are in parallel, divide the resistance of one branch by the number of branches (\(6\ \Omega / 4 = 1.5\ \Omega\)) to save time in exams.

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Question 1. Define electrical power. What information does the power rating of an electrical appliance convey?
Answer: Electrical power is defined as the rate at which electrical energy is transformed or mechanical work is performed, representing either the work executed or the energy converted each second.
Most electrical devices are marked with specific wattage and voltage ratings. This power rating provides two key details:
- It indicates the maximum safe current limit that the device can handle.
- It specifies the operational voltage limit above which the device should not be operated.
In simple words: Electrical power is how fast an appliance uses up electricity. The power rating tells us the maximum voltage it can handle and the safest amount of current that should flow through it.

Exam Tip: Examiners often look for the phrase 'work done per second' or 'rate of consumption of energy' when defining power.

Question 2. What is electrical energy? Define electrical power in terms of electrical energy, state its SI unit, and give the alternative name of kilowatt-hour (kWh).
Answer: Electrical energy represents the total cumulative work done to keep a steady electric current flowing through a circuit for a specified duration. Its standard SI unit is the **Joule** (\(\text{J}\)).
The rate at which this electrical energy is consumed over a given period is termed electrical power (\(P\\)), which has the SI unit of **watt** (\(\text{W}\)): \[ \text{Electrical power } (P) = \frac{\text{Electrical energy}}{\text{time}} \] Another common designation for the kilowatt-hour (\(\text{kWh}\)) is the **Board of Trade Unit** (\(\text{BOTU}\)).
In simple words: Electrical energy is the total work needed to keep electricity flowing. Power is how fast that energy is used (Energy divided by Time). Kilowatt-hour is also called the Board of Trade Unit.

Exam Tip: Always remember that the commercial unit of electrical energy is kWh, which is equivalent to \(3.6 \times 10^6\ \text{J}\).

Question 3. Draw a circuit diagram to show a cell, a key, an ammeter, a rheostat, a resistor, and a voltmeter connected properly. State the functions of each of these components.
Answer: A schematic circuit diagram incorporating these essential components is shown below:
The specific functions of these components are:
(A) **Cell**: Serves as the power source that maintains the voltage (potential difference) in the loop.
(B) **Key**: Functions as an on-off switch to complete or break the circuit as desired.
(C) **Ammeter**: Connects in series to measure the electrical current flowing through the circuit.
(D) **Rheostat**: Helps vary the total resistance in the path to regulate the current without changing the main voltage.
(E) **Resistor**: Delivers a fixed, constant electrical resistance.
(F) **Voltmeter**: Connects in parallel to measure the potential difference across the chosen resistor.
In simple words: This diagram shows how components are linked. The cell provides voltage, the key switches it, the ammeter measures current, the rheostat adjusts current, the resistor offers constant opposition, and the voltmeter measures voltage drop.

Exam Tip: Make sure the positive and negative terminals of the ammeter and voltmeter are aligned correctly with the positive and negative terminals of the cell.

Question 4. What is meant by the statement 'the current rating of a fuse is 5 A'?
Answer: The expression indicates that the threshold of current that can safely pass through the fuse without causing its conductive filament to melt and open the circuit is \(5\ \text{A}\). If the current exceeds this limit, the fuse blows.
In simple words: A 5 A fuse rating means that if the current goes above 5 amperes, the fuse wire will melt and safely turn off the electricity.

Exam Tip: Explain that this protects sensitive household appliances from getting damaged during sudden power surges.

Question 5. An electric heater of rating \(5\ \text{kW}\) is operated on a \(200\ \text{V}\) line. Calculate the current drawn by it. Can a fuse of rating \(8\ \text{A}\) be safely used with this appliance?
Answer: Using the power formula: \[ P = V \cdot I \] We can solve for current (\(I\)): \[ I = \frac{P}{V} \] Given values:
- Power, \(P = 5\ \text{kW} = 5000\ \text{W}\)
- Voltage, \(V = 200\ \text{V}\)
Substituting these: \[ I = \frac{5000}{200} = 25\ \text{A} \] Since the working current drawn by the appliance is \(25\ \text{A}\), an \(8\ \text{A}\) fuse is highly inadequate and will blow immediately. Thus, an \(8\ \text{A}\) fuse cannot be used.
In simple words: The heater needs 25 amperes to run. Since 25 A is much larger than 8 A, an 8 A fuse would melt immediately, so it cannot be used.

Exam Tip: Always convert power from kilowatts (\(\text{kW}\)) to watts (\(\text{W}\)) by multiplying by 1000 before substituting into formulas.

Question 6. An electric oven of rating \(3\ \text{kW}\) is connected to a \(250\ \text{V}\) source. Find the current drawn by it. Can this appliance be run in a circuit protected by a \(13\ \text{A}\) fuse?
Answer: Using the relationship for electrical power: \[ I = \frac{P}{V} \] Given values:
- Power, \(P = 3\ \text{kW} = 3000\ \text{W}\)
- Voltage, \(V = 250\ \text{V}\)
Substituting these: \[ I = \frac{3000}{250} = 12\ \text{A} \] Because the current drawn by the oven (\(12\ \text{A}\)) is less than the safe limit of the \(13\ \text{A}\) fuse, it can be safely used in this circuit.
In simple words: The oven draws 12 amperes of current. Since 12 A is safely below the 13 A fuse limit, the appliance can run without blowing the fuse.

Exam Tip: For safety, the fuse rating should always be slightly higher than the normal operating current of the appliance.

Question 7. A power source of \(12\ \text{V}\) is connected to a load of \(2.4\ \text{kW}\). Calculate the current supplied by the source.
Answer: We calculate the current (\(I\)) utilizing the power relation: \[ I = \frac{P}{V} \] Given:
- Power, \(P = 2.4\ \text{kW} = 2400\ \text{W}\)
- Voltage, \(V = 12\ \text{V}\)
Substituting these values: \[ I = \frac{2400}{12} = 200\ \text{A} \] Thus, the current drawn from the battery is \(200\ \text{A}\).
In simple words: Dividing the 2400 watts of power by the 12-volt battery gives 200 amperes of current.

Exam Tip: Double-check your arithmetic, especially when converting powers of ten in kilowatt conversions.

Question 8. An electrical appliance rated \(720\ \text{W}\) is run on a \(220\ \text{V}\) supply. Find: (a) the electrical energy consumed in \(3\ \text{hours}\), and (b) the cost of this energy consumption at the rate of 60 paise per kWh.
Answer: Given parameters:
- Power rating, \(P = 720\ \text{W} = 0.72\ \text{kW}\)
- Supply voltage, \(V = 220\ \text{V}\)

**(a) Electrical energy consumed in \(3\ \text{hours}\):** Using the energy formula: \[ \text{Energy} = \text{Power } (P) \times \text{Time } (t) \] \[ \text{Energy} = 0.72\ \text{kW} \times 3\ \text{h} = 2.16\ \text{kWh} \]
**(b) Cost of energy consumed:** At a tariff of 60 paise per kilowatt-hour (which is \(\text{Rs. } 0.60\)): \[ \text{Cost} = 2.16 \times 0.6 = \text{Rs. } 1.30 \]
In simple words: (a) Running the 0.72 kW appliance for 3 hours uses 2.16 units (kWh) of electricity. (b) At 60 paise per unit, the total cost comes to Rs. 1.30.

Exam Tip: Make sure to state energy in kWh when calculating the cost of electricity, as standard rates are always given per kWh (unit).

Question 9. What is an electric fuse? Explain its purpose and state the materials used to construct it.
Answer: An electric fuse is a safety component integrated into an electrical system to restrict the current to a safe level. By doing so, it protects the wiring network and any connected electronics from severe damage caused by overloading.
Typically, the fuse wire is manufactured using a specialized alloy of lead and tin, which has a low melting point.
In simple words: A fuse is a safety wire that melts and stops the flow of electricity if the current gets too high. This prevents house fires and saves appliances from breaking. It is made from lead and tin.

Exam Tip: Mention the low melting point and high resistance as the core physical properties of a fuse wire to score full marks.

Question 10. State two essential characteristics of a fuse wire.
Answer: A proper fuse wire must possess the following two properties:
(i) A low melting temperature so that it melts quickly under excessive heating.
(ii) A high electrical resistance relative to the rest of the copper connecting wires in the household network.
In simple words: A fuse wire must have a low melting point so it breaks easily when overheated, and high resistance.

Exam Tip: These two properties ensure that when current surges, the fuse wire heats up faster than the household wiring and breaks first.

Question 11. Fill in the blanks: (a) A fuse wire should have high _____ and low _____. (b) A fuse is made of an alloy with _____ melting point, consisting of _____ and _____, which _____ when current exceeds the safe limit. (c) Fuses are always connected in _____ with the _____ wire.
Answer: The completed sentences are:
(a) A fuse wire should have high **resistance** and a low **melting point**.
(b) A fuse is made of an alloy with a **low** melting point, consisting of **lead** and **tin**, which **melts** when current exceeds the safe limit.
(c) Fuses are always connected in **series** with the **live** wire.
In simple words: (a) Fuses need high resistance and a low melting point. (b) They are made of low-melting-point lead and tin, which melts easily. (c) Fuses are connected in series with the live wire.

Exam Tip: Fuses must never be connected to the neutral wire because if they blow, high voltage would still reach the appliance, presenting a shock hazard.

Page 202

Question 12. Calculate the heat produced in a resistor of \(5\ \Omega\) when a current of \(10\ \text{A}\) flows through it for \(15\ \text{minutes}\). State the law used.
Answer: According to Joule's Law of Heating, the thermal energy (\(H\)) generated is given by: \[ H = I^2 \cdot R \cdot t \] Given values:
- Current, \(I = 10\ \text{A}\)
- Resistance, \(R = 5\ \Omega\)
- Time, \(t = 15\ \text{minutes} = 15 \times 60 = 900\ \text{s}\)
Substituting these parameters into the formula: \[ H = (10)^2 \times 5 \times 900 \] \[ H = 100 \times 5 \times 900 = 4.5 \times 10^5\ \text{J} \] Thus, the total heat produced is \(4.5 \times 10^5\ \text{J}\).
In simple words: We find heat using the formula \(I^2 R t\). Squaring the 10 A current, multiplying by 5 ohms, and multiplying by 900 seconds gives 450,000 Joules of heat.

Exam Tip: Always convert time from minutes to seconds before plugging it into the Joule's heating formula.

Question 13. An electrical appliance connected across a \(200\ \text{V}\) line draws a current of \(11\ \text{A}\). Calculate its resistance.
Answer: Using Ohm's law: \[ V = I \cdot R \implies R = \frac{V}{I} \] Given:
- Supply voltage, \(V = 200\ \text{V}\)
- Current, \(I = 11\ \text{A}\)
Substituting these: \[ R = \frac{200}{11} \approx 18.2\ \Omega \] Therefore, the electrical resistance of the appliance is \(18.2\ \Omega\).
In simple words: Dividing the 200 volts by the 11 amperes of current gives a resistance of about 18.2 ohms.

Exam Tip: Round off your final decimal answers to one or two decimal places, and always include the ohm (\(\Omega\)) unit symbol.

Question 14. An electric iron of power rating \(1200\ \text{W}\) draws a current of \(5\ \text{A}\). Find: (i) the resistance of the electric iron, and (ii) the potential difference across it.
Answer: We utilize the relationships for electrical power: \[ P = I^2 \cdot R \quad \text{and} \quad P = V \cdot I \] Given:
- Power, \(P = 1200\ \text{W}\)
- Current, \(I = 5\ \text{A}\)

**(i) Resistance (\(R\)):** \[ R = \frac{P}{I^2} = \frac{1200}{5 \times 5} = \frac{1200}{25} = 48\ \Omega \]
**(ii) Potential difference (\(V\)):** \[ V = \frac{P}{I} = \frac{1200}{5} = 240\ \text{V} \]
In simple words: (i) Resistance is power divided by current squared, which is 48 ohms. (ii) Voltage is power divided by current, which is 240 volts.

Exam Tip: This problem can be solved in multiple ways; you can also find voltage first (\(V = P/I\)) and then use Ohm's law (\(R = V/I\)) to get the resistance.

Question 15. Where is the main fuse connected in a household circuit? In which wire is it placed?
Answer: The main fuse is positioned between the commercial kilowatt-hour (\(\text{kWh}\)) meter and the primary distribution box of the home electrical system. It is always wired in series with the live conductor.
In simple words: The main fuse is placed on the live wire, right after the electricity meter and before the main switchboard.

Exam Tip: Fuses are always wired in series with the live wire so that if they melt, the high-voltage line is immediately disconnected from the entire house.

Question 16. Is kilowatt-hour (kWh) a unit of electrical energy?
Answer: Yes, the kilowatt-hour (\(\text{kWh}\)) represents a standard commercial unit of electrical energy.
In simple words: Yes, kWh is the unit we use to measure how much electrical energy we consume.

Exam Tip: One kilowatt-hour is commonly called 'one unit' on household electricity bills.

Question 17. What is a switch? State its main function and explain why switches are always connected in the live wire.
Answer: A switch serves as an on-off controller to regulate current flow in a circuit or device. Its primary role is to establish or break the electrical connection as required.
Switches must always be wired into the live line. When the switch is flipped to the 'off' state, the appliance is completely isolated from the high-voltage supply. Both the live and neutral wiring within the appliance then remain at safe, zero potential. This ensures a user will not receive an electric shock even if they touch the internal live wires, and even if the fuse has blown.
In simple words: A switch is an on-off device. It must be connected to the live wire so that when it is turned off, no electricity can reach the appliance, making it safe to touch.

Exam Tip: If a switch is wrongly placed in the neutral wire, the appliance remains connected to the high-voltage live wire even when turned off, which is a major safety hazard.

Question 18. Why should an electrical switch never be touched with wet hands?
Answer: An electrical switch must never be operated with wet hands because moisture on the skin forms an active conducting channel. If any current leaks from the live switch terminals, it will flow easily through this water layer to the body, delivering a highly dangerous and potentially fatal shock.
In simple words: Water conducts electricity. If you touch a switch with wet hands, electricity can easily pass through the water and give you a severe shock.

Exam Tip: Pure water is a poor conductor, but dissolved salts on human skin instantly turn water into a highly efficient conductor of electric current.

Question 19. Explain the meaning of the rating '250 W - 230 V' marked on an electric bulb. Calculate the safe current limit for this bulb. Also, find the current through a \(60\ \text{W}\) lamp rated for \(250\ \text{V}\), and determine its power consumption if the line voltage drops to \(200\ \text{V}\).
Answer: A rating of '250 W - 230 V' means that if the bulb is operated on a standard \(230\ \text{V}\) line, it will consume electric power of \(250\ \text{W}\), meaning \(250\ \text{J}\) of electrical energy is transformed into light and heat energy every single second.
- The safe current threshold (\(I\)) for this bulb is: \[ I = \frac{P}{V} = \frac{250}{230} \approx 1.1\ \text{A} \]
- For a \(60\ \text{W}\) lamp designed for a \(250\ \text{V}\) line, the operating current is: \[ I_2 = \frac{P_2}{V_2} = \frac{60}{250} = 0.24\ \text{A} \] Since resistance remains constant, the resistance (\(R\)) of this second lamp is: \[ R = \frac{V_2^2}{P_2} = \frac{250 \times 250}{60} = \frac{62500}{60} \approx 1041.7\ \Omega \] If the voltage drops to \(200\ \text{V}\), the new power consumption is: \[ P_{\text{new}} = \frac{V_{\text{new}}^2}{R} = \frac{200 \times 200}{1041.7} \approx 38.4\ \text{W} \] Thus, the lamp's output power drops to \(38.4\ \text{W}\).
In simple words: (i) The rating means the bulb uses 250 Joules of energy per second when run on 230 V. Its safe current is 1.1 A. (ii) A 60 W, 250 V lamp draws 0.24 A. If the voltage drops to 200 V, its output drops to 38.4 W.

Exam Tip: Remember that the electrical resistance of a filament lamp is treated as constant under normal working conditions. Use \(R = V^2/P\) to link power and voltage directly.

Question 20. Why is the metallic body of an electrical appliance earthed?
Answer: The metallic outer case of a device is connected to the ground so that if a live wire accidentally touches the metal body, the resulting current is instantly diverted away. The earth acts as a massive sink of charge, offering an extremely low-resistance route. This protects the user from receiving a lethal electric shock.
In simple words: Connecting the metal case to the ground (earthing) ensures that if a loose wire touches the case, the electricity flows safely into the ground instead of shocking you.

Exam Tip: Use keywords like 'low-resistance path' and 'sink of charge' to ensure your answer aligns with what examiners expect.

Question 21. An electric bulb is connected to a \(12\ \text{V}\) battery and consumes \(24\ \text{W}\) of power. If it is kept on for \(20\ \text{minutes}\), calculate: (i) the current drawn, and (ii) the energy liberated by it.
Answer: Given:
- Potential difference, \(V = 12\ \text{V}\)
- Power, \(P = 24\ \text{W}\)
- Time, \(t = 20\ \text{minutes} = 20 \times 60 = 1200\ \text{s}\)

**(i) Current (\(I\)):** \[ I = \frac{P}{V} = \frac{24}{12} = 2\ \text{A} \]
**(ii) Energy Liberated (\(H\)):** Using the energy formula: \[ H = V \cdot I \cdot t = 12 \times 2 \times 1200 = 28,800\ \text{J} \] Therefore, the bulb draws \(2\ \text{A}\) of current and releases \(28,800\ \text{J}\) of energy.
In simple words: (i) Dividing 24 watts by 12 volts gives 2 amperes of current. (ii) Multiplying 12 volts, 2 amperes, and 1200 seconds together gives 28,800 Joules of energy.

Exam Tip: You can also calculate energy directly by multiplying power and time in seconds (\(P \times t\)), which is \(24 \times 1200 = 28,800\ \text{J}\).

Question 22. State the international color coding convention for the live, neutral, and earth wires.
Answer: The modern international standard for wiring insulation color codes is:
1. **Live wire**: Brown
2. **Neutral wire**: Light blue
3. **Earth wire**: Green or yellow
In simple words: The international wire colors are brown for live, light blue for neutral, and green or yellow for earth.

Exam Tip: Do not confuse old wiring colors (red, black, green) with the new international standards. Always write the modern standard unless specifically asked otherwise.

Question 23. A resistor draws a current of \(0.2\ \text{A}\) when connected to a potential difference of \(15\ \text{V}\). Find: (i) its resistance, and (ii) the energy produced by it in \(1\ \text{minute}\).
Answer: Given:
- Current, \(I = 0.2\ \text{A}\)
- Potential difference, \(V = 15\ \text{V}\)
- Time, \(t = 1\ \text{minute} = 60\ \text{s}\)

**(i) Resistance (\(R\)):** Using Ohm's law: \[ R = \frac{V}{I} = \frac{15}{0.2} = 75\ \Omega \]
**(ii) Energy Produced (\(H\)):** Using the formula: \[ H = I^2 \cdot R \cdot t \] \[ H = (0.2)^2 \times 75 \times 60 = 0.04 \times 75 \times 60 = 180\ \text{J} \] Thus, the resistance is \(75\ \Omega\) and the energy produced is \(180\ \text{J}\).
In simple words: (i) The resistance is 15 V divided by 0.2 A, which is 75 ohms. (ii) The heat energy made in one minute (60 seconds) is 180 Joules.

Exam Tip: Make sure to write the formula used clearly before performing calculations to secure step marks.

Question 24. An electric appliance rated \(1.5\ \text{kW}\) is operated on a \(220\ \text{V}\) line. Calculate the current drawn. Suggest a suitable fuse rating and state where it should be connected.
Answer: Given:
- Power, \(P = 1.5\ \text{kW} = 1500\ \text{W}\)
- Potential difference, \(V = 220\ \text{V}\)
The current (\(I\)) drawn by the device is: \[ I = \frac{P}{V} = \frac{1500}{220} \approx 6.8\ \text{A} \] Based on this current, a fuse rated at **7 A** is most appropriate. This safety device must be connected in series along the live wire before the line enters the appliance.
In simple words: The appliance draws about 6.8 amperes of current, so a 7 A fuse is the perfect match. This fuse must be connected directly to the live wire.

Exam Tip: Always choose the nearest integer fuse rating that is slightly greater than the calculated operating current.

Question 25. Explain how an electric fuse protects an electrical circuit from damage.
Answer: A fuse is designed to control and limit the maximum current flowing through an electrical circuit. It offers protection by melting and breaking the path whenever the current climbs above its designated safe threshold. Because the fuse wire is engineered with a low melting point, the excess heat generated during a current surge causes it to liquefy quickly, thereby isolating the circuit and halting the current.
In simple words: A fuse stops too much electricity from flowing. If there is a dangerous current surge, the fuse wire gets hot, melts, and cuts off the power, keeping the circuit safe.

Exam Tip: Mention 'Joule heating' (\(H \propto I^2\)) as the physical cause behind the melting of the fuse wire under high current.

Question 26. Why is earthing necessary in electrical wiring? How does it protect a circuit?
Answer: Earthing is implemented to prevent damage to electrical circuits and appliances, as well as to protect human life. If a malfunction like a short circuit causes a massive current to flow, the earthing system safely redirects this excess charge directly to the ground. Without this pathway, the line wires would overheat severely, which could trigger a dangerous fire.
In simple words: Earthing is a safety feature that redirects any leaked or short-circuited electricity into the ground, stopping wires from overheating and catching fire.

Exam Tip: Clearly connect 'earthing' with the mitigation of 'fire hazards' and 'electric shocks' to provide a comprehensive answer.

Question 27. Draw a labeled diagram of a three-pin socket, indicating the Earth, Live, and Neutral terminals.
Answer: The standard socket layout and terminal assignments are shown in the diagram below:
In simple words: This diagram shows a standard wall plug socket. The top hole is for Earth (E), the bottom left is for Neutral (N), and the bottom right is for Live (L).

Exam Tip: Remember that the Earth pin hole (E) at the top is always made larger and thicker than the other two pin holes.

Question 28. Explain the circumstances under which a user can receive an electric shock from an appliance with a metallic body.
Answer: A user can suffer a severe electric shock if the protective insulation surrounding the internal live wire cracks or degrades over time, allowing the bare live wire to directly touch the appliance's metallic outer body. If a person touches this charged casing without adequate earthing, the current passes through their body to reach the ground.
In simple words: If the insulation inside an appliance wears out and the live wire touches the metal casing, touching the casing will pass the electricity through your body to the ground, giving you a shock.

Exam Tip: Emphasize that earthing is the primary countermeasure against this type of shock hazard.

Question 29. State the SI units of electrical energy and power. Also: (i) Define kilowatt-hour (kWh) in terms of power and time, and (ii) State the typical household voltage supplied to residential areas.
Answer: The standard SI unit for electrical energy is the **Joule** (\(\text{J}\)), and for electrical power, it is the **watt** (\(\text{W}\)).
(i) **Kilowatt-hour (kWh)**: This represents the commercial unit of electrical energy. It is defined as the total electrical energy consumed over a span of exactly one hour when the rate of consumption is maintained at 1000 watts (or \(1000\ \text{J/s}\)).
(ii) Residential homes are generally provided with a supply voltage of **220 V**.
In simple words: Energy is measured in Joules, power in watts. A kilowatt-hour is the energy consumed by a 1000-watt appliance running for one hour. Household voltage in India is 220 V.

Exam Tip: Know the exact conversion: \(1\ \text{kWh} = 1\ \text{kW} \times 1\ \text{h} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^6\ \text{J}\).

Question 30. An electric bulb is rated \(100\ \text{W} - 220\ \text{V}\). Calculate: (i) its resistance, and (ii) the current drawn by it.
Answer: Given parameters:
- Power rating, \(P = 100\ \text{W}\)
- Potential difference, \(V = 220\ \text{V}\)

**(i) Resistance (\(R\)):** Using the relation: \[ P = \frac{V^2}{R} \implies R = \frac{V^2}{P} \] \[ R = \frac{(220)^2}{100} = \frac{48400}{100} = 484\ \Omega \]
**(ii) Current Drawn (\(I\)):** Using the power formula: \[ P = V \cdot I \implies I = \frac{P}{V} \] \[ I = \frac{100}{220} \approx 0.45\ \text{A} \] Thus, the bulb's resistance is \(484\ \Omega\) and the current drawn is \(0.45\ \text{A}\).
In simple words: (i) Resistance is \(V^2 / P\), which works out to 484 ohms. (ii) Current is \(P / V\), which is approximately 0.45 amperes.

Exam Tip: Always calculate the resistance first when evaluating bulb parameters, as the hot resistance of the filament remains constant during operations.

Question 31. Two resistors of \(4\ \Omega\) and \(6\ \Omega\) are connected in parallel across a \(6\ \text{V}\) battery. Calculate: (i) the total power supplied by the battery, and (ii) the power dissipated in each resistor.
Answer: Given:
- Parallel resistances, \(r_1 = 4\ \Omega\) and \(r_2 = 6\ \Omega\)
- Battery potential difference, \(V = 6\ \text{V}\)
First, calculate the equivalent resistance (\(R_p\)) of the parallel network: \[ \frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \] \[ R_p = \frac{12}{5} = 2.4\ \Omega \]
**(i) Total power supplied by the battery (\(P_{\text{total}}\)):** \[ P_{\text{total}} = \frac{V^2}{R_p} = \frac{6 \times 6}{2.4} = \frac{36}{2.4} = 15\ \text{W} \]
**(ii) Power dissipated in each resistor:** Since the resistors are connected in parallel, both experience the full supply voltage of \(6\ \text{V}\).
- Power dissipated in the \(4\ \Omega\) resistor: \[ P_{4\Omega} = \frac{V^2}{r_1} = \frac{6 \times 6}{4} = 9\ \text{W} \]
- Power dissipated in the \(6\ \Omega\) resistor: \[ P_{6\Omega} = \frac{V^2}{r_2} = \frac{6 \times 6}{6} = 6\ \text{W} \] Note that the sum of the individual powers (\(9\ \text{W} + 6\ \text{W} = 15\ \text{W}\)) matches the total power supplied by the battery.
In simple words: (i) Parallel resistance is 2.4 ohms. Total power supplied by the battery is 15 watts. (ii) Since both branches get 6 volts, the 4-ohm resistor uses 9 watts and the 6-ohm resistor uses 6 watts.

Exam Tip: In parallel circuits, use \(P = V^2/R\) to find individual power since voltage is the same across all parallel components.

Question 32. List the three terminals of a three-pin plug and explain the purpose of the earth pin.
Answer: A standard three-pin plug is equipped with the following three terminals:
(i) **Earth pin**: This terminal establishes a ground connection to route leakages away.
(ii) **Live pin**: Connects the appliance to the high-voltage supply wire.
(iii) **Neutral pin**: Completes the circuit loop back to the supply station.
The primary function of the **Earth pin** is to connect the metallic frame of the appliance to the ground, protecting the user from electric shock in case of insulation failure.
In simple words: A three-pin plug contains Earth, Live, and Neutral terminals. The Earth pin links the appliance to the ground to keep you safe from shocks.

Exam Tip: The Earth pin is always designed to be longer than the live and neutral pins so that the ground connection is made first when plugging in the appliance.

Question 33. In what electrical unit does a consumer pay their electricity bill?
Answer: A consumer pays their domestic electricity bill based on the total units of energy consumed, measured in **kilowatt-hours** (\(\text{kWh}\)).
In simple words: Electricity bills are charged in kilowatt-hours (often called 'units').

Exam Tip: One 'unit' of electricity on a standard utility bill corresponds directly to 1 kWh.

Question 34. Two resistors of \(4\ \Omega\) and \(6\ \Omega\) are connected to a cell of emf \(15\ \text{V}\) and internal resistance \(2\ \Omega\). Calculate the electrical energy consumed per minute in the \(6\ \Omega\) resistor when the two resistors are connected: (i) in series, and (ii) in parallel.
Answer: Given data:
- emf, \(e = 15\ \text{V}\)
- Internal resistance, \(r = 2\ \Omega\)
- Resistors, \(r_1 = 4\ \Omega\) and \(r_2 = 6\ \Omega\)
- Time, \(t = 1\ \text{minute} = 60\ \text{s}\)

**(i) Resistors connected in series:** The equivalent external series resistance is: \[ R_s = 4\ \Omega + 6\ \Omega = 10\ \Omega \] The total current (\(I_s\)) in the circuit is: \[ I_s = \frac{e}{R_s + r} = \frac{15}{10 + 2} = 1.25\ \text{A} \] Since the current is identical through all series elements, the energy (\(H_s\)) spent per minute in the \(6\ \Omega\) resistor is: \[ H_s = I_s^2 \cdot r_2 \cdot t = (1.25)^2 \times 6 \times 60 = 1.5625 \times 360 = 562.5\ \text{J} \]
**(ii) Resistors connected in parallel:** The equivalent external parallel resistance is: \[ R_p = \left( \frac{1}{4} + \frac{1}{6} \right)^{-1} = 2.4\ \Omega \] The total current (\(I_p\)) drawn from the cell is: \[ I_p = \frac{e}{R_p + r} = \frac{15}{2.4 + 2} = \frac{15}{4.4} \approx 3.41\ \text{A} \] The terminal potential difference (\(V\)) across the parallel branches is: \[ V = e - I_p \cdot r = 15 - (3.41 \times 2) = 15 - 6.82 = 8.18\ \text{V} \] Since both parallel branches share this terminal potential difference, the energy (\(H_p\)) consumed per minute by the \(6\ \Omega\) resistor is: \[ H_p = \frac{V^2}{r_2} \cdot t = \frac{(8.18)^2}{6} \times 60 = 66.91 \times 10 = 669.1\ \text{J} \]
In simple words: (i) In series, the total resistance is 12 ohms, drawing 1.25 A. The 6-ohm resistor consumes 562.5 Joules per minute. (ii) In parallel, the circuit draws 3.41 A, making the terminal voltage 8.18 V. The 6-ohm resistor then consumes 669.1 Joules per minute.

Exam Tip: In parallel circuits, the voltage across external resistors is the terminal potential difference (\(V = E - Ir\)), not the EMF (\(E\)) of the cell. Keep this distinction in mind.

Question 35. What is the commercial unit of electricity?
Answer: The standard commercial unit of electrical energy is the **kilowatt-hour** (\(\text{kWh}\)).
In simple words: The commercial unit of electricity is the kilowatt-hour.

Exam Tip: A consumer's monthly energy consumption is recorded and billed in kilowatt-hours.

Question 36. Write down three different mathematical expressions for electrical power.
Answer: Electrical power (\(P\)) can be expressed using three primary algebraic formulations: \[ P = V \cdot I \] \[ P = I^2 \cdot R \] \[ P = \frac{V^2}{R} \] where \(V\) is potential difference, \(I\) is current, and \(R\) is resistance.
In simple words: Power can be written as Voltage times Current (\(VI\)), Current squared times Resistance (\(I^2R\)), or Voltage squared divided by Resistance (\(V^2/R\)).

Exam Tip: Choose the expression that matches the quantities known in the problem (e.g., use \(I^2R\) for series components and \(V^2/R\) for parallel components).

Question 37. What does the power-voltage rating of an electrical appliance indicate? How is this information useful to the consumer?
Answer: The power-voltage specification marked on an appliance indicates the maximum threshold voltage at which the device is designed to run safely.
This information is highly valuable to the customer in two ways:
- It acts as a safety warning, notifying the user not to run the device on a higher voltage line.
- It enables the consumer to calculate the safe operating current limit to select an appropriate fuse rating.
In simple words: The rating tells you the maximum voltage safe for the appliance. It helps you calculate the safest current and choose the right fuse to protect your home.

Exam Tip: Explain that if an appliance is connected to a voltage higher than its rated value, it will draw excessive current and burn out.

Question 38. Which physical quantity has 'watt' as its SI unit?
Answer: The physical quantity whose SI unit is the watt is **electric power**.
In simple words: Watt is the unit of electrical power.

Exam Tip: Remember that kilowatt (\(\text{kW}\)) and megawatt (\(\text{MW}\)) are larger multiple units of electric power.

Question 39. What is the meaning of the statement 'the bulb is rated 60 W - 200 V'?
Answer: The statement means that when the bulb is operated on a \(200\ \text{V}\) line, it will consume electrical energy at a rate of \(60\ \text{J}\) per second, transforming it into heat and light.
In simple words: This rating means that when connected to a 200 V line, the bulb uses 60 Joules of electrical energy every second to produce light and heat.

Exam Tip: Silently correct textbook typos regarding energy consumption rate: a \(60\ \text{W}\) bulb converts exactly \(60\ \text{J}\) of energy per second, not \(200\ \text{J}\).

Question 40. A student states that a 100 W (220 V) bulb has a greater resistance than a 60 W (220 V) bulb. Is this statement true or false? Explain.
Answer: The resistance (\(R\)) of an appliance is related to its power rating by the equation: \[ R = \frac{V^2}{P} \] When operated on a constant supply voltage (\(V\)), the resistance of the device is inversely proportional to its rated power (\(R \propto \frac{1}{P}\)). Consequently, a lower-power appliance will have a higher electrical resistance. Therefore, the \(60\ \text{W}\) bulb possesses a greater resistance than the \(100\ \text{W}\) bulb, rendering the student's statement **false**.
In simple words: The statement is false. Resistance is inversely proportional to power when voltage is constant. A 60 W bulb has more resistance than a 100 W bulb.

Exam Tip: Keep this inverse relationship in mind: higher wattage appliances draw more current because their heating filaments are designed to have lower resistance.

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Question 41. An electric bulb is rated \(40\ \text{W} - 220\ \text{V}\). Calculate the current passing through it.
Answer: Using the electrical power relation: \[ P = V \cdot I \implies I = \frac{P}{V} \] Given values:
- Power, \(P = 40\ \text{W}\)
- Voltage, \(V = 220\ \text{V}\)
Substituting these: \[ I = \frac{40}{220} \approx 0.11\ \text{A} \] Thus, the current through the bulb is \(0.11\ \text{A}\).
In simple words: Dividing the 40 watts of power by the 220-volt supply line gives approximately 0.11 amperes of current.

Exam Tip: In decimal divisions, calculate your answer up to two decimal places to ensure accuracy.

Question 42. Define the unit watt-hour (Wh). Express its value in Joules.
Answer: A watt-hour (\(\text{Wh}\)) is a unit utilized to measure electrical energy. It is defined as the total quantity of electrical energy consumed by an appliance of power \(1\ \text{W}\) when operated continuously for a duration of \(1\ \text{hour}\).
Its value in Joules is: \[ 1\ \text{Wh} = 1\ \text{W} \times 1\ \text{hour} = 1\ \text{J/s} \times 3600\ \text{s} = 3600\ \text{J} \] Thus, \(1\ \text{Wh} = 3600\ \text{J}\).
In simple words: One watt-hour is the energy consumed by a 1-watt appliance running for one hour. It is equal to 3600 Joules of energy.

Exam Tip: Do not confuse watt-hour (energy unit) with watt (power unit). The presence of the time unit (hour) turns it into an energy measurement.

Question 43. Name the commercial unit of electricity.
Answer: The commercial unit utilized for registering electrical energy consumption is the **kilowatt-hour** (\(\text{kWh}\)).
In simple words: The commercial unit of electricity is the kilowatt-hour.

Exam Tip: Electricity meters measure power consumption in units of kWh, where 1 unit equals 1 kWh.

Question 44. Calculate the electrical energy consumed by a \(60\ \text{W}\) bulb in \(50\ \text{hours}\) in kilowatt-hours (kWh).
Answer: The formula for electrical energy consumption is: \[ E = P \cdot t \] Given:
- Power, \(P = 60\ \text{W} = 0.06\ \text{kW}\)
- Time, \(t = 50\ \text{hours}\)
Substituting the values: \[ E = 0.06\ \text{kW} \times 50\ \text{h} = 3\ \text{kWh} \] Therefore, the energy consumed is \(3\ \text{kWh}\).
In simple words: Converting 60 watts to 0.06 kW and multiplying by 50 hours gives 3 kilowatt-hours of energy used.

Exam Tip: Converting watts to kilowatts before multiplying by hours is the most direct way to obtain energy consumption in kWh.

Question 45. A household uses two \(60\ \text{W}\) bulbs for \(4\ \text{hours}\) daily and three \(100\ \text{W}\) bulbs for \(5\ \text{hours}\) daily. Calculate the total electrical energy consumed in the month of June (30 days).
Answer: We calculate the daily energy consumption in separate steps:
- **For the two \(60\ \text{W}\) bulbs:** \[ E_1 = 2 \times \frac{60}{1000}\ \text{kW} \times 4\ \text{h} = 0.48\ \text{kWh/day} \]
- **For the three \(100\ \text{W}\) bulbs:** \[ E_2 = 3 \times \frac{100}{1000}\ \text{kW} \times 5\ \text{h} = 1.50\ \text{kWh/day} \]
- **Total daily energy consumption:** \[ E_{\text{daily}} = 0.48 + 1.50 = 1.98\ \text{kWh/day} \]
- **Total energy consumed in June (30 days):** \[ E_{\text{June}} = 1.98\ \text{kWh/day} \times 30\ \text{days} = 59.4\ \text{kWh} \] Thus, the total electrical energy consumed in June is \(59.4\ \text{kWh}\).
In simple words: The two 60 W bulbs use 0.48 kWh per day, and the three 100 W bulbs use 1.5 kWh per day, totaling 1.98 units daily. In 30 days of June, they consume 59.4 units in total.

Exam Tip: Be mindful of the calendar month mentioned: June has 30 days, while months like May or July have 31 days. Using the wrong number of days will result in a penalty.

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Question 1. Can electricity be produced from magnetism?
Answer: Yes, electrical currents can be generated from magnetic fields through the process of electromagnetic induction.
In simple words: Yes, we can generate electricity using magnets.

Exam Tip: This discovery by Michael Faraday forms the working principle behind modern electrical generators.

Question 2. What is electromagnetic induction?
Answer: Electromagnetic induction is defined as the generation of an electric current inside a straight conductor when it is moved relative to a magnetic field.
In simple words: Moving a wire through a magnetic field creates an electric current inside the wire. This process is called electromagnetic induction.

Exam Tip: Be sure to mention 'relative motion' between the conductor and the magnetic field as the trigger for generating current.

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Question 3. State Fleming's Right-Hand Rule.
Answer: Fleming's Right-Hand Rule is formulated as follows: Stretch the index finger, the middle finger, and the thumb of your right hand so that they are mutually perpendicular to one another. If the index finger represents the direction of the external magnetic field, and the thumb points in the direction of the conductor's movement, then the middle finger will indicate the direction of the induced electrical current within that conductor.
In simple words: Hold the thumb, index finger, and middle finger of your right hand so they point in three different directions at right angles. The index finger points with the magnetic field, the thumb points with the wire's motion, and the middle finger will show which way the induced current flows.

Exam Tip: Make sure you clearly specify the 'right hand' when writing this rule, as confusing it with the left hand is a very common exam mistake.

Question 4. Describe the observations in a galvanometer connected to a coil when: (a) the north pole of a bar magnet is pushed into the coil, (b) the magnet is held stationary inside the coil, and (c) the north pole is pulled out of the coil.
Answer:
(a) Pushing the north pole of the bar magnet into the coil produces a brief, momentary deflection in the galvanometer needle, indicating that a temporary current has been induced in the coil.
(b) Keeping the magnet completely stationary inside the coil results in zero deflection in the galvanometer, which shows that no electric current is generated when there is no relative movement.
(c) Pulling the north pole rapidly out of the coil causes the galvanometer needle to deflect in the opposite direction, demonstrating that an induced current is generated in the reverse direction.
In simple words: (a) Pushing a magnet into a coil makes the meter needle jump. (b) Holding the magnet still makes the needle go back to zero. (c) Pulling the magnet out makes the needle jump in the opposite direction.

Exam Tip: Emphasize that the deflection is only momentary and exists only while there is active relative motion between the magnet and the coil.

Question 5. State the respective uses of: (i) Fleming's Left-Hand Rule, and (ii) Fleming's Right-Hand Rule.
Answer:
(i) **Fleming's Left-Hand Rule**: This rule is employed to determine the direction of the magnetic force (Lorentz force) acting on a current-carrying conductor placed inside a magnetic field.
(ii) **Fleming's Right-Hand Rule**: This rule is used to find the direction of the induced electric current generated in a conductor moving through a magnetic field.
In simple words: Use your left hand to find the direction of force (movement) on a wire carrying current in a magnetic field. Use your right hand to find the direction of the induced current when you move a wire through a magnetic field.

Exam Tip: A handy way to remember the difference is: Left hand is for motors (force output), and Right hand is for generators (current output).

Question 6. Can Fleming's Right-Hand Rule be used to find the magnitude of the induced current? Explain.
Answer: No, Fleming's Right-Hand Rule cannot be used to determine the magnitude of the induced current. It is strictly a directional rule designed to identify the orientation of the induced current, not its numerical value.
In simple words: No, this hand rule only shows which way the current flows. It cannot tell you how many amperes of current are produced.

Exam Tip: To find the magnitude of the induced electromotive force and current, you must apply Faraday's Law of Electromagnetic Induction (\(e = -N \frac{d\Phi}{dt}\)) instead.

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Question 7. What is direct current (D.C.)?
Answer: Direct current (D.C.) is an electric current that flows continuously in a single, constant direction over time.
In simple words: Direct current is electricity that always travels in one direction, like the current from a battery.

Exam Tip: Direct current has a constant polarity, meaning the positive and negative terminals do not switch places.

Question 8. What is alternating current (A.C.)?
Answer: Alternating current (A.C.) is an electric current that periodically reverses its direction of flow at regular, equal intervals of time.
In simple words: Alternating current is electricity that rapidly switches its direction back and forth at regular intervals, like the power in household wall outlets.

Exam Tip: Standard household A.C. is sinusoidal, completing many complete back-and-forth cycles every second.

Question 9. State the frequency of direct current (D.C.).
Answer: The frequency of a direct current (D.C.) supply is exactly zero hertz (\(0\ \text{Hz}\)) because the current does not undergo any periodic cycles or direction changes.
In simple words: Since direct current never changes direction, its frequency is zero.

Exam Tip: A frequency of zero means the time period of D.C. is mathematically infinite.

Question 10. State the frequency of alternating current (A.C.) supplied for household use in India.
Answer: The frequency of the alternating current (A.C.) distributed for residential use in India is standardized at \(50\ \text{Hz}\) (cycles per second).
In simple words: The electricity in Indian homes changes direction back and forth 50 times every second.

Exam Tip: A frequency of \(50\ \text{Hz}\) implies that the current changes direction 100 times each second.

Question 11. Distinguish between alternating current and direct current based on their direction of flow.
Answer: An alternating current systematically reverses its direction of flow after equal intervals of time, whereas a direct current maintains a single, unchanging direction of flow indefinitely.
In simple words: Alternating current constantly changes its direction of travel, while direct current always flows in the same direction.

Exam Tip: When plotting these currents over time, D.C. appears as a flat horizontal line, while A.C. appears as a repeating wave.

Question 12. Which device is used to generate direct current? What component distinguishes it?
Answer: A D.C. generator, which features a split-ring commutator connected to its rotating armature, is used to generate direct current.
In simple words: A D.C. generator uses a special split ring called a commutator to produce current that flows in only one direction.

Exam Tip: The commutator acts as a mechanical rectifier, converting the internal alternating current of the coils into a direct current output.

Question 13. What energy conversion takes place in an electric dynamo?
Answer: An electric dynamo converts kinetic mechanical energy into electrical energy through electromagnetic induction.
In simple words: A dynamo turns movement (mechanical energy) into electrical energy.

Exam Tip: A dynamo operates on the principal of rotating a coil inside a magnetic field to induce an electromotive force.

Question 14. How is a fuse connected in a household circuit?
Answer: A safety fuse is always connected in series along the live wire path within a household electrical installation.
In simple words: A fuse is always connected in series with the live wire of a circuit.

Exam Tip: Connecting a fuse in series ensures that if the current gets too high, the fuse wire melts and immediately interrupts the entire flow of electricity.

Question 15. State the three colors of wires used in household electrical wiring according to standard color coding.
Answer: The three standard functional colors used in domestic electrical installations are:
(i) **Red** (representing the live wire)
(ii) **Green** (representing the earth safety wire)
(iii) **Black** (representing the neutral return wire)
In simple words: In older standard color codes, the live wire is red, the earth wire is green, and the neutral wire is black.

Exam Tip: Note that these represent the traditional color codes; be prepared to specify the modern international codes (Brown, Green-Yellow, Blue) if asked.

Question 16. What is short-circuiting? How does it affect an electrical circuit?
Answer: Short-circuiting happens when the live wire accidentally comes into direct, physical contact with the neutral wire. This bypasses the appliance, creating an alternate path with almost zero electrical resistance. As a result, an exceptionally high current surges through the wires, which can cause overheating and fires.
In simple words: A short circuit occurs when a live wire touches a neutral wire directly. This creates a shortcut with no resistance, causing a huge surge of electricity that can start a fire.

Exam Tip: Under short-circuit conditions, the current rises to an extremely high level because the load resistance is bypassed (\(I = V/R\) where \(R \to 0\)).

Question 17. State two important safety precautions to be taken when using household electricity.
Answer: To ensure safety when working around electrical systems, the following measures must be observed:
(i) Always employ wires and cables that have a current-carrying capacity higher than the maximum potential current expected in the circuit.
(ii) Never touch, turn on, or operate any electrical switch or appliance with wet hands.
In simple words: (i) Use thick enough wires that can safely carry the current your appliances need. (ii) Never touch switches or appliances if your hands are wet.

Exam Tip: Wires with insufficient carrying capacity will overheat, melt their insulation, and potentially cause electrical fires.

Question 18. What is earthing? Explain how it acts as a safety device to prevent electric shocks.
Answer: Earthing is a critical protective measure designed to prevent electric shocks caused by current leakages or short circuits. It involves linking the metallic body of an appliance (or the main home utility meter) to a thick copper conductor buried deep in the ground. The terminal of this underground conductor is attached to a copper plate encased in a protective blend of charcoal and common salt. This provides an alternative zero-resistance path, safely draining any stray electricity into the earth.
In simple words: Earthing connects the metal frame of an appliance to a copper wire buried deep in the ground. If electricity leaks onto the metal frame, it flows safely into the ground instead of shocking you when you touch it.

Exam Tip: The charcoal and salt mixture around the underground copper plate retains moisture, maintaining a highly conducting connection with the earth.

Question 19. How does a fuse protect household circuits and appliances from damage?
Answer: An electric fuse serves as a safety cut-off device designed to restrict current flow in a circuit. It consists of a short, thin segment of wire made from a lead-tin alloy. If the electrical current flowing through the circuit exceeds the rated safe limit, the heat generated by the current melts the low-melting-point alloy, immediately breaking the circuit and protecting the connected appliances from overloading.
In simple words: A fuse is a safety wire that melts and cuts the power if too much current flows through it, protecting appliances from burning out.

Exam Tip: Remember that the fuse is a sacrificial device - once it melts and breaks the circuit, it must be replaced before the circuit can function again.

Question 20. Identify the colors associated with the live, neutral, and earth wires in older domestic color coding schemes.
Answer: In traditional color coding schemes, the insulation colors are:
- **Live wire**: Red
- **Neutral wire**: Black
- **Earth wire**: Green
In simple words: The traditional colors for home wiring are red for live, black for neutral, and green for earth.

Exam Tip: Always read the question carefully to see if the examiner is asking for the traditional (Red/Black/Green) or modern international (Brown/Blue/Green-Yellow) color scheme.

Question 21. Why are electrical circuits and appliances protected by an earthing connection?
Answer: This safety measure is implemented to defend electrical circuits and appliances from short-circuit damage, preventing accidents such as electrical fires and lethal shocks. Earthing offers an exceptionally low-resistance pathway that immediately directs any excess current safely into the ground.
In simple words: Earthing protects you and your appliances by sending accidental electrical leakages safely into the ground, preventing shocks and fires.

Exam Tip: Explain that by providing a low-resistance path, the fault current rises high enough to blow the fuse instantly, isolating the faulty appliance.

Question 22. In which wire is a switch always connected in a household circuit?
Answer: A switch must always be wired in series with the **live wire** of a domestic circuit.
In simple words: Switches must always be connected to the live wire.

Exam Tip: Putting a switch in the neutral wire is extremely dangerous because the appliance remains live (at 220 V) even when the switch is turned off.

Question 23. State Fleming's Right-Hand Rule and explain what each finger represents.
Answer: Fleming's Right-Hand Rule is formulated as follows: Align the thumb, index finger, and middle finger of your right hand so they are perpendicular to one another. Under this configuration:
- The index finger points in the direction of the magnetic field.
- The thumb indicates the direction of motion of the conductor.
- The middle finger indicates the direction of the induced electrical current.
In simple words: Stretch the thumb, index, and middle fingers of your right hand at right angles. The index finger points with the magnetic field, the thumb shows the wire's motion, and the middle finger shows the direction of the induced current.

Exam Tip: This rule is fundamental for explaining the working of AC and DC generators.

Question 24. State Fleming's Left-Hand Rule and explain what each finger represents.
Answer: Fleming's Left-Hand Rule is formulated as follows: Stretch the thumb, index finger, and middle finger of your left hand so they are mutually perpendicular to one another. Under this configuration:
- The index finger indicates the direction of the magnetic field.
- The middle finger indicates the direction of the electric current flowing in the conductor.
- The thumb points in the direction of the mechanical force or motion exerted on the conductor.
In simple words: Stretch the thumb, index, and middle fingers of your left hand at right angles. The index finger points with the magnetic field, the middle finger with the current, and your thumb will point in the direction of the force (movement).

Exam Tip: This rule is used to find the direction of rotation in electric motors.

Question 25. What is a solenoid?
Answer: A solenoid is defined as a cylindrical coil made by winding a long insulated conducting wire into a series of tight loops, where the length of the cylinder is significantly larger than its diameter.
In simple words: A solenoid is a long wire wound into a tight coil that looks like a cylinder. When electricity flows through it, it acts like a magnet.

Exam Tip: The magnetic field inside a long, current-carrying solenoid is uniform and parallel to its axis.

Question 26. Can a copper wire be used as a fuse wire? Explain why.
Answer: No, a copper wire cannot be used as a fuse wire. This is because copper possesses a very low electrical resistance and an exceptionally high melting point, which prevents it from heating up and melting when a current overload occurs, risking a fire.
In simple words: No, copper wire cannot be used as a fuse because it does not melt easily. If too much current flows, it will let the electricity pass through, which can overheat your house wires and cause a fire.

Exam Tip: A good fuse wire must have high resistance and a low melting point so that it melts before the copper home wires can overheat.

Question 27. What type of electric current is supplied to our homes?
Answer: The electricity supplied to residential households is in the form of **alternating current** (A.C.).
In simple words: Alternating current (A.C.) is the type of electricity supplied to our homes.

Exam Tip: Alternating current is preferred for power transmission because its voltage can be easily stepped up or down using transformers, minimizing energy loss over long distances.

Question 28. Which alloy is used to manufacture fuse wires?
Answer: A fuse wire is manufactured using a specialized alloy of **lead and tin** due to its favorable thermal and electrical properties.
In simple words: An alloy made of lead and tin is used to make fuse wires.

Exam Tip: This alloy is chosen because it combines a low melting point with a relatively high resistivity.

Question 29. Explain the safety function of earthing in electrical circuits.
Answer: Earthing acts as a vital safety system to protect users from dangerous electrical shocks caused by short circuits. It involves linking the metallic chassis of an appliance to a thick copper cable buried deep underground, ending in a copper plate surrounded by charcoal and salt. This setup establishes an alternative low-resistance path, safely draining leaked currents to the ground.
In simple words: Earthing connects the metal body of an appliance to the ground. If a wire breaks and touches the metal, the electricity flows safely into the ground instead of shocking you.

Exam Tip: Grounding/earthing works by keeping the metal body of the appliance at zero potential, matching the earth's potential.

Question 30. Identify and compare series and parallel circuits. Explain why a parallel connection is preferred for household wiring.
Answer:
- **Arrangement (a)** represents a series circuit, where components are connected end-to-end in a single path.
- **Arrangement (b)** represents a parallel circuit, where components are connected across the same common terminals.
In domestic electrical wiring, a **parallel configuration** is strongly preferred due to the following advantages:
(i) Every household appliance operates independently across the same standard supply voltage.
(ii) If one appliance or bulb is switched off, is disconnected, or fails, the remaining appliances in the circuit continue to function normally without interruption.
In simple words: Series circuits connect everything in one single loop, while parallel circuits connect each appliance to its own path. Parallel is better for houses because every appliance gets the same voltage, and turning off one light won't turn off all the other lights in the house.

Exam Tip: In a series circuit, if one component breaks, the entire circuit opens and stops working. This is why series wiring is unsuitable for homes.

Question 31. Identify the terminals labeled E, N, and L in a three-pin plug. Explain why the earth pin is designed to be longer and thicker than the other two pins.
Answer: In a three-pin plug, the terminals correspond to:
- **E**: Earth pin
- **N**: Neutral pin
- **L**: Live pin
The **earth pin** is designed to be longer and thicker than the other two pins for crucial safety reasons:
- It is made **longer** so that when inserting the plug, the earth connection is established first, before the live and neutral pins make contact. When unplugging, it is disconnected last.
- It is made **thicker** to prevent the user from accidentally inserting the earth pin into the live or neutral slots of the wall socket.
Additionally, the terminal pins are slightly split at their tips to provide a spring action, ensuring a tight and secure fit inside the socket holes.
In simple words: E is for Earth, N is for Neutral, and L is for Live. The earth pin is longer so that the appliance is grounded first for safety before any electricity flows, and it is thicker so you can't accidentally plug it into the wrong slot.

Exam Tip: The 'safety-first' principle of the longer earth pin is a highly frequent question in board exams. Be sure to describe the sequence of connection clearly.

Question 32. Answer the following questions regarding a three-pin plug:
**(i) What is the purpose of terminal E?**
**(ii) To which pin of the plug is terminal E connected?**
**(iii) How is a fuse connected in relation to wire L, and how does it protect the appliance?**
Answer:
(i) The purpose of terminal E is to establish a secure safety ground or earth connection for the appliance.
(ii) Terminal E is directly wired to the earth pin of the three-pin plug.
(iii) The safety fuse is connected in series along the live wire (L). If a fault causes an excessive current to flow, the fuse wire heats up rapidly and melts, immediately opening the circuit to cut off power and protect the appliance from damage.
In simple words: (i) Terminal E provides the earth connection. (ii) It is connected to the top earth pin of the plug. (iii) The fuse is connected in series with the live wire so that if too much current flows, it melts and turns off the electricity to protect the appliance.

Exam Tip: Fuses must always be placed on the live side (L), never on the neutral side (N), to ensure that the appliance is completely disconnected from high potential when the fuse blows.

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Question 1. Answer the following multiple choice questions:
**(a) Find the current flowing in the given circuit.**
**(b) If the voltage across a conductor is doubled while resistance is kept constant, the current will...**
**(c) Materials that obey Ohm's law are classified as...**
**(d) What is the equivalent resistance of three \(18\ \Omega\) resistors connected in parallel?**
**(e) Kilowatt-hour is the unit of...**
**(f) What is the main function of earthing an appliance?**
**(g) Suggest a suitable fuse rating for an appliance drawing \(11\ \text{A}\) of current.**
Answer:
(a) **(i) 2/3 A** - The current is calculated by dividing the total potential difference by the equivalent resistance of the circuit.
(b) **(ii) double in value** - According to Ohm's law, current is directly proportional to voltage, so doubling the voltage doubles the current.
(c) **(iv) ohmic conductors** - Materials that show a linear relationship between voltage and current are called ohmic conductors.
(d) **(iv) 6 \(\Omega\)** - Three identical \(18\ \Omega\) resistors in parallel yield an equivalent resistance of \(18 / 3 = 6\ \Omega\).
(e) **(ii) electrical energy** - The kilowatt-hour (kWh) is the standard commercial unit of electrical energy.
(f) **(i) protect the user from electric shock by short circuiting and consequently breaking the circuit** - Earthing provides a safe route to ground, blowing the fuse and protecting the user.
(g) **(iii) A 13 ampere fuse is the most suitable rating to use** - A \(13\ \text{A}\) fuse is the closest safe standard rating above the operating current of \(11\ \text{A}\).
In simple words: This set of multiple-choice answers covers basic concepts of current calculation, Ohm's law, resistor networks, units of energy, earthing protection, and choosing correct fuse ratings.

Exam Tip: When selecting a fuse rating, always pick the next standard size above the normal operating current (e.g., choose 13 A for an 11 A load).

Question 2. Define the following electrical terms and state their units:
**(a) Potential difference**
**(b) (i) Coulomb, (ii) Ohm**
**(c) Electromotive force (e.m.f.)**
**(d) Semiconductors (with examples)**
**(e) Superconductors (with examples)**
Answer:
(a) **Potential difference**: This represents the amount of work required to transport a unit positive charge from one point to another in an electric field.
(b) **Units**:
(i) **Coulomb**: This is the standard SI unit of electric charge.
(ii) **Ohm**: This is the SI unit of electrical resistance. A conductor has a resistance of \(1\ \Omega\) if a current of \(1\ \text{A}\) flows through it under a potential difference of \(1\ \text{V}\).
(c) **Electromotive force (e.m.f.)**: When no current is being drawn from an electric cell (meaning the cell is in an open circuit), the potential difference across its terminals is defined as its electromotive force.
(d) **Semiconductors**: These are substances whose electrical resistance decreases as their temperature increases. Examples include Germanium and Silicon.
(e) **Superconductors**: These are materials whose electrical resistance decreases dramatically with falling temperature, reaching practically zero as they approach absolute zero. Examples include lead, tin, and mercury at extremely low temperatures.
In simple words: (a) Potential difference is the work needed to move a charge. (b) Coulomb measures charge, and Ohm measures resistance. (c) EMF is the voltage of a battery when it is not connected to any circuit. (d) Semiconductors conduct better when they get hotter (like Germanium). (e) Superconductors lose all their resistance when cooled close to absolute zero (like lead or tin).

Exam Tip: Make sure you clearly distinguish between EMF (open circuit voltage) and terminal voltage (closed circuit voltage) in your descriptive answers.

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Question 3. State Ohm's law. What are its main limitations?
Answer: Ohm's law states that the electrical current flowing through a conductor is directly proportional to the potential difference applied across its two ends, provided its temperature and other physical conditions remain constant.
The primary limitations of Ohm's law are:
1. It does not apply to non-ohmic conductors, such as diodes, vacuum tubes, and metal rectifiers, where current does not vary linearly with voltage.
2. It holds true only as long as the physical conditions of the conductor do not change.
3. It is applicable only when the temperature of the conducting material is maintained constant.
In simple words: Ohm's law says current increases with voltage. It doesn't work for non-ohmic parts like diodes, and it only works if the temperature and physical shape of the wire stay exactly the same.

Exam Tip: Non-ohmic conductors show a curved V-I graph instead of a straight line, which is a classic exam question topic.

Question 4. List the factors on which the resistance of a conductor depends. Distinguish between conductors and insulators with examples.
Answer: The electrical resistance of a conductor is determined by the following four factors:
(i) **Nature of the material**: Different substances have different concentrations of free conduction electrons.
(ii) **Length**: The resistance of a conductor is directly proportional to its length (\(R \propto l\)).
(iii) **Cross-sectional area**: The resistance is inversely proportional to the cross-sectional area of the conductor (\(R \propto 1/A\)).
(iv) **Temperature**: For metals, resistance generally increases as the temperature rises.

- **Conductors**: Materials that allow electrical charges to flow through them easily are called conductors (e.g., metals like copper, aluminium).
- **Insulators**: Materials that do not allow electrical charges to flow through them are called insulators (e.g., rubber, dry wood, plastic).
In simple words: Resistance depends on: what the wire is made of, its length (longer is more), its thickness (thinner is more), and its temperature (hotter is more). Conductors let electricity flow easily (like copper), while insulators block it (like rubber).

Exam Tip: Remember that resistivity (\(\rho\)) depends only on the material type and temperature, not on the length or area of the wire.

Question 5. Draw a circuit diagram containing a cell, key, ammeter, rheostat, resistor, and voltmeter, and state their functions.
Answer: A complete schematic diagram of this circuit configuration is illustrated below:
The respective components perform the following roles:
(A) **Cell**: Delivers voltage to maintain potential difference.
(B) **Key**: Controls the current path as an electrical switch.
(C) **Ammeter**: Gauges the current magnitude in series.
(D) **Rheostat**: Alters circuit resistance to adjust the current.
(E) **Resistor**: Introduces a fixed resistance value.
(F) **Voltmeter**: Measures the voltage drop across the resistor in parallel.
In simple words: This diagram shows how components are linked. The cell provides voltage, the key switches it, the ammeter measures current, the rheostat adjusts current, the resistor offers constant opposition, and the voltmeter measures voltage drop.

Exam Tip: Make sure the positive and negative terminals of the ammeter and voltmeter are aligned correctly with the positive and negative terminals of the cell.

Question 6. Find the total equivalent resistance between points A and B for each of the following resistor networks:
**(a) Two resistors of \(5\ \Omega\) and \(3\ \Omega\) in series, connected in parallel with an \(8\ \Omega\) resistor.**
**(b) A parallel combination of \(9\ \Omega\) and \(18\ \Omega\) resistors connected in series with a \(2\ \Omega\) resistor.**
**(c) Three \(2\ \Omega\) resistors connected in series, and this group connected in parallel with a fourth \(2\ \Omega\) resistor, with the entire arrangement flanked by two series \(2\ \Omega\) resistors.**
Answer:
**(a) First combination:**
- The series equivalent of the \(5\ \Omega\) and \(3\ \Omega\) resistors is: \[ R_s = 5\ \Omega + 3\ \Omega = 8\ \Omega \]
- This is connected in parallel with an \(8\ \Omega\) resistor: \[ R = \left( \frac{1}{8} + \frac{1}{8} \right)^{-1} = 4\ \Omega \]
**(b) Second combination:**
- The parallel combination of \(9\ \Omega\) and \(18\ \Omega\) is: \[ R_p = \left( \frac{1}{9} + \frac{1}{18} \right)^{-1} = \left( \frac{2+1}{18} \right)^{-1} = \frac{18}{3} = 6\ \Omega \]
- This is in series with a \(2\ \Omega\) resistor, so: \[ R_{\text{total}} = R_p + 2 = 6 + 2 = 8\ \Omega \]
**(c) Third combination:**
- The three \(2\ \Omega\) resistors in series have resistance: \[ R_{\text{group}} = 2 + 2 + 2 = 6\ \Omega \]
- This group is in parallel with the fourth \(2\ \Omega\) resistor: \[ R_p = \left( \frac{1}{6} + \frac{1}{2} \right)^{-1} = \left( \frac{1+3}{6} \right)^{-1} = \frac{6}{4} = 1.5\ \Omega \]
- Now, the total equivalent resistance of the network (with two more \(2\ \Omega\) resistors in series) is: \[ R_{\text{total}} = 2 + 1.5 + 2 = 5.5\ \Omega \]
In simple words: (a) First, combine series parts (5 + 3 = 8 ohms), then parallel them with 8 ohms to get 4 ohms. (b) Find parallel value of 9 and 18 ohms (which is 6 ohms), then add 2 ohms in series to get 8 ohms. (c) Simplify the series branch (6 ohms) and parallel it with 2 ohms to get 1.5 ohms, then add the outer series resistors (2 + 1.5 + 2) to get 5.5 ohms.

Exam Tip: Always draw intermediate simplified steps for complex resistor network questions to prevent calculation errors.

Question 7. Solve the following electrical charge problems:
**(a) Calculate the current flowing when a charge of \(80\ \text{C}\) passes through a point in \(2\ \text{minutes}\).**
**(b) Find the charge passing in \(8\ \text{seconds}\) if the current is \(4\ \text{A}\).**
Answer:
**(a) Calculation of Current (\(I\)):** Given:
- Charge, \(q = 80\ \text{C}\)
- Time, \(t = 2\ \text{minutes} = 2 \times 60 = 1200\ \text{s}\) \[ I = \frac{q}{t} = \frac{80}{1200} \approx 0.67\ \text{A} \]
**(b) Calculation of Charge (\(q\)):** Given:
- Current, \(I = 4\ \text{A}\)
- Time, \(t = 8\ \text{s}\) \[ q = I \cdot t = 4 \times 8 = 32\ \text{C} \]
In simple words: (a) Divide the charge (80 coulombs) by the time in seconds (1200 seconds) to get 0.67 amperes of current. (b) Multiply the current of 4 amperes by the time of 8 seconds to get 32 coulombs of charge.

Exam Tip: Ensure time is always in seconds before using the \(q = It\) formula to calculate charge or current.

Question 8. (a) Calculate the current flowing through a \(3\ \Omega\) resistor connected across a potential difference of \(6\ \text{V}\). (b) Determine the resistance of a filament lamp that draws \(0.5\ \text{A}\) of current under a potential difference of \(3\ \text{V}\).
Answer:
**(a) Calculation of Current (\(I\)):** Given:
- Voltage, \(V = 6\ \text{V}\)
- Resistance, \(R = 3\ \Omega\) \[ I = \frac{V}{R} = \frac{6}{3} = 2\ \text{A} \]
**(b) Calculation of Resistance (\(R\)):** Given:
- Voltage, \(V = 3\ \text{V}\)
- Current, \(I = 0.5\ \text{A}\) \[ R = \frac{V}{I} = \frac{3}{0.5} = 6\ \Omega \]
In simple words: (a) Divide 6 volts by 3 ohms to get 2 amperes of current. (b) Divide 3 volts by 0.5 amperes to find that the lamp's filament resistance is 6 ohms.

Exam Tip: These are direct applications of Ohm's law. Always write out the formula used clearly to gain partial marks in case of calculation errors.

Question 9. Two resistors P and Q are made of the same material and have the same length. The cross-sectional area of P is twice that of Q. Find:
**(a) the ratio of the resistance of P to the resistance of Q.**
**(b) the ratio of the current in P to the current in Q when connected across the same voltage supply.**
Answer: Given that resistors P and Q are composed of identical materials and share equal lengths, their resistivities (\(\rho\)) and lengths (\(l\)) are identical:
- Let the cross-sectional area of Q be \(a\).
- The cross-sectional area of P is \(2a\).

**(a) Ratio of Resistances (\(R_P / R_Q\)):** Using the formula \(R = \rho \frac{l}{A}\): \[ R_P = \rho \frac{l}{2a} \quad \text{and} \quad R_Q = \rho \frac{l}{a} \] \[ \frac{R_P}{R_Q} = \frac{\rho \frac{l}{2a}}{\rho \frac{l}{a}} = \frac{1}{2} \] Thus, the ratio of their resistances is **1:2**.

**(b) Ratio of Currents (\(I_P / I_Q\)):** Since the resistors are connected across the same potential difference \(V\), from Ohm's law (\(I = V/R\)), current is inversely proportional to resistance: \[ \frac{I_P}{I_Q} = \frac{R_Q}{R_P} = \frac{2}{1} \] Thus, the ratio of currents is **2:1**.
In simple words: (a) Since P is twice as thick as Q, it offers half the resistance, making the resistance ratio 1:2. (b) Under the same voltage, thinner wire Q has more resistance, so P draws twice as much current as Q, making the current ratio 2:1.

Exam Tip: A thicker wire has lower resistance and will therefore carry a higher current when connected across the same potential difference.

Question 10. Show how you would connect three \(6\ \Omega\) resistors to get an equivalent resistance of: (a) \(9\ \Omega\), and (b) \(4\ \Omega\). Draw the circuit diagrams and show your calculations.
Answer:
**(a) To obtain \(9\ \Omega\):** Connect two \(6\ \Omega\) resistors in parallel, and then connect this combination in series with the third \(6\ \Omega\) resistor.
- First, the parallel combination equivalent resistance is: \[ R_{\text{parallel}} = \left( \frac{1}{6} + \frac{1}{6} \right)^{-1} = 3\ \Omega \]
- Now, adding the third resistor in series: \[ R_{\text{total}} = 3\ \Omega + 6\ \Omega = 9\ \Omega \]
**(b) To obtain \(4\ \Omega\):** Connect two \(6\ \Omega\) resistors in series, and then connect this combination in parallel with the third \(6\ \Omega\) resistor.
- First, the series combination equivalent resistance is: \[ R_{\text{series}} = 6\ \Omega + 6\ \Omega = 12\ \Omega \]
- Now, putting this \(12\ \Omega\) equivalent resistance in parallel with the remaining \(6\ \Omega\) resistor: \[ R_{\text{total}} = \left( \frac{1}{12} + \frac{1}{6} \right)^{-1} = \left( \frac{1+2}{12} \right)^{-1} = \frac{12}{3} = 4\ \Omega \]
In simple words: (a) Putting two 6-ohm resistors in parallel gives 3 ohms. Adding the third 6-ohm resistor in series gives 9 ohms in total. (b) Putting two 6-ohm resistors in series gives 12 ohms. Connecting the third 6-ohm resistor in parallel with this 12-ohm branch yields 4 ohms.

Exam Tip: Practice these standard combination questions; they are highly popular in exams and test your conceptual understanding of series and parallel networks.

Question 11. Analyze the given circuit diagram where a \(3\ \text{V}\) cell is connected to a network with parallel \(3\ \Omega\) and \(6\ \Omega\) resistors connected in series with a \(10\ \Omega\) resistor. Calculate:
**(a) the equivalent resistance of the parallel combination.**
**(b) the total current flowing through the ammeter.**
**(c) the potential drop across the \(10\ \Omega\) resistor.**
**(d) the current passing through each of the parallel resistors.**
Answer: Given circuit parameters:
- Cell voltage, \(V = 3\ \text{V}\)
- Parallel branch resistors, \(r_1 = 3\ \Omega\) and \(r_2 = 6\ \Omega\)
- Series resistor, \(R = 10\ \Omega\)

**(a) Resistance of the parallel combination (\(R_p\)):** \[ \frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} \implies R_p = 2\ \Omega \]
**(b) Total resistance and current (\(I\)):** The total resistance of the entire circuit is: \[ R_{\text{total}} = R_p + R = 2\ \Omega + 10\ \Omega = 12\ \Omega \] The current recorded by the series-connected ammeter is: \[ I = \frac{V}{R_{\text{total}}} = \frac{3}{12} = 0.25\ \text{A} \]
**(c) Potential drop across the \(10\ \Omega\) resistor (\(V_1\)):** \[ V_1 = I \times 10\ \Omega = 0.25 \times 10 = 2.5\ \text{V} \]
**(d) Currents in individual parallel resistors:** The remaining potential difference across the parallel combination is: \[ V_2 = V - V_1 = 3\ \text{V} - 2.5\ \text{V} = 0.5\ \text{V} \]
- Current through the \(3\ \Omega\) resistor: \[ I_{3\Omega} = \frac{V_2}{3} = \frac{0.5}{3} \approx 0.17\ \text{A} \]
- Current through the \(6\ \Omega\) resistor: \[ I_{6\Omega} = \frac{V_2}{6} = \frac{0.5}{6} \approx 0.08\ \text{A} \]
In simple words: (a) The parallel 3 and 6-ohm resistors equal 2 ohms. (b) The total circuit resistance is 12 ohms, drawing 0.25 amperes of current. (c) The 10-ohm resistor drops 2.5 volts. (d) This leaves 0.5 volts for the parallel section, sending 0.17 A through the 3-ohm resistor and 0.08 A through the 6-ohm resistor.

Exam Tip: Always cross-check that the sum of the currents in the parallel branches (\(0.17\ \text{A} + 0.08\ \text{A} = 0.25\ \text{A}\)) equals the total circuit current.

Question 12. Find the total equivalent resistance between terminals P and Q for each of the following networks:
**(a) Five \(6\ \Omega\) resistors connected as shown, with two series \(6\ \Omega\) resistors flanking a parallel combination of two \(6\ \Omega\) resistors in series parallel with a third.**
**(b) A network consisting of parallel resistors of \(40\ \Omega\) and \(120\ \Omega\) connected in series with \(10\ \Omega\) and \(20\ \Omega\) resistors.**
Answer:
**(a) First Network:**
- The parallel section consists of a series branch of two \(6\ \Omega\) resistors in parallel with a third \(6\ \Omega\) resistor: \[ R_{\text{series}} = 6\ \Omega + 6\ \Omega = 12\ \Omega \] \[ R_{\text{parallel}} = \left( \frac{1}{12} + \frac{1}{6} \right)^{-1} = 4\ \Omega \]
- The total resistance between terminals P and Q (including the two outer series \(6\ \Omega\) resistors) is: \[ R_{\text{total}} = 6\ \Omega + 4\ \Omega + 6\ \Omega = 16\ \Omega \]
**(b) Second Network:**
- First, calculate the parallel equivalent of the \(40\ \Omega\) and \(120\ \Omega\) resistors: \[ \frac{1}{R_p} = \frac{1}{40} + \frac{1}{120} = \frac{3+1}{120} = \frac{4}{120} \implies R_p = 30\ \Omega \]
- This parallel combo is connected in series with the \(10\ \Omega\) and \(20\ \Omega\) resistors: \[ R_{\text{total}} = 10\ \Omega + 30\ \Omega + 20\ \Omega = 60\ \Omega \] This matches option (C).
In simple words: (a) Simplify the parallel section to get 4 ohms. Adding the two outer 6-ohm resistors in series gives 16 ohms total. (b) The parallel part yields 30 ohms. Adding 10 and 20 ohms in series gives a total resistance of 60 ohms, matching choice C.

Exam Tip: When solving series-parallel networks, scan the circuit and identify the innermost parallel loops to solve first.

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Question 13. Two identical cells, each of emf \(2\ \text{V}\) and internal resistance \(2\ \Omega\), are connected to an external resistor of \(4\ \Omega\). Find the total current in the circuit if the cells are connected: (a) in series, and (b) in parallel.
Answer: Given:
- Individual cell emf, \(e = 2\ \text{V}\)
- Individual cell internal resistance, \(r = 2\ \Omega\)
- External resistance, \(R = 4\ \Omega\)

**(a) Cells connected in series:**
- Total emf: \[ E_{\text{series}} = 2 \times 2\ \text{V} = 4\ \text{V} \]
- Total internal resistance: \[ r_{\text{series}} = 2 \times 2\ \Omega = 4\ \Omega \]
- Total current in the circuit: \[ I = \frac{E_{\text{series}}}{R + r_{\text{series}}} = \frac{4}{4 + 4} = 0.5\ \text{A} \]
**(b) Cells connected in parallel:**
- Total equivalent emf: \[ E_{\text{parallel}} = e = 2\ \text{V} \]
- Total equivalent internal resistance: \[ r_{\text{parallel}} = \left( \frac{1}{2} + \frac{1}{2} \right)^{-1} = 1\ \Omega \]
- Total current in the circuit: \[ I = \frac{E_{\text{parallel}}}{R + r_{\text{parallel}}} = \frac{2}{4 + 1} = 0.4\ \text{A} \]
In simple words: (a) In series, the total EMF is 4 V and total internal resistance is 4 ohms. Dividing 4 V by the 8-ohm total resistance gives 0.5 amperes. (b) In parallel, the EMF is still 2 V but the internal resistance drops to 1 ohm. Dividing 2 V by 5 ohms total resistance gives 0.4 amperes of current.

Exam Tip: Connecting identical cells in parallel does not increase the EMF, but it reduces the internal resistance, which can be useful for low-resistance loads.

Question 14. In the given circuit, a cell of voltage \(10\ \text{V}\) is connected in series with a \(5\ \Omega\) resistor and a parallel combination of a switch S and a \(20\ \Omega\) resistor. Calculate the current recorded by the ammeter when: (a) switch S is open, and (b) switch S is closed.
Answer: Given:
- Supply voltage, \(V = 10\ \text{V}\)
- Series resistor, \(R = 5\ \Omega\)

**(a) When switch S is open:** Since the switch is open, the \(60\ \Omega\) resistor is completely disconnected from the circuit path. The current flows solely through the \(20\ \Omega\) and \(5\ \Omega\) series resistors:
- Total circuit resistance: \[ R_{\text{total}} = 20\ \Omega + 5\ \Omega = 25\ \Omega \]
- Current in the circuit: \[ I = \frac{V}{R_{\text{total}}} = \frac{10}{25} = 0.4\ \text{A} \] The ammeter reads **0.4 A**.

**(b) When switch S is closed:** Closing the switch introduces the \(60\ \Omega\) resistor in parallel with the \(20\ \Omega\) resistor:
- Equivalent parallel resistance (\(R_p\)): \[ R_p = \left( \frac{1}{60} + \frac{1}{20} \right)^{-1} = \left( \frac{1+3}{60} \right)^{-1} = 15\ \Omega \]
- Total circuit resistance: \[ R_{\text{total}} = R_p + 5 = 15\ \Omega + 5\ \Omega = 20\ \Omega \]
- Current in the circuit: \[ I = \frac{V}{R_{\text{total}}} = \frac{10}{20} = 0.5\ \text{A} \] The ammeter reads **0.5 A**.
In simple words: (a) When open, only the 20-ohm and 5-ohm resistors are active, giving 25 ohms total. 10 V divided by 25 ohms is 0.4 A. (b) When closed, the 60-ohm and 20-ohm resistors are in parallel, which equals 15 ohms. Adding the 5-ohm resistor gives 20 ohms total. 10 V divided by 20 ohms is 0.5 A.

Exam Tip: Closing a switch that adds a parallel branch always reduces the total equivalent resistance of the circuit, which increases the total current drawn from the source.

Question 15. In a given circuit, a potential difference of \(1.8\ \text{V}\) is maintained across a \(6\ \Omega\) resistor, which is connected in parallel with a \(9\ \Omega\) resistor. This parallel combination is connected in series with a \(2.4\ \Omega\) resistor and a cell. Calculate:
**(a) the potential difference across the parallel combination.**
**(b) the current flowing through the \(9\ \Omega\) resistor.**
**(c) the potential difference across the \(2.4\ \Omega\) resistor.**
**(d) the electromotive force (emf) of the cell.**
Answer:
**(a) Potential difference across the parallel combination:** Since parallel branches share the same potential difference, the voltage drop across the entire parallel combination (including the \(9\ \Omega\) resistor) is equal to that across the \(6\ \Omega\) resistor: \[ V_{\text{parallel}} = 0.3\ \text{A} \times 6\ \Omega = 1.8\ \text{V} \]
**(b) Current through the \(9\ \Omega\) resistor (\(I_2\)):** Using Ohm's law with the parallel voltage of \(1.8\ \text{V}\): \[ I_2 = \frac{V_{\text{parallel}}}{9\ \Omega} = \frac{1.8}{9} = 0.2\ \text{A} \]
**(c) Potential difference across the \(2.4\ \Omega\) resistor (\(V_{2.4}\)):** The total current (\(I_{\text{total}}\)) flowing through the series part of the circuit is the sum of the branch currents: \[ I_{\text{total}} = 0.3\ \text{A} + 0.2\ \text{A} = 0.5\ \text{A} \] The potential drop across the \(2.4\ \Omega\) resistor is: \[ V_{2.4} = I_{\text{total}} \times 2.4\ \Omega = 0.5 \times 2.4 = 1.2\ \text{V} \]
**(d) Electromotive force (emf) of the cell:** Assuming negligible internal resistance, the total emf of the cell is the sum of the series potential drops: \[ \text{emf} = V_{\text{parallel}} + V_{2.4} = 1.8\ \text{V} + 1.2\ \text{V} = 3.0\ \text{V} \]
In simple words: (a) Since the resistors are in parallel, they share the same voltage of 1.8 V. (b) Dividing 1.8 V by 9 ohms gives 0.2 amperes through that resistor. (c) The total current is 0.5 A, so the 2.4-ohm resistor drops 1.2 V. (d) The battery's voltage is the sum of both drops, which is 3.0 V.

Exam Tip: When solving multi-part circuit problems, carry your results (like total current) forward systematically to solve the subsequent parts.

Question 16. A cell of emf \(6\ \text{V}\) and internal resistance \(0.8\ \Omega\) is connected to a parallel combination of two resistors \(2\ \Omega\) and \(3\ \Omega\). Find:
**(a) the equivalent resistance of the parallel combination and total circuit resistance.**
**(b) the potential drop across each parallel branch.**
**(c) the current flowing through each of the parallel resistors.**
**(d) the terminal potential difference across the battery.**
Answer: Given:
- emf, \(e = 6\ \text{V}\)
- Internal resistance, \(r = 0.8\ \Omega\)
- Resistors, \(r_1 = 2\ \Omega\) and \(r_2 = 3\ \Omega\)

**(a) Resistance calculations:** The parallel equivalent of the \(2\ \Omega\) and \(3\ \Omega\) resistors is: \[ R_p = \left( \frac{1}{2} + \frac{1}{3} \right)^{-1} = 1.2\ \Omega \] The total resistance of the circuit including the \(3\ \Omega\) series resistor and internal resistance is: \[ R_{\text{total}} = R_{\text{series}} + R_p + r = 3\ \Omega + 1.2\ \Omega + 0.8\ \Omega = 5\ \Omega \] The total current flowing through the ammeter is: \[ I = \frac{e}{R_{\text{total}}} = \frac{6}{5} = 1.2\ \text{A} \]
**(b) Potential drop across the \(3\ \Omega\) series resistor:** The voltage drop across the series resistor is: \[ V = 1.2\ \text{A} \times 3\ \Omega = 3.6\ \text{V} \]
**(c) Current in each parallel branch (C and D):** The total current (\(1.2\ \text{A}\)) splits between the parallel branches:
- Current in the \(2\ \Omega\) resistor (C): \[ I_C = \frac{1.2 \times 3}{2 + 3} = \frac{3.6}{5} = 0.72\ \text{A} \]
- Current in the \(3\ \Omega\) resistor (D): \[ I_D = \frac{1.2 \times 2}{2 + 3} = \frac{2.4}{5} = 0.48\ \text{A} \]
**(d) Terminal potential difference of the battery:** The terminal potential difference across the external circuit is: \[ V_{\text{terminal}} = e - (I \cdot r) = 6 - (1.2 \times 0.8) = 6 - 0.96 = 5.04\ \text{V} \]
In simple words: (a) The parallel part equals 1.2 ohms. Adding the 3-ohm series resistor and 0.8-ohm internal resistance gives 5 ohms total, drawing 1.2 A. (b) The series resistor drops 3.6 V. (c) The current splits: 0.72 A goes through the 2-ohm resistor, and 0.48 A goes through the 3-ohm resistor. (d) The terminal potential difference of the battery is 5.04 V.

Exam Tip: Terminal potential difference can also be found directly by multiplying total current by external resistance: \(V = I \times R_{\text{ext}} = 1.2 \times 4.2 = 5.04\ \text{V}\).

Question 17. Match the following physical quantities with their respective SI units: 1. Electrical potential, 2. Resistance, 3. Power, 4. Energy, 5. Resistivity.
Answer: The correct mapping of the electrical quantities to their standard units is:

In simple words: This table links common electrical terms with the correct units we use to measure them.

Exam Tip: Make sure to capitalize 'Ohm-meter' correctly and remember that resistivity is never written with 'per' (like ohm/meter).

Question 18. Four appliances are rated as follows: (i) a television of \(150\ \text{W}\), (ii) an electric iron of \(750\ \text{W}\), (iii) an immersion heater of \(3000\ \text{W}\), and (iv) a hair dryer of \(500\ \text{W}\). All are connected to a \(240\ \text{V}\) supply. Calculate the current drawn by each and suggest a suitable fuse rating for each appliance. Also, find the cost of running the television for \(4\ \text{hours}\) daily for \(100\ \text{days}\) at the rate of Rs. 0.6 per unit.
Answer: Using the power formula \(P = V \cdot I\), we calculate the operating current (\(I = P/V\)):

**(i) Television (\(150\ \text{W}\)):** \[ I = \frac{150}{240} = 0.625\ \text{A} \] A fuse rated at **2 A** is suitable.

**(ii) Electric iron (\(750\ \text{W}\)):** \[ I = \frac{750}{240} = 3.125\ \text{A} \] A fuse rated at **5 A** is suitable.

**(iii) Immersion heater (\(3000\ \text{W}\)):** \[ I = \frac{3000}{240} = 12.5\ \text{A} \] A fuse rated at **13 A** is suitable.

**(iv) Hair dryer (\(500\ \text{W}\)):** \[ I = \frac{500}{240} = 2.08\ \text{A} \] A fuse rated at **5 A** is suitable.

**(v) Energy cost calculation for television:** The total energy consumed in \(100\ \text{days}\) is: \[ E = \frac{150}{1000}\ \text{kW} \times 4\ \text{h/day} \times 100\ \text{days} = 60\ \text{kWh} \] At the tariff of Rs. 0.6 per unit (kWh): \[ \text{Cost} = 60 \times 0.6 = \text{Rs. } 36 \]
In simple words: The currents drawn are: TV (0.625 A, needs 2 A fuse), iron (3.125 A, needs 5 A fuse), heater (12.5 A, needs 13 A fuse), and dryer (2.08 A, needs 5 A fuse). Running the TV for 4 hours daily for 100 days uses 60 units of power, costing Rs. 36.

Exam Tip: Fuses must be chosen with a rating slightly above the operating current to prevent them from blowing during normal usage.

Question 19. (a) What is an electric fuse? (b) State two essential characteristics of a fuse wire. (c) Calculate the maximum power that can be drawn from a \(240\ \text{V}\) supply using a \(5\ \text{A}\) fuse. (d) What is the hazard of replacing a blown fuse with one of a higher rating?
Answer:
(a) An electric fuse is a safety device designed to restrict current flow in a circuit to prevent fire hazards and protect connected appliances.
(b) A fuse wire must possess **high electrical resistance** and a **low melting point**.
(c) The maximum power (\(P\)) that can be safely drawn is: \[ P = V \cdot I = 240\ \text{V} \times 5\ \text{A} = 1200\ \text{W} \]
(d) Replacing a blown fuse with a higher-rating fuse is highly dangerous. It allows a much larger, unsafe current to flow through the circuit without blowing, which can cause the home wiring to overheat, melt its insulation, and start a fire.
In simple words: (a) A fuse is a safety wire that cuts power under high current. (b) It needs high resistance and a low melting point. (c) At 240 V, a 5 A fuse allows a maximum power of 1200 W. (d) Using a higher-rated fuse is dangerous because it won't break when it should, which can cause wires to melt and catch fire.

Exam Tip: Always emphasize that a fuse is designed as the weakest link in a circuit, specifically to protect the expensive copper wiring of the house.

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Question 20. An electric heater of power \(2\ \text{kW}\) is connected to a \(250\ \text{V}\) supply. Calculate: (a) the current drawn, (b) the resistance of the heating element, (c) the heat produced in \(1\ \text{minute}\), and (d) the cost of running it for \(10\ \text{minutes}\) daily for \(30\ \text{days}\) at the rate of Rs. 3 per unit.
Answer: Given values:
- Power, \(P = 2\ \text{kW} = 2000\ \text{W}\)
- Supply voltage, \(V = 250\ \text{V}\)

**(a) Current drawn (\(I\)):** \[ I = \frac{P}{V} = \frac{2000}{250} = 8\ \text{A} \]
**(b) Resistance of the heating element (\(R\)):** \[ R = \frac{V^2}{P} = \frac{(250)^2}{2000} = \frac{62500}{2000} = 31.25\ \Omega \]
**(c) Heat produced in \(1\ \text{minute}\) (\(t = 60\ \text{s}\)):** \[ H = \frac{V^2}{R} \cdot t = \frac{(250)^2}{31.25} \times 60 = 2000 \times 60 = 1.2 \times 10^5\ \text{J} \]
**(d) Cost of running the heater:**
- Daily operating duration: \(t = 10\ \text{minutes} = \frac{10}{60}\ \text{h}\)
- Daily energy consumption: \[ E_{\text{daily}} = P \times t = 2\ \text{kW} \times \frac{10}{60}\ \text{h} = 0.33\ \text{kWh} \]
- Energy consumed in \(30\ \text{days}\): \[ E_{\text{total}} = 0.33\ \text{kWh} \times 30 = 10\ \text{kWh} \]
- Total cost at the rate of Rs. 3 per unit (kWh): \[ \text{Cost} = 10\ \text{units} \times \text{Rs. } 3 = \text{Rs. } 30 \]
In simple words: (a) The heater draws 8 amperes. (b) Its element resistance is 31.25 ohms. (c) It generates 120,000 Joules of heat in one minute. (d) Running it for 10 minutes daily for a month uses 10 units of electricity, costing Rs. 30.

Exam Tip: Convert power to kW to directly calculate energy in kWh, making the cost computation much simpler.

Question 21. An electric kettle rated \(3\ \text{kW}\) is run on a \(250\ \text{V}\) line. Calculate the current drawn and suggest a suitable fuse rating.
Answer: Given values:
- Power, \(P = 3\ \text{kW} = 3000\ \text{W}\)
- Voltage, \(V = 250\ \text{V}\)
Using the power relation: \[ I = \frac{P}{V} = \frac{3000}{250} = 12\ \text{A} \] Since the operational current is \(12\ \text{A}\), a standard fuse of **13 A** is recommended to protect this appliance.
In simple words: The kettle draws 12 amperes of current, so a 13 A fuse is the correct safety rating to use.

Exam Tip: Always choose a fuse rating slightly higher than the device's normal current so it does not blow under standard loads.

Question 22. A moment of force of \(10\ \text{N m}\) is applied by a force of \(20\ \text{N}\). Calculate the perpendicular distance from the line of action of the force to the pivot.
Answer: We use the definition of torque (moment of force): \[ \text{Moment of force} = \text{Force } (F) \times \text{Perpendicular distance } (d) \] Given:
- Moment of force \(= 10\ \text{N m}\)
- Applied force \(= 20\ \text{N}\)
Substituting these values: \[ 10 = 20 \times d \implies d = \frac{10}{20} = 0.5\ \text{m} \] Therefore, the perpendicular distance from the pivot is \(0.5\ \text{m}\).
In simple words: Divide the torque of 10 N m by the force of 20 N to find that the distance to the pivot is 0.5 meters.

Exam Tip: Torque or moment of force is a rotational physical quantity, so ensure you state the distance in meters.

Question 23. (a) Identify the proper terminal connections for the Brown, Blue, and Green wires in a three-pin socket labeled A, B, and C. (b) Does current flow through the earth terminal under normal circumstances? (c) How does earthing protect a user from fatal shocks?
Answer:
(a) The wire connections are assigned as follows:
- The **Brown wire** (live conductor) must be linked to terminal C.
- The **Blue wire** (neutral conductor) must be linked to terminal B.
- The **Green wire** (earth safety conductor) must be linked to terminal A.
(b) No, under standard operating conditions, no current passes through the earth safety terminal (A).
(c) If a live wire accidentally contacts the metal body of an appliance, the low-resistance earth wire provides a safe, alternative pathway for the leakage current to drain into the ground (which acts as an infinite charge sink). This protects the user from receiving a lethal electric shock.
In simple words: (a) Connect Brown (live) to C, Blue (neutral) to B, and Green (earth) to A. (b) No current flows through the earth wire normally. (c) Earthing sends accidental electrical leaks safely into the ground so they won't shock you.

Exam Tip: Remember the standard socket terminal letters: L is live, N is neutral, and E/A is earth.

Question 24. What is a kilowatt-hour (kWh)? Define it in terms of power and time.
Answer: A kilowatt-hour (\(\text{kWh}\)) is the commercial unit used to measure electrical energy. One kilowatt-hour represents the electrical energy consumed by a device of power rating \(1\ \text{kW}\) when operated continuously for a duration of \(1\ \text{hour}\).
In simple words: A kilowatt-hour is the amount of electricity used by a 1000-watt appliance running for one hour.

Exam Tip: Be sure to mention both '1 kW' and '1 hour' in your definition to secure full marks.

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Question 25. Describe the construction and working of an incandescent filament light bulb, and list the materials used.
Answer: An incandescent bulb consists of a thin metal filament connected to two support wires, which terminate at two metal contacts at the base. The filament is positioned in the center of the bulb and held up by a glass mount. All of these components are sealed inside a glass envelope filled with an inert gas like argon.
When connected to a voltage source, electric current flows through the filament. As electrons travel through the metal lattice, they collide with tungsten atoms, converting electrical energy into heat. This thermal agitation raises the temperature to around 4,000 degrees Fahrenheit, causing the filament to glow and emit visible light.
Tungsten is chosen for the filament due to its high melting point and strength. The inert argon gas surrounding it prevents the tungsten from evaporating easily and stops combustion reactions.
In simple words: A light bulb has a thin tungsten filament held in place by wires inside a glass bulb filled with argon gas. When electricity flows, it heats the tungsten until it glows white-hot, making light. The argon stops the filament from burning up.

Exam Tip: Always mention 'Tungsten' as the filament material and 'Argon' as the inert gas when explaining bulb construction.

Question 26. Calculate the household electricity bill for the month of April (30 days) if the following appliances are used daily: six \(100\ \text{W}\) bulbs for \(4\ \text{hours}\), four \(60\ \text{W}\) bulbs for \(8\ \text{hours}\), one \(2.5\ \text{kW}\) immersion heater for \(1\ \text{hour}\), and one \(800\ \text{W}\) electric iron for \(2\ \text{hours}\). The rate of electricity is Rs. 0.8 per unit.
Answer: First, evaluate the daily energy consumption in separate steps:
- **Six \(100\ \text{W}\) bulbs:** \[ E_1 = 6 \times \frac{100}{1000}\ \text{kW} \times 4\ \text{h} = 2.4\ \text{kWh/day} \]
- **Four \(60\ \text{W}\) bulbs:** \[ E_2 = 4 \times \frac{60}{1000}\ \text{kW} \times 8\ \text{h} = 1.92\ \text{kWh/day} \]
- **One \(2.5\ \text{kW}\) immersion heater:** \[ E_3 = 1 \times 2.5\ \text{kW} \times 1\ \text{h} = 2.5\ \text{kWh/day} \]
- **One \(800\ \text{W}\) electric iron:** \[ E_4 = 1 \times \frac{800}{1000}\ \text{kW} \times 2\ \text{h} = 1.6\ \text{kWh/day} \]
- **Total daily energy consumption:** \[ E_{\text{daily}} = 2.4 + 1.92 + 2.5 + 1.6 = 8.42\ \text{kWh/day} \]
- **Total energy consumed in April (30 days):** \[ E_{\text{total}} = 8.42\ \text{kWh/day} \times 30\ \text{days} = 252.6\ \text{kWh} \]
- **Total cost of electricity at Rs. 0.8 per unit:** \[ \text{Cost} = 252.6 \times 0.8 = \text{Rs. } 202.08 \]
In simple words: The bulbs, heater, and iron use a total of 8.42 units of electricity every day. In the 30 days of April, they consume 252.6 units. At 80 paise per unit, the total bill comes to Rs. 202.08.

Exam Tip: Carefully check the number of days in the specified month; April always has exactly 30 days.

Question 27. State the required physical properties and typical materials used for the following electrical components: (a) Connecting wires, (b) Fuse wires, (c) Heating elements, (d) Connecting wires of a power line, (e) Earthing elements.
Answer: The physical properties and typical materials are:
(a) **Connecting wires**: Must have low resistance, low resistivity, and a high melting point (e.g., copper, aluminium).
(b) **Fuse wires**: Must have high resistance and a low melting point (e.g., solder, an alloy of lead and tin).
(c) **Heating elements**: Must have high resistivity and a high melting point (e.g., nichrome, tungsten).
(d) **Power line wires**: Must have low resistance and high corrosion resistance (e.g., high-tension steel-reinforced wires).
(e) **Earthing elements**: Must be excellent conductors of electricity (e.g., copper wire, thick copper plates buried with salt and charcoal).
In simple words: (a) Hookup wires need to conduct well, like copper. (b) Fuses need to melt easily, like lead-tin solder. (c) Heating coils need high resistance and high melting points, like nichrome. (d) Power lines must be strong and carry current with low loss. (e) Ground wires must easily dump electricity into the earth.

Exam Tip: Make sure to match the specific properties (melting point and resistivity) with each component type to earn full marks.

Question 28. Find the ratio of the resistances of a \(60\ \text{W} - 220\ \text{V}\) bulb to a \(60\ \text{W} - 110\ \text{V}\) bulb.
Answer: We know that electrical resistance is given by the relation: \[ R = \frac{V^2}{P} \] For two bulbs of the same power rating (\(P = 60\ \text{W}\)):
- For the \(220\ \text{V}\) bulb: \[ R_{220} = \frac{(220)^2}{60} \]
- For the \(110\ \text{V}\) bulb: \[ R_{110} = \frac{(110)^2}{60} \] The ratio of their resistances is: \[ \frac{R_{220}}{R_{110}} = \frac{\frac{220^2}{60}}{\frac{110^2}{60}} = \left( \frac{220}{110} \right)^2 = (2)^2 = 4 \] Thus, the ratio of their resistances is **4:1**.
In simple words: Since resistance is proportional to the square of voltage, doubling the operational voltage (from 110 V to 220 V) increases the resistance by four times, giving a ratio of 4:1.

Exam Tip: When calculating ratios, simplify the common terms (like the power \(P = 60\ \text{W}\)) first to make the division extremely simple.

Question 29. An electric oven of power rating \(1.5\ \text{kW}\) is connected to a \(250\ \text{V}\) line. Find: (a) the current drawn, (b) the energy consumed in \(20\ \text{hours}\) in kWh, and (c) the cost of this energy at Rs. 1 per unit.
Answer: Given:
- Power, \(P = 1.5\ \text{kW} = 1500\ \text{W}\)
- Potential difference, \(V = 250\ \text{V}\)

**(a) Current drawn (\(I\)):** \[ I = \frac{P}{V} = \frac{1500}{250} = 6\ \text{A} \]
**(b) Energy consumed in \(20\ \text{hours}\) (\(t = 20\ \text{h}\)):** \[ E = P \times t = 1.5\ \text{kW} \times 20\ \text{h} = 30\ \text{kWh} \]
**(c) Cost of energy at Rs. 1 per unit:** \[ \text{Cost} = 30\ \text{units} \times \text{Rs. } 1 = \text{Rs. } 30 \]
In simple words: (a) The oven draws 6 amperes of current. (b) Running it for 20 hours uses 30 units of electricity. (c) At Rs. 1 per unit, the total cost is Rs. 30.

Exam Tip: Energy is calculated as Power (in kW) multiplied by Time (in hours) to directly find the cost-billing units.

Question 30. A bulb of resistance \(R\) is connected to a battery of emf \(4\ \text{V}\) and internal resistance \(2.5\ \Omega\). If the current in the circuit is \(0.5\ \text{A}\), find: (a) the resistance of the bulb, and (b) the heat energy dissipated in the bulb in \(10\ \text{minutes}\).
Answer: Given data:
- emf, \(e = 4\ \text{V}\)
- Internal resistance, \(r = 2.5\ \Omega\)
- Current, \(I = 0.5\ \text{A}\)

**(a) Resistance of the bulb (\(R\)):** Using Ohm's law for a complete circuit: \[ I = \frac{e}{R + r} \implies R + r = \frac{e}{I} \] \[ R + 2.5 = \frac{4}{0.5} \implies R + 2.5 = 8 \] \[ R = 8 - 2.5 = 5.5\ \Omega \]
**(b) Heat dissipated in the bulb in \(10\ \text{minutes}\) (\(t = 600\ \text{s}\)):** \[ H = I^2 \cdot R \cdot t = (0.5)^2 \times 5.5 \times 600 \] \[ H = 0.25 \times 5.5 \times 600 = 825\ \text{J} \]
In simple words: (a) The total resistance is 8 ohms, so subtracting the 2.5-ohm internal resistance leaves 5.5 ohms for the bulb. (b) The heat generated in 10 minutes (600 seconds) is 825 Joules.

Exam Tip: Make sure you only use the bulb's resistance (\(5.5\ \Omega\)) and not the total circuit resistance (\(8\ \Omega\)) when calculating the heat dissipated in the bulb itself.

Question 31. (a) What is the ring system of household wiring? (b) List two main advantages of a parallel connection over a series connection.
Answer:
(a) The **ring system** of wiring consists of a closed loop of three wires (live, neutral, and earth) that originates from the main fuse box, runs around all the rooms of a house, and returns to the fuse box. Separate parallel taps are taken from this ring for individual appliances.
(b) The main advantages of parallel connections are:
(i) Every appliance operates at the same supply voltage, ensuring they work at full rated capacity.
(ii) The operation of any appliance is completely independent of the others; switching off one device does not interrupt the current to any other device.
In simple words: (a) The ring system is a circular loop of wires that goes around the house and back to the fuse box, letting you tap electricity anywhere. (b) Parallel connections are better because they give every appliance the same voltage and keep other appliances running even if one is turned off.

Exam Tip: The ring system reduces the required wire thickness and cost because current can reach any socket from two opposite directions around the loop.

Question 32. (a) Explain the electrical hazard when a person touches an un-earthed appliance. (b) Why are switches and fuses always connected to the live wire rather than the neutral wire?
Answer:
(a) Touching an appliance whose live wire is in accidental contact with its metal body causes current to flow through the person to the ground. This occurs because the appliance is not earthed, resulting in a severe electric shock. If the metal body is earthed, the leakage current safely drains to the ground, causing the fuse to melt and break the circuit.
(b) Switches and fuses must always be connected directly to the live wire. This ensures that when a switch is turned off or a fuse blows, the appliance is completely isolated from the high potential of the live line, making it safe to touch or service.
In simple words: (a) Touching a faulty metal appliance that has no earth connection causes electricity to shock you on its way to the ground. (b) Connecting switches and fuses to the live wire ensures that turning them off completely cuts the high-voltage supply, keeping you safe.

Exam Tip: If a switch is placed on the neutral side, the appliance remains at a dangerous 220 V potential relative to the earth even when switched off.

Question 33. Define electromotive force (e.m.f.) of a cell.
Answer: The electromotive force (e.m.f.) is defined as the potential difference maintained across the terminals of an electric cell when no current is being drawn from it (meaning the cell is in an open circuit).
In simple words: Electromotive force is the maximum voltage a battery can supply when it is not connected to any external circuit.

Exam Tip: e.m.f. is a property of the cell itself, depending on its chemical components, and is independent of any external resistance.

Question 34. A cell of emf \(10\ \text{V}\) and internal resistance \(2.5\ \Omega\) is connected to two identical resistors \(r_1 = r_2 = 50\ \Omega\). Calculate the power dissipated in each resistor when they are connected: (a) in series, and (b) in parallel.
Answer: Given data:
- emf, \(e = 10\ \text{V}\)
- Internal resistance, \(r = 2.5\ \Omega\)
- Resistors, \(r_1 = r_2 = 50\ \Omega\)

**(a) Resistors connected in series:** The external resistance is: \[ R_s = 50\ \Omega + 50\ \Omega = 100\ \Omega \] The total current flowing in the circuit is: \[ I = \frac{e}{R_s + r} = \frac{10}{100 + 2.5} \approx 0.098\ \text{A} \] The power dissipated in each resistor is: \[ P_1 = I^2 \cdot r_1 = (0.098)^2 \times 50 \approx 0.48\ \text{W} \]
**(b) Resistors connected in parallel:** The external parallel resistance is: \[ R_p = \left( \frac{1}{50} + \frac{1}{50} \right)^{-1} = 25\ \Omega \] The total current in the circuit is: \[ I = \frac{e}{R_p + r} = \frac{10}{25 + 2.5} = 0.364\ \text{A} \] The terminal potential difference across the parallel combination is: \[ V = e - I \cdot r = 10 - (0.364 \times 2.5) = 10 - 0.91 = 9.09\ \text{V} \] The power dissipated in each of the parallel resistors is: \[ P_{\text{parallel}} = \frac{V^2}{r_1} = \frac{(9.09)^2}{50} \approx 1.65\ \text{W} \]
In simple words: (a) In series, the total resistance is 102.5 ohms, drawing 0.098 A. Each resistor uses 0.48 watts of power. (b) In parallel, the external resistance is 25 ohms, drawing 0.364 A. This drops the terminal voltage to 9.09 V, so each resistor uses 1.65 watts.

Exam Tip: Silently correct textbook printing errors (like the original 4140.5 W value) and carry out correct divisions on scrap paper before finalized writing.

Question 35. The following table lists the power and voltage ratings of five household appliances. Find the resistance of each appliance and identify which one has the largest resistance:

Answer: Using the resistance-power relation: \[ R = \frac{V^2}{P} \] We evaluate each appliance systematically:
- **(a) washing machine:** \[ R_a = \frac{250^2}{3000} \approx 20.8\ \Omega \]
- **(b) Television:** \[ R_b = \frac{240^2}{160} = 360\ \Omega \]
- **(c) Electric iron:** \[ R_c = \frac{240^2}{1500} \approx 38.4\ \Omega \]
- **(d) Hair curler:** \[ R_d = \frac{250^2}{20} = 3125\ \Omega \]
- **(e) Car head lamp:** \[ R_e = \frac{12^2}{36} = 4\ \Omega \] Comparing these values, the **hair curler** clearly possesses the largest electrical resistance (\(3125\ \Omega\)).
In simple words: By calculating \(V^2/P\) for each, we find the resistances: machine (20.8 ohms), TV (360 ohms), iron (38.4 ohms), curler (3125 ohms), and car lamp (4 ohms). The hair curler has the highest resistance.

Exam Tip: Write out the formula and show at least one sample calculation clearly in your exam to secure method marks.

Question 36. Calculate the energy transferred when a charge of \(20\ \text{C}\) is moved across a potential difference of \(10^8\ \text{V}\).
Answer: The relation for energy transferred (\(E\)) is given by: \[ E = \text{Potential difference } (V) \times \text{Charge } (Q) \] Given values:
- Potential difference, \(V = 10^8\ \text{V}\)
- Charge, \(Q = 20\ \text{C}\)
Substituting these values: \[ E = 10^8 \times 20 = 2 \times 10^9\ \text{J} \] This corresponds to option (d).
In simple words: Multiplying the charge of 20 coulombs by the potential difference of \(10^8\) volts gives \(2 \times 10^9\) Joules of energy.

Exam Tip: High voltage operations (such as lightning strikes) involve massive energy transfers, which are naturally calculated in scientific exponential notation.

Question 37. Which of the following statements is incorrect for a household circuit containing appliances that draw a maximum combined current of \(11\ \text{A}\)?
Answer: (a) A 13A fuse is the most appropriate value to use
In simple words: This answer identifies the incorrect option from the multiple-choice question.

Exam Tip: Standard domestic fuses come in specific ratings (like 3A, 5A, 13A). Always choose the closest size above the normal operating current of the circuit.

Question 38. (i) State the working principle of a transformer. (ii) What is the main function of a step-up transformer? (iii) Can a transformer work on a direct current (d.c.) source? Explain why. (iv) Sketch a labeled diagram of a step-down transformer and (v) a step-up transformer.
Answer:
(i) A transformer works on the principle of **mutual electromagnetic induction** between two coils (primary and secondary).
(ii) The function of a **step-up transformer** is to increase the alternating potential difference (voltage) while proportionally decreasing the alternating current.
(iii) No, a transformer cannot operate on a direct current (d.c.) source. This is because d.c. provides a constant magnetic field, meaning there is no change in the magnetic flux linked with the secondary coil to induce an electromotive force.
(iv) **Step-down transformer representation:**
(v) **Step-up transformer representation:**
In simple words: (i) Transformers work by transferring electricity from one coil to another using changing magnetic fields. (ii) Step-up transformers increase AC voltage while decreasing current. (iii) They cannot work on DC because constant current doesn't create the changing magnetic fields needed for induction.

Exam Tip: A step-up transformer has more turns in its secondary coil than its primary coil, whereas a step-down transformer has the opposite.

Question 39. Two coaxial coils X and Y are placed near each other. (i) If a current flows anticlockwise at end B of coil X, what is the magnetic polarity at this end? (ii) What polarity is induced at end C of coil Y when the key in coil X is closed? How does this change when the current in coil X becomes steady? (iii) Describe the induced polarity at end C of coil Y when: (a) coil Y is moved toward coil X, and (b) coil Y is moved away from coil X.
Answer:
(i) An anticlockwise current at end B of coil X indicates that this face behaves as a **North pole** according to the clock rule.
(ii) While closing the key of coil X, the rising magnetic flux induces a **North pole** at end C of the adjacent coil Y to oppose the change (Lenz's law). Once the current in coil X reaches a steady value, the magnetic flux stops changing, and coil Y loses all induced polarity.
(iii)
- (a) When coil Y is moved closer to coil X, a **North pole** is induced at end C of coil Y to repel the incoming coil.
- (b) When coil Y is moved away from coil X, a **South pole** is induced at end C of coil Y to attract and oppose its motion.
In simple words: (i) Anticlockwise current means the end acts as a North pole. (ii) Closing the switch induces a North pole in coil Y to push back, but once current stops changing, the polarity vanishes. (iii) Moving Y closer induces a North pole to repel it, while pulling Y away induces a South pole to pull it back.

Exam Tip: Use Lenz's law ("the induced current always opposes the change that produces it") to determine the induced magnetic polarities.

Question 40. The diagram shows a fixed coil of several turns connected to a center-zero galvanometer G and a magnet NS which can move in the direction shown. Describe the observation in the galvanometer if:
(a) (i) The magnet is moved rapidly in the direction of the arrow,
(ii) The magnet is kept still inside the coil,
(iii) The magnet is then rapidly pulled out of the coil.
(b) How would the observation in (i) of part (a) change if a more powerful magnet is used?
Answer:
(a) (i) Rapidly pushing the magnet toward the coil alters the magnetic flux passing through it, producing a momentary deflection on the galvanometer scale which shows that an induced current is flowing.
(ii) If the magnet is held completely still inside the coil, the magnetic flux remains constant, resulting in zero deflection because no electromotive force is induced.
(iii) Pulling the magnet quickly out of the coil causes another change in flux, deflecting the galvanometer needle in the reverse direction to indicate that current is now flowing the other way.
(b) Employing a stronger magnet increases the rate of change of magnetic flux, leading to a much larger deflection on the galvanometer due to a higher induced current.
In simple words: Electricity is only made when the magnet is moving. Moving it in makes the needle go one way, holding it still does nothing, and pulling it out makes the needle go the other way.

Exam Tip: Emphasize that the deflection is temporary and only lasts as long as there is relative motion between the magnet and the coil. Use the term "rate of change of magnetic flux" to explain why a faster or stronger magnet causes a bigger deflection.

Question 41. (a) A strong short bar magnet is dropped into a solenoid connected to a galvanometer. The galvanometer shows a deflection to the left when the current in the coil as seen from above is counter-clockwise. Which of the following shows the correct deflection of the galvanometer when the magnet is at positions A, B, and C?
(i) Left, left, left
(ii) Left, zero, zero
(iii) Left, zero, right
(iv) Left, right, left
(v) Left, right, right

(b) Consider two cases of two parallel current-carrying conductors. Current in the same direction and currents in the opposite directions will produce:
(i) Attraction and repulsion respectively
(ii) Repulsion and attraction respectively
(iii) Attraction in both cases
(iv) Repulsion in both cases
(v) Oscillation in both cases

(c) Which of the following changes would enable a milliammeter to be used as a voltmeter?
(i) connecting a large resistor in series
(ii) connecting a large resistor in parallel
(iii) connecting a small resistor in series
(iv) connecting a small resistor in parallel

(d) XY is at right angles to the magnetic field. The direction in which XY should be moved to induce an electron flow in the direction Y to X is:
(i) Along the direction of XY
(ii) Towards N and perpendicular to XY
(iii) Towards S and perpendicular to XY
(iv) Upwards and perpendicular to XY

(e) There are 500 turns and 2000 turns in the primary and secondary coil of a transformer respectively. If the output voltage is 1000 V, how large is the input voltage?
(i) 250 V
(ii) 500 V
(iii) 1000 V
(iv) 2000 V
(v) 4000 V
Answer:
(a) (v) Left, left, left
(b) (ii) Repulsion and attraction respectively
(c) (i) connecting a large resistor in series
(d) (iv) Upwards and perpendicular to XY
(e) (i) 250 V
In simple words: These multiple-choice questions cover several key ideas: magnet movement, how parallel currents attract or repel, converting a meter into a voltmeter, the direction of induced current, and calculating transformer voltages.

Exam Tip: For conversion of a galvanometer, always remember that a voltmeter needs a high resistance in series (multiplier), whereas an ammeter needs a low resistance in parallel (shunt).

Question 42. What would be the effect of using the following materials for the core of an electric bell?
(a) Plastic
(b) Steel
(c) Copper
Answer:
(a) Since plastic is a non-magnetic substance, it cannot concentrate or strengthen the magnetic field, making it useless as a core material.
(b) Steel possesses high magnetic retentivity, meaning it holds onto its magnetism even when the current is cut off. This would cause the bell's armature to remain permanently attracted, stopping the ringing action.
(c) A copper core would lead to strong eddy currents forming within it, which creates an opposing magnetic effect and wastes energy as heat, disrupting the proper function of the bell.
In simple words: An electric bell needs a core that magnetizes and demagnetizes instantly. Plastic cannot magnetize, steel stays magnetized too long, and copper creates unwanted currents that interfere with the bell.

Exam Tip: Explain the magnetic properties clearly - use terms like "non-magnetic" for plastic, "high retentivity" for steel, and "eddy currents" for copper to ensure full marks.

Question 43. (a) Draw the magnetic field lines due to a loop of wire carrying current.
(b) Explain why two sections of the loop behave as they do when current flows.
(c) How can the repulsive force between the sections be increased?
Answer:
(a) The magnetic field lines are illustrated in the diagram below: (b) Because the magnetic flux lines traverse the interior of the loop in an identical direction, the two segments experience mutual repulsion.
(c) Increasing the electric current flowing through the loop will amplify the magnetic field strength, thereby boosting the repulsive force.
In simple words: The magnetic lines of force push in the same direction inside the loop, causing its parts to repel. If you run more current through the loop, this push becomes even stronger.

Exam Tip: When sketching the magnetic field of a loop, ensure that the field lines pass through the center in one direction and form concentric loops around the wires on either side.

Question 44. (a) Why does the coil of a d.c. motor rotate?
(b) How can the direction of rotation of the armature of a d.c. motor be reversed?
(c) State three ways to increase the speed of rotation of a d.c. motor.
Answer:
(a) The rotation of the armature coil is driven by the equal and opposite electromagnetic forces that act on its parallel sides, creating a torque or couple. (b) To reverse the direction of rotation, one can simply swap the terminal connections of the battery connected to the brushes of the motor.
(c) The rotational speed can be enhanced by:
1. Increasing the electric current passing through the armature.
2. Winding more turns of wire on the coil.
3. Increasing the strength of the magnetic field from the poles.
In simple words: A motor spins because the magnetic field pushes on the wires carrying current. Swapping the battery wires reverses the spin, and using more current or turns makes it spin faster.

Exam Tip: Use the term "couple" or "deflecting torque" when explaining why the coil rotates. Clearly list the three factors affecting motor speed as separate points.

Question 45. When the north pole of a magnet is pushed into a coil connected to a galvanometer, a deflection is observed.
(i) Why is a deflection observed in the galvanometer?
(ii) What is the direction of current when viewed from the end where the magnet enters?
(iii) What happens to the deflection if the magnet is pulled out of the coil?
(iv) What is observed if the magnet is held stationary inside the coil?
Answer:
(i) Pushing the magnet into the coil changes the magnetic flux linked with it, which induces an electromotive force (e.m.f.) and causes a current to register on the galvanometer.
(ii) As viewed from end A, the current moves counter-clockwise because Lenz's law dictates that end A must develop a North pole to repel the incoming North pole of the magnet.
(iii) Pulling the magnet out causes the galvanometer needle to swing to the left (the opposite direction).
(iv) Since there is no relative movement when the magnet is stationary, the magnetic flux does not change, resulting in zero deflection.
In simple words: Moving a magnet inside a coil creates a changing magnetic field that produces electricity. The direction of the current depends on which way the magnet is moving, and stops completely if the magnet is still.

Exam Tip: Always explain the direction of the induced current using Lenz's Law - the induced pole must oppose the motion of the magnet.

Question 46. A primary coil of a step-down transformer has 800 turns and is connected to a 220 V A.C. supply. If the secondary coil has 8 turns, calculate the output voltage.
Answer:
We are given the following values:
Number of primary turns (\(N_p\)) = 800
Number of secondary turns (\(N_s\)) = 8
Primary voltage (\(E_p\)) = \(220\text{ V}\)
Using the transformer turn ratio relation: \[ \frac{N_s}{N_p} = \frac{E_s}{E_p} \] Solving for the output voltage of the secondary coil (\(E_s\)): \[ E_s = \frac{N_s}{N_p} \times E_p \] \[ E_s = \frac{8}{800} \times 220 \] \[ E_s = \frac{1}{100} \times 220 = 2.2\text{ V} \] Therefore, the secondary output voltage is \(2.2\text{ V}\).
In simple words: Since the secondary coil has one-hundredth the number of turns as the primary coil, the output voltage will also be one-hundredth of the input voltage, which is 2.2 volts.

Exam Tip: Always state the formula clearly before performing the calculation. Keep track of your units (\(V\) for volts) in the final step to secure full marks.

Question 47. How can a moving coil galvanometer be converted into:
(i) an ammeter,
(ii) a voltmeter?
Answer:
(i) To convert a galvanometer into an ammeter, connect a low-resistance wire (known as a shunt) in parallel across the galvanometer coil.
(ii) To convert a galvanometer into a voltmeter, connect a high-resistance resistor in series with the galvanometer coil.
In simple words: A shunt (low resistance) in parallel lets most current bypass the meter so it can act as an ammeter. A large resistance in series limits the current so the meter can safely measure voltage.

Exam Tip: Use key terms like "shunt" and "multiplier". Be sure to specify the connection type - parallel for the ammeter shunt and series for the voltmeter resistor.

Question 48. Why is an ammeter connected in series and a voltmeter connected in parallel in an electric circuit?
Answer:
An ammeter has very low resistance so that it does not decrease the current flowing through the circuit, which is why it must be connected in series. Conversely, a voltmeter has extremely high resistance so that it draws negligible current from the circuit, which is why it is connected in parallel across the components.
In simple words: An ammeter must not block the current it is measuring, so it has low resistance and goes in the main path (series). A voltmeter must not steal any current, so it has high resistance and goes across the device (parallel).

Exam Tip: Explain this in terms of circuit impact: connecting an ammeter in parallel would cause a short circuit, while connecting a voltmeter in series would stop the current flow.

Question 49. Describe two experiments to demonstrate the phenomenon of electromagnetic induction.
Answer:
Experiment 1 (Single Coil and Magnet):
1. Wind a coil of insulated wire \(AB\) with many turns and connect its ends to a sensitive galvanometer.
2. Thrust the North pole of a strong bar magnet quickly into end \(B\) of the coil. A temporary deflection will be observed on the galvanometer, showing an induced current.
3. If you pull the magnet out of the coil, the needle deflects in the opposite direction.
4. Keeping the magnet stationary inside the coil results in no deflection, showing that relative motion is essential.

Experiment 2 (Two Coils):
1. Place two independent coils of copper wire, Coil 1 (primary) and Coil 2 (secondary), wound on a non-conducting hollow tube.
2. Connect Coil 1 in series with a battery and a tap key, and connect Coil 2 to a galvanometer.
3. Upon closing the key in Coil 1, a momentary deflection is observed in Coil 2. When the current becomes steady, the deflection returns to zero.
4. Opening the key causes a brief deflection in the opposite direction.

To find the direction of the induced current, we use Fleming's Right-Hand Rule:
Stretch the thumb, forefinger, and middle finger of your right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the thumb indicates the direction of motion of the conductor, then the middle finger shows the direction of the induced current.
In simple words: We can create current by moving a magnet in and out of a coil, or by switching the current on and off in a nearby coil. Fleming's right-hand rule tells us which way the induced current will flow.

Exam Tip: When describing these experiments, clearly state the observations (momentary deflection) and the conclusions (flux change causes induction). Ensure you describe Fleming's Right-Hand Rule correctly.

Question 50. (a) What is observed in the galvanometer of the secondary coil when current is switched on in the primary coil?
(b) What will be the effect of introducing a soft iron bar in the tube?
Answer:
(a) The galvanometer needle will momentarily deflect to one side and then quickly return to the zero position, which indicates a brief, temporary flow of current in the secondary coil.
(b) Placing a soft iron rod inside the tube concentrates the magnetic field lines, increasing the magnetic flux linkage and leading to a larger induced current and a larger galvanometer deflection.
In simple words: Switching the current on creates a quick burst of electricity in the second coil. Putting an iron bar inside helps guide the magnetism, making the electrical pulse much stronger.

Exam Tip: Emphasize that the deflection is "momentary" or "transient" because induction only occurs when the magnetic field is changing (increasing or decreasing), not when it is steady.

Question 51. (a) Name the material used for the core of a transformer and describe its structure.
(b) A transformer has an input voltage of 250 V and output voltage of 50 V. Calculate its turns ratio and state whether it is a step-up or step-down transformer.
(c) If the current in the secondary coil is 2 A, calculate the current in the primary coil.
Answer:
(a) The core of a transformer is constructed from soft iron. To minimize energy loss due to eddy currents, the core is made of thin, laminated sheets of soft iron (or silicon steel) that are insulated from one another and stacked together. (b) We are given:
Primary voltage (\(E_p\)) = \(250\text{ V}\)
Secondary voltage (\(E_s\)) = \(50\text{ V}\)
The turns ratio of the transformer is: \[ \text{Turns Ratio} = \frac{N_s}{N_p} = \frac{E_s}{E_p} \] \[ \frac{N_s}{N_p} = \frac{50}{250} = \frac{1}{5} \] Since the turns ratio is less than 1, it is a step-down transformer.
(c) We are given:
Secondary current (\(I_s\)) = \(2\text{ A}\)
Using the current relationship for an ideal transformer: \[ \frac{I_p}{I_s} = \frac{E_s}{E_p} = \frac{1}{5} \] \[ I_p = \frac{1}{5} \times I_s \] \[ I_p = \frac{1}{5} \times 2 = 0.4\text{ A} \] Thus, the current in the primary coil is \(0.4\text{ A}\).
In simple words: A soft iron core helps transfer magnetism efficiently. Since the voltage drops from 250 to 50 volts, the number of turns decreases by five times, and the primary current is only 0.4 amperes.

Exam Tip: Keep in mind that for a step-down transformer, the voltage decreases but the current increases. The primary current must always be smaller than the secondary current in a step-down transformer.

Question 52. (a) Draw a labeled diagram of an AC generator.
(b) State one way to increase the current produced by the generator.
(c) Which terminal of the generator will be positive when rotating?
(d) Under what position of the coil is the magnetic flux linked maximum and the potential difference zero?
(e) Draw a graph showing the variation of induced emf with the angle of rotation of the coil.
Answer:
(a) The generator diagram is shown below: (b) The current generated can be increased by winding more turns of wire on the coil.
(c) Under the given conditions, terminal X will act as the positive terminal.
(d) When the armature coil's plane lies perpendicular (normal) to the magnetic field lines, the magnetic flux passing through the coil is at its maximum, but the rate of change of flux is zero, so the induced potential difference is zero.
(e) The wave graph showing the induced e.m.f. variation is shown below: In simple words: An AC generator makes current that changes direction back and forth. When the coil is perpendicular to the magnets, the magnetic link is biggest, but since it is not changing at that split second, the voltage is zero.

Exam Tip: Clearly distinguish between the positions of maximum flux (normal to field) and maximum induced voltage (parallel to field). The voltage depends on the rate of change of flux, not the flux itself.

Question 53. The diagram shows a transformer connected to an AC source and a bulb.
(a) What happens to the brightness of the bulb if the number of turns in the secondary coil is increased? Give a reason.
(b) What happens if the core of the transformer is removed?
(c) Why is the core of a transformer not made of copper?
(d) Why cannot a transformer be used with a direct current (D.C.) source?
Answer:
This is a diagram of a step-down transformer.
(a) The bulb will glow more brightly because a greater number of turns in the secondary winding increases the magnetic flux linkage, which raises the induced output voltage across the lamp.
(b) If the soft iron core (Part X) is taken out, it turns into an open-core system. This leads to a severe leakage of magnetic flux, meaning many of the magnetic field lines from the primary coil will fail to reach the secondary coil.
(c) If copper is used to make the core, large eddy currents will be induced in it, leading to massive power losses in the form of heat.
(d) Transformers require alternating current to function. Since direct current (D.C.) is steady and unchanging, the magnetic flux it creates is also constant. Without a changing magnetic flux, no voltage can be induced in the secondary coil.
In simple words: More turns in the output coil make a brighter bulb. Taking the core out lets the magnetism leak away, and using copper instead of iron creates heat-wasting currents. Direct current cannot be used because it doesn't change.

Exam Tip: Use scientific terms such as "flux leakage" when explaining the effect of removing the core, and "eddy currents" to describe the losses in a copper core.

Question 54. A transformer has 400 turns in the primary coil and 10 turns in the secondary coil. The primary is connected to a 250 V AC supply, and the current in the primary is 2 A.
(a) Calculate the output voltage.
(b) Calculate the current in the secondary coil.
Answer:
Since the primary winding has more turns than the secondary winding (\(N_p > N_s\)), this device is a step-down transformer.
Given data:
\(N_p = 400\)
\(N_s = 10\)
\(E_p = 250\text{ V}\)
\(I_p = 2\text{ A}\)

(a) Calculating the secondary voltage (\(E_s\)): Using the turns ratio equation: \[ \frac{N_s}{N_p} = \frac{E_s}{E_p} \] \[ E_s = \frac{10}{400} \times 250 \] \[ E_s = \frac{1}{40} \times 250 = 6.25\text{ V} \]
(b) Calculating the secondary current (\(I_s\)): Using the current inverse relation with turns ratio: \[ \frac{N_s}{N_p} = \frac{I_p}{I_s} \] \[ I_s = \frac{N_p}{N_s} \times I_p \] \[ I_s = \frac{400}{10} \times 2 = 40 \times 2 = 80\text{ A} \]
In simple words: The transformer steps down the voltage by forty times, so the voltage drops to 6.25 volts. However, this raises the current by forty times, boosting it from 2 to 80 amperes.

Exam Tip: Keep in mind that as voltage goes down, current goes up in a step-down transformer. Always write down the separate equations for voltage and current to get step-by-step marks.

Question 55. State two features of a transformer that help in increasing its efficiency.
Answer:
The efficiency of a transformer is optimized by the following design features:
(i) Using a laminated core, which breaks up path loops and significantly reduces energy losses from induced eddy currents.
(ii) Using a closed soft-iron core, which provides a continuous path of high permeability to minimize both magnetic flux leakage and energy loss from magnetic hysteresis.
In simple words: To stop energy from being wasted, the transformer's iron core is built out of thin insulated layers (which stops heating) and is closed in a loop (which keeps the magnetic field inside).

Exam Tip: Explain both features using their respective physical causes: laminations combat eddy currents, while a soft iron material minimizes hysteresis losses.

Question 56. Fill in the blanks by writing (i) Only soft iron, (ii) Only steel, (iii) Both soft-iron and steel for the material of core and/or magnet:
(a) Electromagnet ______
(b) Electric bell ______
(c) D.C. motor ______
(d) A.C. generator ______
(e) Transformer ______
Answer:
(a) magnet - soft iron (due to low retentivity)
(b) core - soft iron (needs temporary magnetization)
(c) core - soft iron, magnet - steel (permanent field needed)
(d) core - soft iron, magnet - steel
(e) core - soft iron (to guide changing magnetic fields)
In simple words: Cores always use soft iron because it demagnetizes quickly when current is turned off. Permanent magnets use steel because they need to hold their magnetism forever.

Exam Tip: Match the magnetic material with its function: soft iron is used for temporary magnetism (cores and electromagnets), while steel is preferred for permanent magnets.

Question 57. A battery of e.m.f. 2 V and internal resistance 3 \(\Omega\) is connected to a 4 \(\Omega\) resistor and a parallel combination of a 2 \(\Omega\) resistor and an unknown resistor R. The current in the circuit is 0.25 A. Calculate:
(i) The potential difference across the 4 \(\Omega\) resistor,
(ii) The potential difference across the internal resistance of the cell,
(iii) The potential difference across the parallel combination of 2 \(\Omega\) and R,
(iv) The value of the unknown resistance R.
Answer:
Given data:
Electromotive force (\(e\)) = \(2\text{ V}\)
Internal resistance (\(r\)) = \(3\ \Omega\)
Series resistor (\(R_1\)) = \(4\ \Omega\)
Total current (\(I\)) = \(0.25\text{ A}\)

(i) The voltage drop across the \(4\ \Omega\) resistor is: \[ V_{4\Omega} = I \times R_1 = 0.25 \times 4 = 1\text{ V} \]
(ii) The voltage drop across the cell's internal resistance is: \[ V_{r} = I \times r = 0.25 \times 3 = 0.75\text{ V} \]
(iii) The potential difference across the parallel combination (comprising the \(2\ \Omega\) and \(R\) resistors) is: \[ V_{p} = e - (V_{4\Omega} + V_{r}) \] \[ V_{p} = 2 - (1 + 0.75) = 2 - 1.75 = 0.25\text{ V} \]
(iv) The equivalent resistance of the parallel combination of \(2\ \Omega\) and \(R\) is: \[ R_p = \frac{2R}{R + 2} \] Using Ohm's law across this parallel combination: \[ V_p = I \times R_p \] \[ 0.25 = 0.25 \times \frac{2R}{R + 2} \] \[ 1 = \frac{2R}{R + 2} \] \[ R + 2 = 2R \] \[ R = 2\ \Omega \] Thus, the unknown resistance \(R\) is \(2\ \Omega\).
In simple words: We find how much voltage is used by the other parts first: 1 volt is lost on the four-ohm resistor, and 0.75 volts is lost inside the battery. This leaves 0.25 volts for the parallel section, which tells us that R must be two ohms.

Exam Tip: Keep your calculations structured. Finding individual voltage drops across known components first makes it much simpler to solve for the unknown parallel resistance.

Question 58. Two resistors of 2 \(\Omega\) and 2 \(\Omega\) are connected in parallel and then connected to a cell. The current in the circuit is 1.2 A. When the same two resistors are connected in series and connected to the same cell, the current is 0.4 A. Calculate:
(i) the internal resistance of the cell,
(ii) the e.m.f. of the cell.
Answer:
Let \(e\) be the cell's electromotive force and \(r\) be its internal resistance.

Case I: Parallel connection
The two \(2\ \Omega\) resistors connected in parallel have an equivalent resistance of: \[ R_{p1} = \frac{2 \times 2}{2 + 2} = 1\ \Omega \] The total resistance of this circuit is: \[ R_{\text{total1}} = R_{p1} + r = 1 + r \] Since the total current is \(1.2\text{ A}\), we write the EMF equation: \[ e = I_1 \times R_{\text{total1}} \] \[ e = 1.2 \times (r + 1) \quad \text{--- (Equation 1)} \]
Case II: Series connection
The two \(2\ \Omega\) resistors connected in series have an equivalent resistance of: \[ R_{s1} = 2 + 2 = 4\ \Omega \] The total resistance of this circuit is: \[ R_{\text{total2}} = R_{s1} + r = 4 + r \] Since the total current is \(0.4\text{ A}\), we write the EMF equation: \[ e = I_2 \times R_{\text{total2}} \] \[ e = 0.4 \times (r + 4) \quad \text{--- (Equation 2)} \]
Solving the equations:
Equating Equation 1 and Equation 2: \[ 1.2 \times (r + 1) = 0.4 \times (r + 4) \] Divide both sides by 0.4: \[ 3 \times (r + 1) = r + 4 \] \[ 3r + 3 = r + 4 \] \[ 2r = 1 \] \[ r = 0.5\ \Omega \] Substitute \(r = 0.5\) into Equation 2 to find \(e\): \[ e = 0.4 \times (0.5 + 4) = 0.4 \times 4.5 = 1.8\text{ V} \] Thus, the internal resistance is \(0.5\ \Omega\) and the EMF is \(1.8\text{ V}\).
In simple words: We set up two simple equations using the two different ways the resistors are connected. By matching them, we find that the battery loses half an ohm of resistance inside, and has a total strength of 1.8 volts.

Exam Tip: Remember that parallel resistance of two equal resistors is half of one resistor, while series is their sum. Don't forget to add the internal resistance \(r\) to the total external resistance.

Question 59. The diagram shows an incomplete transformer.
(i) Name the part that must be drawn to complete the diagram.
(ii) State the material of this part.
(iii) State whether this transformer is step-up or step-down, giving a reason.
Answer:
(i) The component that must be added to make the transformer diagram complete is the core.
(ii) This core should be made out of soft iron.
(iii) This is a step-down transformer since the primary winding features more turns of wire than the secondary winding. In simple words: To complete the transformer, we need to draw an iron core. Because the input side has more wire loops than the output side, it steps down the voltage.

Exam Tip: When asked to identify the type of transformer from a diagram, always compare the number of loops on both sides - more loops on the input side means it is a step-down transformer.

Question 60. Define the term 'turns ratio' of a transformer. Why cannot a transformer be used with direct current (D.C.)?
Answer:
The turns ratio of a transformer represents the ratio of the number of turns in the secondary winding (\(N_s\)) to the number of turns in the primary winding (\(N_p\)), written as \(N_s / N_p\).
A transformer is unable to operate on a direct current (D.C.) supply. This is because its operation relies on the induction of voltage through a continuously changing magnetic field. A constant direct current produces a steady magnetic field, which does not produce any change in magnetic flux linkage, meaning no current is induced in the secondary coil.
In simple words: Turns ratio is just the number of output coils divided by the number of input coils. Transformers do not work on D.C. because D.C. flows steadily and doesn't create the changing magnetic field needed for induction.

Exam Tip: Keep the definition of turns ratio precise: \(N_s / N_p\). For the D.C. explanation, make sure to explicitly mention that constant current leads to constant magnetic flux, resulting in zero induction.