Maharashtra Board Class 9 Science Chapter 3 Current Electricity Solutions

Question 1. The accompanying figure shows some electrical appliances connected in a circuit in a house. Answer the following questions.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक घर में विभिन्न विद्युत उपकरण समांतर क्रम में जुड़े हुए दिखाए गए हैं, जैसे कि लैंप, टेलीविजन, और पंखा, जो सभी एक मुख्य सर्किट से जुड़े हैं।
A. By which method are the appliances connected?
Answer: Appliances are connected in parallel.
B. What must be the potential difference across individual appliances?
Answer: The potential difference across all appliances is same in parallel connection.
C. Will the current passing through each appliance be the same? Justify your answer.
Answer: No, as every appliance has a different load (resistance), the current flowing through each appliance will be different.
D. Why are the domestic appliances connected in this way?
Answer: The domestic appliances are connected in parallel as the potential difference remains same.
E. If the T.V. stops working, will the other appliances also stop working? Explain your answer.
Answer: No, the other devices will not stop working as the current flowing through them is along different paths.
In simple words: Household appliances are connected in parallel to ensure each receives the same voltage, allowing them to operate independently and draw appropriate current based on their resistance. If one appliance fails, others continue to function because each has its own path for current.

🎯 Exam Tip: Understanding parallel connections in domestic circuits is crucial for explaining independent operation and constant voltage distribution among appliances.

Question 2. The following figure shows the symbols for components used in the accompanying electrical circuit.
ℹ️ चित्र व्याख्या (Diagram Explanation): विद्युत परिपथ में उपयोग होने वाले विभिन्न घटकों के प्रतीक दिखाए गए हैं, जैसे कि प्रतिरोधक, कुंजी, और बैटरी।
ℹ️ चित्र व्याख्या (Diagram Explanation): एक विद्युत परिपथ का आरेख जिसमें एक बैटरी, एमीटर, वोल्टमीटर, प्रतिरोधक और कुंजी जुड़े हुए हैं, ओम के नियम को सिद्ध करने के लिए दर्शाया गया है।
Which law can you prove with the help of the above circuit?
Answer:
(b) This circuit can be used to prove Ohm's law.
(c) V = 1R is the expression of Ohm's law
In simple words: The circuit shown, containing a voltmeter, ammeter, and resistance, is designed to demonstrate Ohm's law, which states that voltage (V) is directly proportional to current (I) and resistance (R), expressed as V = IR.

🎯 Exam Tip: Be able to identify circuit components and their symbols, and explain how a simple circuit can be used to verify Ohm's Law.

Question 3. Umesh has two bulbs having resistances of 15 W and 30 W. He wants to connect them in a circuit, but if he connects them one at a time the filament gets burnt. Answer the following.
A. Which method should he use to connect the bulbs?
B. What are the characteristics of this way of connecting the bulbs depending on the answer of A above?
C. What will be the effective resistance in the above circuit?
In simple words: To prevent bulbs from burning out when connected individually, Umesh needs to choose a connection method that appropriately manages current and resistance, likely aiming to reduce the total effective resistance if the individual resistances are too low for the voltage source.

🎯 Exam Tip: This question tests understanding of series and parallel connections, particularly how they affect total resistance and current flow to prevent damage to components.

Question 4. The following table shows current in Amperes and potential difference in Volts.
a. Find the average resistance.
b. What will be the nature of the graph between the current and potential difference? (Do not draw a graph.)
c. Which law will the graph prove? Explain the law.
In simple words: This question requires applying Ohm's law to calculate resistance from given current and potential difference values, determining the graphical relationship between these quantities, and identifying the law it represents.

🎯 Exam Tip: Focus on using Ohm's Law \( R = V/I \) to find resistance and describing the linear V-I relationship characteristic of ohmic conductors, which visually confirms Ohm's law.

Question 5. Match the pairs
'A' Group - 'B' Group
1. Free electrons a. V/ R
2. Current - b. Increases the resistance in the circuit
3. Resistivity - c. Weakly attached
4. Resistances in series - d. VA/LI
In simple words: This matching exercise connects fundamental electrical concepts like free electrons, current, resistivity, and series resistance to their corresponding definitions, formulas, or effects in a circuit.

🎯 Exam Tip: Memorize the definitions, units, and effects of key electrical terms such as free electrons, current, resistivity, and resistance combinations for quick recall in matching questions.

Question 6. The resistance of a conductor of length x is r. If its area of crosssection is a, what is its resistivity? What is its unit?
In simple words: This question asks for the formula for resistivity, which relates a conductor's resistance to its physical dimensions (length and cross-sectional area), and also requires stating the standard unit for resistivity.

🎯 Exam Tip: Remember the resistivity formula \( \rho = R \cdot \frac{A}{L} \) and its SI unit, Ohm-meter \( (\Omega \cdot m) \). This is a common conceptual and formula-based question.

Question 7. Resistances R1, R2, R3 and R4 are connected as shown in the figure. S₁ and S2 are two keys. Discuss the current flowing in the circuit in the following cases.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक परिपथ आरेख जिसमें प्रतिरोधक R1, R2, R3, और R4 दो कुंजियों S1 और S2 के साथ जुड़े हुए हैं, यह दर्शाता है कि कुंजियों की स्थिति के अनुसार धारा कैसे प्रवाहित होती है।
a. Both S₁ and S2 are closed.
b. Both S₁ and S2 are open.
c. S₁ is closed but S2 is open.
Answer:
(a) When both S₁ and S2 are closed, the effective resistance of the circuit decreases and hence, current will increase.
(b) When both S₁ and S2 are open, the effective resistance of the circuit increases and hence, current will decrease.
(c) When S2 is closed and S2 is open, the effective resistance of the circuit decreases and hence current will increase. [Current will be more than case (b) but less than in case (a)]
In simple words: Closing keys in a circuit generally decreases total resistance, allowing more current to flow, while opening them increases resistance, thereby reducing current flow. The specific impact depends on how the components are arranged relative to the keys.

🎯 Exam Tip: Analyze how closing or opening switches alters the paths for current, effectively changing the total resistance in a circuit and, consequently, the overall current flow according to Ohm's law.

Question 8. Three resistances X1, X2 and x3 are connected in a circuit in different ways. x is the effective resistance. The properties observed for these different ways of connecting X1, X2 and x3 are given below. Write the way in which they are connected in each case. (I-current, V-potential difference, x-effective resistance)
a. Current I flows through X1, X2 and x3
b. x is larger than x1, x2 and x3
c. x is smaller than x1, x2 and x3
d. The potential difference across x1, x2and x3 is the same
e. x = X1 + X2 + X3
f. \( \frac{1}{x} = \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} \)
In simple words: This question explores the characteristics of series and parallel connections for three resistors, asking to identify the connection type based on how current, voltage, and effective resistance behave in each scenario.

🎯 Exam Tip: Crucially differentiate between series (same current, additive resistance, voltage division) and parallel (same voltage, reciprocal resistance sum, current division) connections to accurately identify the setup based on the given properties.

Question 9. Solve the following problems.

A. The resistance of a 1m long nichrome wire is 6Ω. If we reduce the length of the wire to 70 cm. what will its resistance be? (Answer : 4.2Ω)
Answer: The resistance of 70cm wire will be 4.2 Ω
In simple words: The resistance of a wire is directly proportional to its length; therefore, reducing the length of the nichrome wire from 1 meter to 70 centimeters will proportionally decrease its resistance.

🎯 Exam Tip: Remember that resistance is directly proportional to length \( (R \propto L) \). Use ratios or direct proportionality to solve such problems quickly: \( R_2 = R_1 \cdot (L_2/L_1) \).

B. When two resistors are connected in series, their effective resistance is 80Ω. When they are connected in parallel, their effective resistance is 20Ω. What are the values of the two resistances? (Answer : 40Ω, 40Ω)
Answer: The values of the two resistances R₁ and R2 are 40Ω and 40Ω.
In simple words: Given the effective resistance in both series and parallel configurations, this problem requires solving a system of equations to find the individual resistance values of the two resistors.

🎯 Exam Tip: Set up two equations: \( R_s = R_1 + R_2 \) and \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \). Solve these simultaneous equations to find the individual resistances. This is a common numerical problem.

C. If a charge of 420 C flows through a conducting wire in 5 minutes what is the value of the current? (Answer : 1.4 A)
Answer:
Given: Electric charge (Q) = 420 C
Time (t) = 5 min = 5 x 60
= 300 sec.
To find: Electric current (1) = ?
Formula: \( I = \frac{Q}{t} \)
Solution: \( I = \frac{Q}{t} \)
The current in the circuit is 1.4 A.
In simple words: To find the current, simply divide the total electric charge that flows through the wire by the time taken for it to pass, ensuring time is in seconds.

🎯 Exam Tip: Convert time to seconds before calculation. The formula \( I = \frac{Q}{t} \) is fundamental for current calculations. Units (Coulombs, seconds, Amperes) must be consistent.

Can You Recall?

Question 1. You must have seen a waterfall. Which way does the water flow?
Answer: Water flows from a certain height of a mountain towards the ground.
In simple words: Water always flows downhill, from a higher elevation to a lower one, due to gravity.

🎯 Exam Tip: This question sets a simple analogy for understanding potential difference in electricity; just as water flows from high to low physical potential, electric current flows from high to low electrical potential.

Question 2. Material: Copper and aluminium wires, glass rod, rubber. Make connection as shown in figure 3.8. First connect a copper wire between points A and B and measure the current in the circuit. Then in place of the copper wire, connect the aluminium wire, glass rod, rubber, etc one at a time and measure the current each time. Compare the values of the current in different cases.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक साधारण विद्युत परिपथ का आरेख जिसमें बैटरी, एमीटर, कुंजी और परीक्षण के लिए दो बिंदुओं (A और B) के बीच एक सामग्री जोड़ने की व्यवस्था दिखाई गई है।
Also take different metal strips (Iron, Copper, Zinc, and Aluminium) and connect it in slot AB. Now observe the difference in the resistance using Ohm meter.
Answer: When copper and aluminium wires are connected to the circuit, current flows through it, as both are good conductors of electricity. When glass rod or rubber was connected to the circuit, current does not flow through it, as both are bad conductors of electricity. Copper displays lowest resistance while the resistance increases with aluminium, zinc and iron respectively.
In simple words: Conductors like copper and aluminum allow current to flow easily due to low resistance, while insulators like glass and rubber block current due to high resistance. Different metals have varying degrees of conductivity.

🎯 Exam Tip: Understand the difference between conductors (low resistance, free electrons) and insulators (high resistance, no free electrons). Metals vary in conductivity, with copper being an excellent conductor.

Question 3. Set up the experiment as shown in figure. Then remove the clamp from the rubber tube.
ℹ️ चित्र व्याख्या (Diagram Explanation): दो बीकर (बोतलें) एक रबर ट्यूब और क्लैंप से जुड़े हुए हैं, जिनमें पानी का स्तर अलग-अलग है, जो पानी के प्रवाह की दिशा और स्तर को दर्शाता है।
(a) What happens when the clamp is removed?
Answer: When the clamp is removed, water flows from higher level to lower level.
(b) Does the water stop flowing? Why?
Answer: Yes, the water stops flowing. This happens when the level of water becomes equal in both the bottles, i.e., there is no difference in the water levels.
(c) What will you do to keep the water flowing for a longer duration?
Answer: The difference in the water level has to be maintained till that time. The difference must never be zero.
In simple words: Water flows from a higher level to a lower level, stopping when levels equalize. To maintain continuous flow, a constant difference in water levels must be upheld. This demonstrates how a potential difference drives flow.

🎯 Exam Tip: This analogy highlights the concept of potential difference: just as water flows due to a height difference, electric current flows due to an electrical potential difference. Maintaining this difference is key for continuous flow.

Question 4. Point out the mistakes in the figure below:
ℹ️ चित्र व्याख्या (Diagram Explanation): विभिन्न विद्युत परिपथों के आरेख जिनमें एक बैटरी और एक बल्ब जुड़ा हुआ है, यह दिखाने के लिए कि सर्किट में गलतियाँ होने पर बल्ब कैसे नहीं जलता।
Answer: A: Wire is broken at the negative terminal. Bulb will not glow as the circuit is incomplete. B: Wire is disconnected at the negative terminal. Bulb will not glow as the circuit is incomplete. C: The circuit is complete. Therefore, bulb will glow. D: Rubber is a bad conductor of electricity. Hence, it will not allow current to flow and the bulb will not glow.
In simple words: For a bulb to light, the electrical circuit must be complete, providing an uninterrupted path for current to flow from the power source through the bulb and back. Breaks in the wire or the inclusion of insulating materials prevent this flow.

🎯 Exam Tip: Focus on the conditions for a complete circuit: a continuous path from the positive terminal, through the load (bulb), to the negative terminal. Any break or insulator will prevent current flow.

Question 5. Why are the bulbs in Figures B, C and D not lighting up?
ℹ️ चित्र व्याख्या (Diagram Explanation): पांच विद्युत परिपथों के आरेख (A से E) जिनमें प्रत्येक में एक बैटरी और एक बल्ब जुड़ा हुआ है, यह दर्शाते हैं कि सर्किट कनेक्शन में त्रुटियों के कारण कुछ बल्ब क्यों नहीं जलते।
Answer:
• In B, the blue wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
• In C, the red wire is broken. Hence circuit is incomplete and current does not flow. Therefore, bulb will not light up.
• In D, both wires are connected to the same terminal. Hence, there is no potential difference and current does not flow. Therefore, bulb will not light up.
In simple words: Bulbs fail to light up in circuits B and C due to broken wires, which create incomplete circuits, while in circuit D, the bulbs don't light because both wires are connected to the same terminal, eliminating the necessary potential difference for current flow.

🎯 Exam Tip: Remember that a complete circuit with a potential difference across the load is essential for current flow and device operation. Broken wires or incorrect terminal connections prevent this.

Choose And Write The Correct Option:

Question 1. 1mA = ............ A.
(a) 103
(b) 10-3
(c) 106
(d) 10-6
Answer: (b) 10-3
In simple words: One milliampere (mA) is equal to one-thousandth of an ampere (A), so 1 mA = \( 10^{-3} \) A.

🎯 Exam Tip: Know the common prefixes for units: milli- \( (10^{-3}) \), micro- \( (10^{-6}) \), kilo- \( (10^{3}) \), mega- \( (10^{6}) \). This helps in quick conversions.

Question 2. To increase the effective resistance in a circuit the resistors are connected in ................
(a) Series
(b) Parallel
(c) Both ways
(d) None of these
Answer: (a) Series
In simple words: Connecting resistors in series adds up their individual resistances, which increases the total effective resistance of the circuit.

🎯 Exam Tip: Remember that for series connections, total resistance \( R_s = R_1 + R_2 + \dots \), which is always greater than any individual resistance. For parallel, total resistance is always less than the smallest individual resistance.

Question 3. 1 kilowatt hr = ............... joules.
(a) 4.6 x \( 10^6 \)
(b) 3.6 x \( 10^6 \)
(c) 30.6 x \( 10^6 \)
(d) 3.6 x \( 10^5 \)
Answer: (b) 3.6 x \( 10^6 \)
In simple words: One kilowatt-hour, a common unit for electrical energy consumption, is equivalent to 3.6 million joules.

🎯 Exam Tip: This is a conversion factor. \( 1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} \). Memorize this conversion for energy-related problems.

Question 4. The voltage difference in India between the live and neutral wires is about ............... .
(a) 110 V
(b) 220 V
(c) 440 V
(d) 60 V
Answer: (b) 220 V
In simple words: In India, the standard household voltage supplied between the live and neutral wires is approximately 220 Volts.

🎯 Exam Tip: Know the standard voltage levels for domestic electricity in your region, which is 220 V in India. This is a general knowledge question relevant to electrical safety.

Question 5. Resistivity is the specific property of a ................
(a) Area of cross-section
(b) Temperature
(c) Length
(d) Material
Answer: (d) material
In simple words: Resistivity is an intrinsic characteristic of the material itself, determining how strongly it resists electric current, regardless of its shape or size.

🎯 Exam Tip: Resistivity is a material constant (like density) and does not depend on the dimensions (length or cross-sectional area) of the conductor, unlike resistance.

Question 6. If a P.D. of 12 V is applied across a 3Ω resistor then the current passing through it is ................
(a) 36 A
(b) 4 A
(c) 0.25 A
(d) 15 A
Answer: (b) 4 A.
In simple words: Using Ohm's Law (I = V/R), a potential difference of 12 Volts across a 3 Ohm resistor will result in a current of 4 Amperes.

🎯 Exam Tip: Apply Ohm's law: \( I = \frac{V}{R} \). Ensure correct unit usage. This is a direct application of the formula.

Question 7. In order to measure the electric current flowing through a circuit, we connect ................ with the circuit.
(a) a voltmeter in parallel
(b) a voltmeter in series
(c) an ammeter in parallel
(d) an ammeter in series
Answer: (d) an ammeter in series
In simple words: An ammeter, designed to measure current, must be connected in series within a circuit so that all the current flows through it.

🎯 Exam Tip: Remember that ammeters have very low resistance and must be connected in series. Voltmeters have very high resistance and are connected in parallel.

Question 8. P and Q are two wires of same length and different cross-sectional areas and made of same material. Name the property which is same for both the wires.
(a) Resistivity
(b) Resistance
(c) Current
(d) Both (a) and (b)
Answer: (a) Resistivity
In simple words: Since both wires are made of the same material, their resistivity-an inherent property of the material-will be identical, irrespective of their dimensions.

🎯 Exam Tip: Resistivity is a material property. If two objects are made of the same material, their resistivity is the same, even if their dimensions (length, area) differ.

Question 9. The following is true for identical bulbs connected in parallel.
(a) All bulbs glow with unequal brightness.
(b) If one bulb is non-functional, all will stop working.
(c) All bulbs glow with equal brightness.
(d) Bulbs function for longer time.
Answer: (c) All bulbs glow with equal brightness
In simple words: In a parallel circuit, identical bulbs receive the same voltage across them, ensuring they all glow with equal brightness and can operate independently if one fails.

🎯 Exam Tip: In parallel connections, voltage across each component is the same, leading to equal brightness for identical bulbs. Also, parallel circuits allow independent operation, a key advantage.

Question 10. The ................ wire is either yellow or green in colour.
(a) Live
(b) Neutral
(c) Earth
(d) Fuse
Answer: (c) earth
In simple words: The earth wire, an essential safety component in electrical systems, is typically identifiable by its yellow or green insulation.

🎯 Exam Tip: Remember the color coding for household wiring for safety: Live (red/brown), Neutral (blue/black), and Earth (green/yellow).

Question 11. A current flows through a circuit due to the difference in ............... between two points in the conductor.
(a) Gravity
(b) Potential
(c) Resistance
(d) Fuse
Answer: (b) potential
In simple words: Electric current flows in a circuit when there is a potential difference (voltage) between two points, acting as the driving force for charge movement.

🎯 Exam Tip: The concept of potential difference (voltage) is analogous to pressure difference for fluid flow; it is the necessary condition for electric current to flow.

Question 12. ................ is the amount of charge flowing through a particular cross sectional area in unit time.
(a) Electric current
(b) Ampere
(c) Volt
(d) Force
Answer: (a) Electric current
In simple words: Electric current is defined as the rate at which electric charge passes through a specific point or cross-sectional area per unit of time.

🎯 Exam Tip: This is the fundamental definition of electric current \( (I = Q/t) \). Ensure you know both the definition and the formula.

Question 13. The flow of ................ constitutes the electric current in a wire.
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer: (c) electrons
In simple words: In most conductors, the movement of free electrons is what creates an electric current.

🎯 Exam Tip: Clarify that in metals, free electrons are the charge carriers. Conventional current direction is opposite to electron flow.

Question 14. The conventional direction of flow of current is from ............... terminal to ............... terminal.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer: (c) positive, negative
In simple words: Conventionally, electric current is considered to flow from the positive terminal of a voltage source to the negative terminal, opposite to the actual flow of electrons.

🎯 Exam Tip: Distinguish between conventional current (positive to negative) and electron flow (negative to positive). Most circuit analysis uses conventional current.

Question 15. Current stops flowing when potential difference between two ends of a wire becomes ................
(a) Zero
(b) Positive
(c) Negative
(d) Higher
Answer: (a) zero
In simple words: Electric current ceases to flow when there is no longer a difference in electrical potential (voltage) between the two ends of a conductor.

🎯 Exam Tip: Current requires a potential difference to flow. If the potential difference is zero, no work is done on charges, and thus no net current flows.

Question 16. Resistances are connected in ................ so as to pass the same current through them.
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer: (a) series
In simple words: When resistors are connected in series, the same electric current flows sequentially through each resistor in the circuit.

🎯 Exam Tip: Remember the primary characteristic of a series circuit: current is constant throughout all components. Voltage divides across components in series.

Question 17. To decrease the effective resistance in a circuit, the resistances are connected in ................
(a) Series
(b) Parallel
(c) Reversed
(d) Disconnect
Answer: (b) parallel
In simple words: Connecting resistors in parallel provides multiple paths for current, effectively reducing the overall resistance of the circuit to a value lower than any individual resistor.

🎯 Exam Tip: For parallel connections, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances, resulting in a lower overall resistance.

Question 18. 1μV = ................ V
(a) \( 10^2 \)
(b) \( 10^{-6} \)
(c) \( 10^6 \)
(d) \( 10^3 \)
Answer: (b) \( 10^{-6} \)
In simple words: One microvolt (μV) is equivalent to one-millionth of a Volt (V).

🎯 Exam Tip: Understand metric prefixes: 'micro' \( (\mu) \) always means \( 10^{-6} \).

Question 19. Good conductors contain a large number of ................
(a) Protons
(b) Neutrons
(c) Electrons
(d) Gravitons
Answer: (c) free electrons
In simple words: Good electrical conductors are characterized by having an abundance of free electrons that can move easily, facilitating the flow of electric current.

🎯 Exam Tip: The presence of mobile charge carriers, specifically free electrons in metals, is the defining characteristic of a good conductor.

Question 20. Electrons flow from ................ terminal to ................ terminal in a conductor when a potential difference is applied.
(a) Negative to positive
(b) Neutral to positive
(c) Positive to negative
(d) Positive to neutral
Answer: (a) negative, positive
In simple words: When a voltage is applied, electrons, being negatively charged, are repelled from the negative terminal and attracted towards the positive terminal, thus flowing from negative to positive.

🎯 Exam Tip: Remember that electrons are negatively charged and move opposite to the direction of the electric field and conventional current (which flows from positive to negative).

Find The Odd One Out:

Question 1. Voltmeter, Ammeter, Galvanometer, Thermometer
Answer: Thermometer
In simple words: A thermometer measures temperature, whereas a voltmeter, ammeter, and galvanometer are all instruments used to measure various electrical properties.

🎯 Exam Tip: Categorize instruments based on what they measure (e.g., electrical vs. thermal) to quickly identify the odd one out.

Question 2. Rubber, Silver, Copper, Gold
Answer: Rubber
In simple words: Rubber is an electrical insulator, while silver, copper, and gold are all excellent electrical conductors.

🎯 Exam Tip: Classify materials as conductors or insulators. Conductors (metals) allow current, insulators (rubber, glass) block it.

Question 3. Wood, Glass, Steel, Rubber
Answer: Steel
In simple words: Steel is a metallic conductor, whereas wood, glass, and rubber are all common electrical insulators.

🎯 Exam Tip: Distinguish between materials based on their electrical conductivity; most metals are conductors, while organic and ceramic materials are typically insulators.

Question 4. Graphite, Diamond, Fullerenes, Coal
Answer: Fullerenes
In simple words: Fullerenes are a specific allotrope of carbon, while graphite, diamond, and coal represent other forms or a natural aggregate of carbon.

🎯 Exam Tip: Recognize allotropes of carbon (graphite, diamond, fullerenes) and distinguish them from carbon-rich substances like coal. Fullerenes are distinct molecular structures.

Distinguish Between:

Question 1. Voltmeter and Ammeter
Answer:

In simple words: A voltmeter measures potential difference and is connected in parallel with high resistance, while an ammeter measures current and is connected in series with low resistance.

🎯 Exam Tip: Remember the connection type (series/parallel) and resistance characteristics (high/low) for each instrument for full marks.

Question 2. Ohmic conductors and Non-Ohmic conductors
Answer: (The question likely intended to ask for a distinction between conductors and insulators, as 'Ohmic conductors and Non-Ohmic conductors' is a distinction within conductors, and the provided answer table is for 'Conductors' vs 'Insulators'. Given the context, I will provide the table for Conductors and Insulators as given, assuming 'Non-Ohmic conductors' was a typo or a mislabeled section that led to the 'Conductors' table.)

In simple words: Conductors allow electricity to flow easily due to many free electrons and low resistance, while insulators resist electricity flow due to very few free electrons and high resistance.

🎯 Exam Tip: Focus on the presence/absence of free electrons and the level of electrical resistance to distinguish between conductors and insulators.

Question 3. Conductors and Insulators
Answer:

In simple words: Conductors are materials with low electrical resistance, allowing current to flow easily, typically metals with many free electrons. Insulators have extremely high electrical resistance, preventing current flow, and are mostly non-metals with few free electrons.

🎯 Exam Tip: When distinguishing, always mention electron mobility and resistance level as key characteristics for both conductors and insulators.

Question 4. Resistance in Series and Resistance in Parallel
Answer:

In simple words: In series, resistances add up directly, increasing total resistance and keeping current the same. In parallel, inverses of resistances add, decreasing total resistance and allowing current to divide.

🎯 Exam Tip: Remember the primary purpose of each connection type- series for increasing resistance and parallel for decreasing it, and how current behaves in each.

Question 5. Distinguish between Electric current and Potential difference
Answer:

In simple words: Electric current is the rate of charge flow, measured in amperes, while potential difference is the work done per unit charge to move it between two points, measured in volts.

🎯 Exam Tip: Clearly state the definition, SI unit, and the measuring instrument for both current and potential difference for a complete answer.

Make Pair

Question 1. Copper : Conductor :: Rubber :
Answer: Insulator
In simple words: Copper allows electricity to pass through easily (conductor), while rubber blocks it (insulator).

🎯 Exam Tip: Understand the basic electrical properties of common materials to correctly identify them as conductors or insulators.

Question 2. Aluminium : :: Indium oxide : Super Insulator
Answer: Super conductor
In simple words: Aluminium is an excellent conductor, approaching superconductivity at very low temperatures, while indium oxide is a material known for its strong insulating properties.

🎯 Exam Tip: Recognize materials with extreme electrical properties, such as high conductivity (superconductors) and high insulation (super insulators).

Question 3. Parallel Connection : \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \) :: Series Connection :
Answer: \( R_s = R_1 + R_2 \)
In simple words: The formula \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \) calculates effective resistance for parallel circuits, while \( R_s = R_1 + R_2 \) is used for series circuits.

🎯 Exam Tip: Memorize the distinct formulas for calculating equivalent resistance in both series and parallel connections, as they are fundamental to circuit analysis.

Question 4. Electric Current : :: Electric charge : Coulomb
Answer: Ampere
In simple words: Ampere is the unit for electric current, just as Coulomb is the unit for electric charge.

🎯 Exam Tip: Ensure you know the SI units for fundamental electrical quantities like current (Ampere) and charge (Coulomb).

Question 5. Electric resistance : Ohm :: Potential difference :
Answer: Volt
In simple words: Ohm is the unit for electric resistance, similar to how Volt is the unit for potential difference.

🎯 Exam Tip: It is crucial to associate the correct SI unit with each electrical quantity, such as Ohm for resistance and Volt for potential difference.

State Whether The Following Statements Are True Or False. Correct The False Statements:

Question 1. The SI unit of charge is volt.
Answer: False. The SI unit of charge is coulomb.
In simple words: The unit of electric charge is Coulomb, not Volt; Volt measures potential difference.

🎯 Exam Tip: Always double-check the SI units for basic electrical quantities, as incorrect units are a common source of errors.

Question 2. Voltmeter is always connected in series with the device.
Answer: False. Voltmeter is always connected in parallel with the device.
In simple words: Voltmeters are connected in parallel to measure voltage across a component, not in series.

🎯 Exam Tip: Remember that voltmeters have very high resistance and must be connected in parallel to accurately measure potential difference without affecting the circuit current significantly.

Question 3. The conventional direction of flow of current is from positive terminal to negative terminal.
Answer: True
In simple words: By convention, electric current is considered to flow from the positive to the negative terminal of a power source.

🎯 Exam Tip: This is a fundamental convention in circuit diagrams and problem-solving, even though electron flow is in the opposite direction.

Question 4. Silver and copper are good conductors.
Answer: True
In simple words: Silver and copper are known for having many free electrons and low resistance, making them excellent conductors of electricity.

🎯 Exam Tip: Good conductors are essential in electrical wiring due to their low resistance, minimizing energy loss as heat.

Question 5. Resistivity of pure metals is more than alloys.
Answer: False. Resistivity of pure metals is less than alloys.
In simple words: Pure metals generally have lower resistivity than alloys because the presence of different atoms in alloys interferes more with electron flow.

🎯 Exam Tip: Understand that the atomic structure and composition of a material significantly influence its resistivity, with impurities and mixed atoms often increasing it.

Question 6. Resistance in series arrangement is used to decrease resistance of circuit.
Answer: False. Resistance in series arrangement is used to increase resistance of circuit.
In simple words: Connecting resistors in series adds their resistances, thus increasing the total resistance of the circuit.

🎯 Exam Tip: Recall that the total resistance in a series circuit is the sum of individual resistances, always leading to an increase in overall resistance.

Question 7. A conducting wire offers less resistance to flow of electrons.
Answer: True
In simple words: Conducting wires are designed to have very low resistance, allowing electrons to move through them with minimal opposition.

🎯 Exam Tip: The effectiveness of a conductor is directly related to its ability to offer minimal resistance to electric current, ensuring efficient energy transfer.

Question 8. Charges are measured in ampere.
Answer: False. Charges are measured in coulomb.
In simple words: Electric charge is quantified in coulombs, while amperes measure the rate of charge flow (current).

🎯 Exam Tip: Differentiate between electric charge (Coulomb) and electric current (Ampere), as they are distinct but related quantities.

Question 9. The unit of potential difference is ampere.
Answer: False. The unit of potential difference is volt.
In simple words: The unit for potential difference is the Volt, not the Ampere, which is the unit for current.

🎯 Exam Tip: Confusing units is a common mistake; always verify the correct SI unit for each electrical property.

Question 10. Resistance of a conductor is inversely proportional to the length of the conductor.
Answer: False. Resistance of a conductor is directly proportional to the length of the conductor.
In simple words: A longer conductor offers more resistance to current flow because electrons travel a greater distance and encounter more collisions.

🎯 Exam Tip: Remember the relationship \( R = \rho \frac{L}{A} \), which shows resistance is directly proportional to length (L) and inversely proportional to cross-sectional area (A).

Question 11. Ammeter is connected in parallel to the cell to measure current.
Answer: False. Ammeter is connected in series to the cell to measure current.
In simple words: Ammeters are always connected in series within a circuit to measure the total current flowing through that part of the circuit.

🎯 Exam Tip: Ammeters have very low internal resistance and must be connected in series so that all current passes through them for an accurate measurement.

Question 12. Fuse is made of wire having high melting point.
Answer: False. Fuse is made of wire having low melting point.
In simple words: A fuse wire is designed to melt and break the circuit when current exceeds a safe limit, which requires it to have a low melting point.

🎯 Exam Tip: The low melting point of fuse wire is crucial for its protective function, preventing damage to appliances and circuits from overcurrents.

Answer The Following In One Sentence:

Question 1. Which is the unit used to measure large voltages?
Answer: Kilovolts and Megavolts are the units used to measure large voltages.
In simple words: For very large electrical potentials, units like kilovolts (kV) and megavolts (MV) are used.

🎯 Exam Tip: Be aware of the common prefixes (kilo-, mega-) used with SI units to denote larger magnitudes of electrical quantities.

Question 2. What is the SI unit of potential difference?
Answer: The SI unit of potential difference is volt (V).
In simple words: The standard international unit for measuring potential difference, also known as voltage, is the volt.

🎯 Exam Tip: Knowing the SI units for all fundamental electrical quantities is essential for solving problems and understanding concepts.

Question 3. What is lightning?
Answer: Lightning is the electric discharge travelling from clouds at high potential to earth's surface which is at zero potential.
In simple words: Lightning is a massive, sudden electrical discharge that occurs when there is a significant potential difference between charged clouds and the ground.

🎯 Exam Tip: Lightning is a natural phenomenon illustrating the concept of electrical discharge and potential difference on a large scale.

Question 4. What is the unit of resistivity.
Answer: The unit of resistivity is ohm metre (Ωm).
In simple words: Resistivity, a material's inherent resistance to electric flow, is measured in ohm-meters.

🎯 Exam Tip: Distinguish between resistance (Ohm, Ω) and resistivity (Ohm-meter, Ωm), noting that resistivity is a material property while resistance depends on geometry.

Question 5. Which substances are called conductors of electricity?
Answer: Those substances which have very low electrical resistance are called conductors of electricity.
In simple words: Conductors are materials that allow electric current to pass through them easily due to their low electrical resistance.

🎯 Exam Tip: Understand that conductors facilitate current flow, often by having a high density of free electrons, making them crucial for electrical circuits.

Question 6. What is Earth wire?
Answer: Earth wire is generally yellow or green colour, it is connected to a metal plate buried deep underground near the house and is for safety purpose.
In simple words: An earth wire is a safety component, typically green/yellow, connected to the ground to provide a path for fault current, protecting against electrical shocks.

🎯 Exam Tip: The earth wire is a critical safety feature in domestic wiring, ensuring that excess current from faulty appliances safely flows into the ground.

Write Formula:

(1) Electric current \( I = \frac{Q}{t} \) (2) Electric charge \( Q = It \) (3) Potential difference \( V = IR \) (4) Electric resistance \( R = \frac{V}{I} \) (5) Current \( I = \frac{V}{R} \) (6) Resistivity \( \rho = \frac{RA}{L} \)
In simple words: These formulas relate fundamental electrical quantities like current, charge, time, potential difference, resistance, area, and length.

🎯 Exam Tip: Memorizing these core formulas is essential for solving any numerical problems in current electricity. Pay attention to the correct variables and their relationships.

Give Scientific Reasons:

Question 1. Free electrons are required for conduction of electricity.
Answer:

• Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to the nucleus.
• These are called free electrons. These electrons can easily move from one part of a conductor to its other parts. The negative charge of the electrons also gets transferred as a result of this motion.
• The free electrons in a conductor are the carriers of negative charge. Hence, free electrons are required for conduction of electricity.

In simple words: Electricity flows through materials because free electrons can easily move and carry charge, acting as the primary transporters of electrical energy.

🎯 Exam Tip: When explaining conduction, always highlight the role of weakly bound, mobile electrons as the charge carriers. This concept is fundamental to understanding electrical flow in metals.

Question 2. Wood and glass are good insulators.
Answer:

• Those substances which have infinitely high electrical resistance are called insulators.
• Wood and glass have high resistance and negligible free electrons for conduction of electricity.
• Hence, wood and glass are good insulators.

In simple words: Wood and glass are insulators because they have extremely high electrical resistance and almost no free electrons, preventing the flow of electric current.

🎯 Exam Tip: For insulators, emphasize the lack of free electrons and their very high resistance as the reasons for their inability to conduct electricity.

Question 3. Connecting wires in a circuit are made of copper and aluminium.
Answer:

• Copper and aluminum are good conductors of electricity.
• They have low electrical resistance and large number of free electrons.
• As they are malleable and ductile, they can be drawn into thin wires. Hence, connecting wires in a circuit are made of copper or aluminum.

In simple words: Copper and aluminium are used for connecting wires because they are excellent conductors with low resistance, have many free electrons, and can be easily drawn into thin wires.

🎯 Exam Tip: When explaining material choices for wires, mention both electrical properties (conductivity, resistance) and mechanical properties (malleability, ductility).

Question 4. A thick wire has a low resistance.
Answer:

• The resistance (R) of a wire is inversely proportional to the cross-sectional area (A) of a wire. i.e., \( R \propto \frac{1}{A} \)
• Thus, greater is the cross-sectional area of a conductor (wire), lower is its resistance. Hence, a thick wire has a low resistance.

In simple words: Thicker wires have a larger cross-sectional area, which provides more pathways for electrons, thus reducing their resistance to current flow.

🎯 Exam Tip: Remember the inverse relationship between resistance and cross-sectional area; a larger area (thicker wire) means lower resistance, which is why thick wires are preferred for high current applications.

Question 5. A series combination of resistances is used to increase the resistance of a circuit.
Answer:

• When resistances are connected in series, the effective resistance of the resistors is equal to the sum of their individual resistances. \( R_s = R_1 + R_2 + ........ R_n \)
• The effective resistance is larger than each of the individual resistances. Hence, This arrangement is used to increase the resistance in a circuit.

In simple words: Connecting resistors in series adds their individual resistances together, resulting in a higher total resistance for the entire circuit.

🎯 Exam Tip: A key characteristic of series circuits is that the total resistance is always greater than any individual resistance, making it suitable for applications requiring high resistance.

Question 6. A parallel combination of resistances decreases the effective resistance of the circuit.
Answer:

• In a parallel combination, the inverse of the effective resistance is equal to the sum of the inverses of individual resistances. \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ....... + \frac{1}{R_n} \)
• The effective resistance of resistors connected in parallel is less than the individual resistors.
• Due to this, any addition of an individual resistance in parallel combination will decrease the overall resistance of the circuit. Hence, a parallel combination of resistance decreases the effective resistance of the circuit.

In simple words: When resistors are connected in parallel, the total resistance of the circuit becomes lower than the smallest individual resistance, as it provides multiple paths for current.

🎯 Exam Tip: Understand that parallel connections effectively increase the "width" for current flow, thus reducing the overall resistance. This is why domestic circuits use parallel wiring.

Question 7. Lightning occurs from sky to earth.
Answer:

• Lightning is the electric discharge travelling from clouds at high potential to the earth's surface, which is at zero potential.
• The earth is always at lower potential as compared to the clouds.
• Hence, lightning occurs from sky to earth.

In simple words: Lightning is a massive electrical discharge caused by a large potential difference between highly charged clouds and the relatively lower-potential Earth.

🎯 Exam Tip: The core concept here is that electric current (or discharge) flows from a region of higher potential to a region of lower potential, which applies to natural phenomena like lightning.

Question 8. In streetlights, bulbs are connected in parallel.
Answer:
1. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows through the other paths, and the rest of the bulbs light up.
2. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually. Hence, streetlights are connected in parallel.

In simple words: Streetlights use parallel connections so that if one bulb fails, the others remain lit, and all bulbs receive the full voltage, allowing them to glow brightly.

🎯 Exam Tip: Highlight both the functional advantage (independent operation) and the performance advantage (full brightness) of parallel connections for devices like streetlights.

Numerical:

Numericals based on the formula: (1) Q= It (2) W= VQ

Question 1. A current of 0.4 A flows through a conductor for 5 minutes. How much charge would have passed through the conductor?
Answer: Given: Current (I) = 0.4 A Time (t) = 5 min = 5 x 60 = 300 s To find: Charge (Q) =? Formula: Q = I x t Solution: Q = 0.4 x 300 Q= 120 C. Charge passing through the conductor is 120 C.

In simple words: To find the total charge, multiply the current by the time, ensuring time is in seconds.

🎯 Exam Tip: Always convert time to seconds when using formulas involving current and charge to maintain consistency in SI units.

Question 2. Find the amount of work done if 3 C of charge is moved through a potential difference of 9 V.
Answer: Given: Electric charge (Q) = 3 C P. D. = (V) = 9 (V) To find: Work done (W) = ? Formula: \( V = \frac{W}{Q} \) Solution: \( V = \frac{W}{Q} \)
\( \implies \) W = VQ
\( \implies \) W = 9 x 3
\( \implies \) W = 27 J The work done is 27 joule.

In simple words: Work done is calculated by multiplying the charge moved by the potential difference it moved across.

🎯 Exam Tip: Recall the definition of potential difference (\( V = W/Q \)) and rearrange it to find work done. Ensure units are consistent (Coulomb for charge, Volt for potential difference, Joule for work).

Question 3. The resistance of the filament of a bulb is 1000Ω. It is drawing a current from a source of 230 V. How much current is flowing through it?
Answer: Given: Resistance (R) = 1000 Ω P. D. = (V) = 230 (V) To find: Current (I) = ? Formula: \( \frac{V}{I} = R \) Solution: \( \frac{V}{I} = R \)
\( \implies \) \( I = \frac{V}{R} \)
\( \implies \) \( I = \frac{230}{1000} \)
\( \implies \) I = 0.23 A The current flowing through the filament of bulb is 0.23 A.

In simple words: Use Ohm's Law to find the current by dividing the potential difference across the bulb by its resistance.

🎯 Exam Tip: Ohm's law (V=IR) is fundamental; ensure you can rearrange it to find any of the three variables (V, I, R) when the other two are given.

Question 4. The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer: Given: L = 50 cm = 50 x 10^-2m, r = 0.5 mm = 0.5 x 10^-3 m = 5 x 10^-4m R = 30 Ω To find: Resistivity (\( \rho \)) of wire = ? Formula: \( \rho = \frac{RA}{L} \) Solution: \( \rho = \frac{RA}{L} \) and \( A = \pi r^2 \)
\( \implies \) \( \rho = \frac{R \times \pi r^2}{L} \)
\( \implies \) \( \rho = \frac{30 \times 3.14 \times (5 \times 10^{-4})^2}{50 \times 10^{-2}} \)
\( \implies \) \( \rho = \frac{30 \times 3.14 \times 25 \times 10^{-8}}{50 \times 10^{-2}} \)
\( \implies \) \( \rho \) = 47.1 x 10^-6 Ωm.
\( \implies \) \( \rho \) = 4.71 x 10^-5 Ωm The resistivity of the wire is 4.71 x 10^-5 Ωm.

In simple words: To find resistivity, use the formula \( \rho = \frac{RA}{L} \) after converting all given measurements to SI units (meters for length/radius, square meters for area).

🎯 Exam Tip: Always convert all given values to SI units (m, Ω, m²) before performing calculations in resistivity problems to avoid errors. Pay attention to powers of 10 in scientific notation.

Question 5. A current of 0.24 A flows through a conductor when a potential difference of 24 V is applied between its two ends. What is its resistance?
Answer: Given: Current (I) = 0.24 A P. D. (V) = 24 V To find: Resistance (R) = ? Formula: V = IR Solution: V = IR
\( \implies \) \( R = \frac{V}{I} \)
\( \implies \) \( R = \frac{24}{0.24} \)
\( \implies \) R = 100 Ω The resistance of a conductor is 100Ω.

In simple words: The resistance of the conductor can be found by dividing the applied potential difference by the current flowing through it, according to Ohm's Law.

🎯 Exam Tip: Directly apply Ohm's Law (\( R = V/I \)) and ensure you use the correct units (Volts for V, Amperes for I) to get resistance in Ohms.

Question 6. If three resistors 15Ω, 3Ω and 4Ω each are connected in series, what is the effective resistance in the circuit?
Answer: Given: R\( _1 \)=15Ω R\( _2 \)=3Ω R\( _3 \)=4Ω Effective resistance in series (R\( _s \)) = ? Formula: R\( _s \) = R\( _1 \) + R\( _2 \) + R\( _3 \) Solution: R\( _s \) = 15 + 3 + 4 R\( _s \) = 22Ω The effective resistance in the circuit is 22Ω.

In simple words: When resistors are in series, their effective resistance is simply the sum of all individual resistances.

🎯 Exam Tip: For resistors in series, the calculation is straightforward addition; just ensure all resistances are correctly identified and summed up.

Question 7. Three resistances 15Ω, 20Ω and 10Ω are connected in parallel. Find the effective resistance of the circuit.
Answer: Given: R\( _1 \) = 15Ω R\( _2 \) = 20 Ω R\( _3 \) = 10 Ω To find: Effective resistance in parallel (R\( _p \)) = ? Formula: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \) Solution: \( \frac{1}{R_p} = \frac{1}{15} + \frac{1}{20} + \frac{1}{10} \)
\( \implies \) \( \frac{1}{R_p} = \frac{4+3+6}{60} \)
\( \implies \) \( \frac{1}{R_p} = \frac{13}{60} \)
\( \implies \) \( R_p = \frac{60}{13} \)
\( \implies \) R\( _p \) = 4.615 Ω The effective resistance of the circuit is 4.615 Ω. It is less than the least of the three i.e., 10Ω.

In simple words: For parallel resistors, calculate the sum of the inverses of individual resistances, then take the inverse of that sum to find the effective resistance.

🎯 Exam Tip: Always remember that the effective resistance in a parallel circuit is always less than the smallest individual resistance, which can serve as a quick check for your answer.

Write A Note On The Following:

Question 1. Electric current
Answer: An electric current is the flow of electrons through a conductor. Quantitatively, current (I) is defined as the charge passing through a conductor in unit time. \( I = \frac{Q}{t} \)

In simple words: Electric current is simply the movement of electric charge (electrons) through a material over a specific period.

🎯 Exam Tip: Define electric current by both its qualitative (flow of charge) and quantitative (Q/t) aspects, including its formula and SI unit for a comprehensive answer.

Question 2. 1 ampere
Answer: One ampere current is said to flow in a conductor if one coulomb charge flows through it every second. \( 1 \text{ A} = \frac{1 \text{ C}}{1 \text{ s}} \)

In simple words: One ampere is the unit of electric current, signifying that one Coulomb of electric charge passes a point in a conductor every second.

🎯 Exam Tip: When defining fundamental units like the ampere, relate it directly to other SI units (Coulomb and second) to show a clear understanding of its derivation.

Question 3. 1 volt
Answer: The potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to another. \( 1 \text{ V} = \frac{1 \text{ J}}{1 \text{ C}} \)

In simple words: One volt is the unit of potential difference, meaning that one Joule of work is needed to move one Coulomb of charge between two points.

🎯 Exam Tip: Similar to ampere, define volt in terms of work done (Joule) and charge (Coulomb), emphasizing that it represents energy per unit charge.

Question 4. Potential Difference
Answer: The amount of work done to carry a unit positive charge from point A to point B is called the electric potential difference between the two points. \( V = \frac{W}{Q} \)

In simple words: Potential difference, or voltage, is the energy required to move a unit charge between two specific points in an electric field.

🎯 Exam Tip: Explain potential difference as the work done per unit charge, highlighting its role as the driving force for electric current.

Question 5. Conductor
Answer: Those substances which have very low resistance are called conductors. Current can flow easily through such materials.

In simple words: Conductors are materials with low electrical resistance that allow electric current to flow through them easily.

🎯 Exam Tip: Define conductors by their characteristic low resistance and their ability to facilitate current flow, often due to free electrons.

Question 6. Insulators
Answer: Those substances which have extremely high resistance and through which current cannot flow are called insulators.

In simple words: Insulators are materials with extremely high electrical resistance that effectively block the flow of electric current.

🎯 Exam Tip: Clearly state that insulators are defined by their very high resistance and their inability to conduct electricity, contrasting them with conductors.

Question 7. 1 ohm
Answer: If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm. \( \frac{1 \text{ Volt}}{1 \text{ Ampere}} = 1\text{Ohm} \)

In simple words: One Ohm is the unit of electrical resistance, defined as the resistance that allows one Ampere of current to flow when one Volt of potential difference is applied.

🎯 Exam Tip: Define Ohm based on Ohm's Law, relating it to both Volts and Amperes to illustrate the interdependence of these electrical quantities.

Question 8. Potential
Answer: The level of electric charge present is known as potential.

In simple words: Electric potential refers to the amount of electric potential energy per unit charge at a specific point in an electric field.

🎯 Exam Tip: While potential is related to charge, its definition is more precisely about the "level" or "amount of energy per unit charge" at a point, crucial for understanding potential difference.

Question 9. Ohm's Law
Answer: If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends. V = IR

In simple words: Ohm's Law states that for a constant temperature, the current through a conductor is directly proportional to the voltage applied across it.

🎯 Exam Tip: When stating Ohm's Law, always include the condition of "constant physical state" or "constant temperature" as resistance can change with temperature.

Question 10. Superconductors
Answer: The resistance of some conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Such conductors are called superconductors.

In simple words: Superconductors are special materials that lose all electrical resistance when cooled to extremely low temperatures, allowing current to flow indefinitely without energy loss.

🎯 Exam Tip: Highlight the key characteristic of superconductors: zero electrical resistance below a critical temperature, leading to unique applications in technology.

Question 11. Non-ohmic conductors
Answer: Conductors which do not obey Ohm's law are called non-ohmic conductors.

In simple words: Non-ohmic conductors are materials whose resistance changes with voltage or current, meaning they do not follow Ohm's Law.

🎯 Exam Tip: Remember that not all conductors obey Ohm's Law strictly; non-ohmic materials have a non-linear relationship between voltage and current.

Complete The Flow Charts:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ़्लोचार्ट बिजली से सुरक्षा के लिए उपयोग किए जाने वाले विभिन्न घटकों को दर्शाता है, जिसमें फ्यूज, इंसुलेटर, अर्थिंग और एमसीबी शामिल हैं, जो सभी विद्युत सुरक्षा में योगदान करते हैं। छात्र इन घटकों के माध्यम से विद्युत सुरक्षा प्रणालियों को समझ सकते हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ़्लोचार्ट प्रतिरोध (Resistance) के दो मुख्य प्रकार- समानांतर (parallel) और श्रेणी (series) संयोजन को दर्शाता है। इसमें समानांतर संयोजन के लिए \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \) सूत्र और श्रेणी संयोजन के लिए \( R_s = R_1 + R_2 \) सूत्र भी दिए गए हैं, जिससे छात्रों को विभिन्न प्रकार के प्रतिरोध संयोजनों को समझने में मदद मिलेगी।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह फ़्लोचार्ट प्रतिरोधकता (Resistivity) की अवधारणा को समझाता है, जो सामग्री के विद्युत गुणों पर निर्भर करती है। यह कंडक्टर (जैसे लोहा) और इंसुलेटर (जैसे रबर) के उदाहरणों के साथ प्रतिरोधकता के आधार पर सामग्रियों के वर्गीकरण को दर्शाता है।

Write Properties/Characteristics/Advantages Of The Following:

Question. Superconductors
Answer: The resistance of these conductors becomes nearly zero if their temperature is decreased up to a certain value close to 0 K. Aluminium is an example of Super Conductor. Superconductors can be used in space missions to increase/ boost the signal strength. They are also used in the data fibres to increase the speed of data transfer.

In simple words: Superconductors, like aluminium at very low temperatures, offer zero electrical resistance, making them ideal for high-speed data transfer and boosting signal strength in advanced technologies.

🎯 Exam Tip: Focus on the unique property (zero resistance), the conditions required (low temperature), an example, and practical applications (signal boosting, data transfer speed) of superconductors.

Give Explanations Of The Given Statements:

Question 1. Safety precautions are to be taken while using electricity.
Answer:

• Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
• Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
• One's hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles.
• As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
• If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
• If not, then you should use a wooden pole to push the person away from the electric wire.

In simple words: Electrical safety involves correct installation, maintaining dry hands, using insulated footwear, and knowing how to safely disconnect power or assist someone during a shock to prevent injuries.

🎯 Exam Tip: Focus on understanding the practical application of electrical safety rules, as these are common sense practices and frequently tested for awareness.

Question 2. In a domestic circuit colour code is followed while setting up electrical wiring.
Answer:

• The electricity in our homes is brought through the main conducting cable either from the electric pole or from underground cables.
• Usually, there are three wires in the cable.


• (a) Live wire which brings in the current. It has a red or brown insulation.
  (b) Neutral wire through which the current returns. It is blue or black.
  (c) Earth wire is of yellow or green colour. This is connected to a metal plate buried deep underground near the house and is for safety purposes.
• In India, the voltage difference between the live and neutral wires is about 220 V.
• Live and neutral wires are connected to the electric meter through a fuse.
• They are connected through a main switch, to all the conducting wires inside the home so as to provide electricity to every room.
• In each separate circuit, various electrical appliances are connected between the live and neutral wires.

In simple words: Domestic circuits use specific color codes (red/brown for live, blue/black for neutral, yellow/green for earth) for safety and to standardize wiring, ensuring proper current flow and protection from shocks.

🎯 Exam Tip: Memorize the color codes for live, neutral, and earth wires, along with their functions and the typical voltage difference in domestic circuits, as this is a fundamental concept.

Question 3. Fuse used in electrical circuit can save electrical objects from damage.
Answer:

• Fuse wire is used to protect domestic appliances.
• It is made of a mixture of substances and has a specific melting point.
• It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts.
• The circuit gets broken and the flow of current stops, thus protecting the appliance.
• This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1 A, 2 A, 3 A, 4 A, 5 A and 10 A are used.

In simple words: A fuse protects electrical appliances by breaking the circuit when excessive current flows, melting due to its low melting point, and thus preventing damage to the device.

🎯 Exam Tip: Understand the mechanism of a fuse, its series connection, and how its specific melting point is crucial for protecting appliances from overcurrent.

Question 4. Bulbs arranged in parallel glow brighter than bulbs arranged in series.
Answer:

• The amount of light given out by bulbs in parallel combination will be more than that in series combination.
• In parallel combination the resistance of the overall circuit decreases whereas in series it increases, so the current flowing through the bulbs in parallel circuit is more.
• Due to this, intensity of light given out by bulbs in parallel combination is more than the bulbs in series combination.

In simple words: Bulbs in parallel glow brighter because each receives the full voltage, leading to higher current and lower overall circuit resistance compared to series connections where voltage is divided.

🎯 Exam Tip: Focus on the concepts of voltage division in series and constant voltage in parallel circuits to explain brightness differences, as this is a common comparative question.

Complete The Following Table:

🎯 Exam Tip: Thoroughly learn and practice drawing all standard electrical circuit symbols and their corresponding functions, as diagrams and symbol identification are common exam questions.

Solve The Numerical:

Question 1. The length of a conducting wire is 50 cm and its radius is 0.5 mm. If its resistance is 30Ω, what is the resistivity of its material?
Answer:
Given:
Length of wire \(L = 50 \text{ cm} = 50 \times 10^{-2}\text{ m}\)
radius \(r = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m} = 5 \times 10^{-4}\text{ m}\)
Resistance \(R = 30 \Omega\)
To find: Resistivity \((\rho)\) of wire = ?
Formula: \(\rho = \frac{RA}{L}\)
Solution:
Area of cross-section \(A = \pi r^2\)
\(A = 3.14 \times (5 \times 10^{-4})^2\)
\(A = 3.14 \times 25 \times 10^{-8}\)
\(A = 78.5 \times 10^{-8} \text{ m}^2\)
Now, using the formula for resistivity:
\(\rho = \frac{RA}{L}\)
\(\rho = \frac{30 \times 78.5 \times 10^{-8}}{50 \times 10^{-2}}\)
\(\rho = \frac{2355 \times 10^{-8}}{50 \times 10^{-2}}\)
\(\rho = \frac{2355}{50} \times 10^{-8} \times 10^2\)
\(\rho = 47.1 \times 10^{-6} \Omega \text{m}\)
\(\rho = 4.71 \times 10^{-5} \Omega \text{m}\)
The resistivity of the wire is \(4.71 \times 10^{-5} \Omega \text{m}\).
In simple words: To find resistivity, convert all units to SI (meters), calculate the cross-sectional area using the given radius, then apply the formula \(\rho = \frac{RA}{L}\) with the given resistance and length.

🎯 Exam Tip: Always ensure unit consistency (SI units) before solving numerical problems, especially when dealing with length and radius conversions for resistivity calculations.

Question 2. Determine the current that will flow when a potential difference of 33 V is applied between two ends of an appliance having a resistance of 110 Ω. If the same current is to flow through an appliance having a resistance of 500 Ω, how much potential difference should be applied across its two ends?
Answer:
Given:
Potential Difference \(V_1 = 33 \text{ V}\)
Resistance \(R_1 = 110 \Omega\)
Resistance \(R_2 = 500 \Omega\)
To find:
(i) Current \(I = ?\)
(ii) Potential Difference \(V_2\) for Resistance \(R_2 = ?\)
Formula: \(V = IR\)
Solution:
(i) To find current \(I\):
\(V_1 = IR_1\)
\( \implies I = \frac{V_1}{R_1}\)
\(I = \frac{33}{110}\)
\(I = 0.3 \text{ A}\)
(ii) To find potential difference \(V_2\) for Resistance \(R_2\), assuming the same current \(I\):
\(V_2 = IR_2\)
\(V_2 = 0.3 \times 500\)
\(V_2 = 150 \text{ V}\)
The current is 0.3 A and potential difference to be applied is 150 V.
In simple words: First, use Ohm's law (\(V=IR\)) with the given voltage and resistance to find the current. Then, use this same current with the new resistance to calculate the required potential difference.

🎯 Exam Tip: When a numerical problem has multiple parts, break it down, solve each part sequentially using appropriate formulas like Ohm's Law, and ensure intermediate results are carried forward correctly.

Question 3. Determine the resistance of a copper wire having a length of 1 km and diameter of 0.5 mm.
Answer:
Given: Resistivity of copper \((\rho)\)
\(= 1.7 \times 10^{-8} \Omega \text{m}\)
Converting all measures into metres.
Length of wire \(L = 1 \text{ km} = 1000 \text{ m} = 10^3 \text{ m}\)
Diameter of wire \(d = 0.5 \text{ mm} = 0.5 \times 10^{-3}\text{ m}\)
To find: Resistance of wire \((R) = ?\)
Formula: \(R = \rho \frac{L}{A}\)
Solution:
If \(d\) is the diameter of the wire then, its area of cross-section
\(A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}\)
\(A = \frac{\pi}{4} (0.5 \times 10^{-3})^2 = \frac{3.14}{4} \times 0.25 \times 10^{-6} \text{ m}^2\)
\(A = 0.785 \times 0.25 \times 10^{-6} \text{ m}^2\)
\(A = 0.19625 \times 10^{-6} \text{ m}^2\) (approx \(0.2 \times 10^{-6} \text{ m}^2\) as used in OCR)
Using \(A = 0.2 \times 10^{-6} \text{ m}^2\)
\(R = \rho \frac{L}{A}\)
\(R = \frac{1.7 \times 10^{-8} \times 10^3}{0.2 \times 10^{-6}}\)
\(R = \frac{1.7 \times 10^{-5}}{0.2 \times 10^{-6}}\)
\(R = \frac{1.7}{0.2} \times 10^{-5} \times 10^6\)
\(R = 8.5 \times 10^1\)
\(R = 85 \Omega\)
The resistance of a copper wire is \(85 \Omega\).
(Note: The OCR stated area of cross section is \(0.2 \times 10^{-6} \text{ m}^2\), this leads to \(R=85 \Omega\))
In simple words: To calculate wire resistance, first convert length and diameter to meters. Then, find the cross-sectional area using \(A = \frac{\pi d^2}{4}\) and finally apply the resistance formula \(R = \rho \frac{L}{A}\) with copper's resistivity.

🎯 Exam Tip: Double-check unit conversions (e.g., km to m, mm to m) and correctly calculate the cross-sectional area of a circular wire before applying the resistance formula, as these are common sources of error.

Question 5. Two resistors having resistance of 16 and 14 are connected in series. If a potential difference of 18 V Is applied across them, calculate the current flowing through the circuit and the potential difference across each individual resistor.
Answer:
Given:
\(R_1 = 16 \Omega\)
\(R_2 = 14 \Omega\)
\(V = 18 \text{ V}\)
To find: Current \((I)\) = ?
Potential across each resistance i.e., \(V_1, V_2 = ?\)
Formula: \(R_S = R_1 + R_2\)
\(I = \frac{V}{R_S}\)
Solution:
Effective resistance in series: \(R_S = R_1 + R_2\)
\(R_S = 16 \Omega + 14 \Omega\)
\(R_S = 30 \Omega\)
Current flowing through the circuit:
\(I = \frac{V}{R_S}\)
\(I = \frac{18 \text{ V}}{30 \Omega}\)
\(I = 0.6 \text{ A}\)
Potential difference across each resistor:
For \(R_1\): \(V_1 = IR_1\)
\(V_1 = 0.6 \text{ A} \times 16 \Omega\)
\(V_1 = 9.6 \text{ V}\)
For \(R_2\): \(V_2 = IR_2\)
\(V_2 = 0.6 \text{ A} \times 14 \Omega\)
\(V_2 = 8.4 \text{ V}\)
The current in the circuit is 0.6 A and potential across 16 \(\Omega\) resistor is 9.6 V and 14 \(\Omega\) resistor is 8.4 V.
In simple words: First, find the total resistance by adding series resistors, then use Ohm's law to find the total current. Finally, use the total current and individual resistances to find the voltage drop across each resistor.

🎯 Exam Tip: For series circuits, remember that current is the same through all resistors, but potential difference divides. Calculate total resistance first, then total current, and then individual potential drops.

Question 6. If the resistors 5 Ω, 10 Ω and 30 Ω are connected in parallel to battery of 12 V, find the effective resistance in the circuit. Calculate the total current and current in each resistor.
Answer:
Given:
\(R_1 = 5 \Omega\)
\(R_2 = 10 \Omega\)
\(R_3 = 30 \Omega\)
\(V = 12 \text{ V}\)
To find:
(a) Total current and current in each resistor i.e., \(I, I_1, I_2\), and \(I_3 = ?\)
(b) Effective resistance \((R_P) = ?\)
Formula: (a) \(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)
(b) \(V = IR\)
Solution:
(a) Effective resistance in parallel \((R_P)\):
\(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)
\( \implies \frac{1}{R_P} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}\)
To add fractions, find a common denominator, which is 30:
\(\frac{1}{R_P} = \frac{6}{30} + \frac{3}{30} + \frac{1}{30}\)
\(\frac{1}{R_P} = \frac{6+3+1}{30}\)
\(\frac{1}{R_P} = \frac{10}{30}\)
\(\frac{1}{R_P} = \frac{1}{3}\)
\(R_P = 3 \Omega\)
Total current \(I\):
\(I = \frac{V}{R_P}\)
\(I = \frac{12 \text{ V}}{3 \Omega}\)
\(I = 4 \text{ A}\)
Current in each resistor (in parallel, voltage is same):
\(I_1 = \frac{V}{R_1} = \frac{12 \text{ V}}{5 \Omega} = 2.4 \text{ A}\)
\(I_2 = \frac{V}{R_2} = \frac{12 \text{ V}}{10 \Omega} = 1.2 \text{ A}\)
\(I_3 = \frac{V}{R_3} = \frac{12 \text{ V}}{30 \Omega} = 0.4 \text{ A}\)
Check: \(I = I_1 + I_2 + I_3 = 2.4 + 1.2 + 0.4 = 4 \text{ A}\)
(a) The total current is 4 A and current in each resistor is 2.4 A, 1.2 A and 0.4 A respectively.
(b) The effective resistance in the circuit is 3 \(\Omega\).
In simple words: For parallel resistors, calculate the effective resistance using the inverse sum formula. Then, find the total current using Ohm's law with the total voltage. Finally, since voltage is constant across parallel branches, calculate the current through each resistor individually.

🎯 Exam Tip: For parallel circuits, remember that potential difference is the same across all branches, but current divides. Calculate the equivalent resistance using \(\frac{1}{R_P}\) and then find the total current, followed by individual branch currents.

Complete The Diagram And Answer The Questions:

Question 1. The following figure shows the symbols for components used in the accompanying electrical circuit.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक अधूरा विद्युत परिपथ आरेख है जिसमें एक अज्ञात प्रतिरोध R, एक बैटरी (सेल) और एक खुला प्लग कुंजी के लिए प्रतीक दर्शाए गए हैं। इसमें वोल्टमीटर, एमीटर, प्रतिरोध और बैटरी के लिए खाली स्थान हैं जिन्हें भरना है।

Completed Circuit Diagram:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक पूर्ण विद्युत परिपथ आरेख है जिसमें एक बैटरी, एक प्लग कुंजी, एक प्रतिरोधक, एक एमीटर (जो श्रृंखला में जुड़ा है), और एक वोल्टमीटर (जो प्रतिरोधक के समानांतर जुड़ा है) दिखाया गया है। यह ओम के नियम को प्रदर्शित करने के लिए व्यवस्थित है।
(b) This circuit can be used to prove Ohm's law.
(c) \(V = IR\) is the expression of Ohm's law
In simple words: The provided circuit diagram, when completed with a voltmeter (parallel) and an ammeter (series) connected to a resistor and battery, can be used to experimentally verify Ohm's Law, which states that voltage (\(V\)) is directly proportional to current (\(I\)) and resistance (\(R\)) (\(V=IR\)).

🎯 Exam Tip: Be able to identify and correctly draw standard circuit symbols, understand their placement (series/parallel), and know which laws (like Ohm's law) they help demonstrate.

Question 3. Explain with the help of a diagram, what are free electrons and how they move through the conductor?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु के तार में इलेक्ट्रॉनों की गति को दर्शाता है। पहले भाग में, मुक्त इलेक्ट्रॉन बिना किसी बाहरी विद्युत क्षेत्र के परमाणुओं के बीच यादृच्छिक रूप से घूमते हुए दिखाए गए हैं। दूसरे भाग में, जब एक संभावित अंतर लागू होता है, तो इलेक्ट्रॉन बैटरी के नकारात्मक टर्मिनल से सकारात्मक टर्मिनल की ओर एक विशिष्ट दिशा में बहने लगते हैं, जिससे विद्युत प्रवाह होता है।

• Every atom of a metallic conductor has one or more outermost electrons which are very weakly bound to nucleus.
• These are called free electrons.
• These electrons can easily move from one part of a conductor to its other parts.

In simple words: Free electrons are loosely bound outermost electrons in a metallic conductor that can move easily. When a potential difference is applied, these electrons drift in a specific direction, forming an electric current.

🎯 Exam Tip: Understand that free electrons are the charge carriers in metals; their directed motion under a potential difference constitutes electric current. A simple diagram illustrating random vs. directed motion is beneficial.

Complete The Paragraph:

Question 1. If resistors are connected in series,
Answer:
The same current flows through each resistor. The effective resistance of the resistors is equal to the sum of their individual resistances. The potential difference between the two extremes of the arrangement is equal to the sum of the potential differences across individual resistors. The effective resistance is larger than each of the individual resistances. This arrangement is used to increase the resistance in a circuit. This type of connection is used in electrical heating equipment like geysers, iron, and hair dryers.
In simple words: In a series connection, the current remains constant through all resistors, the total resistance is the sum of individual resistances, and the total voltage is divided among them, making it suitable for increasing circuit resistance.

🎯 Exam Tip: Remember the key characteristics of series connections: constant current, additive resistances, and voltage division. Be prepared to list applications like heating elements.

Question 2. If a number of resistors are connected in parallel,
Answer:
The inverse of the effective resistance is equal to the sum of the inverses of individual resistances. The current flowing through an individual resistor is proportional to the inverse of its resistance and the total current flowing through the circuit is the sum of the currents flowing through individual resistors. The potential difference across all resistors is the same. The effective resistance of resistors connected in parallel is less than the least resistance of individual resistors.

This arrangement is used to reduce the resistance in a circuit. Even if any one of the several bulbs connected in parallel becomes non-functional because of some damage to its filament, the circuit does not break as the current flows, through the other paths, and the rest of the bulbs light up. When several bulbs are connected in parallel, they emit the same amount of light as when they are connected individually in the circuit, while bulbs connected in series emit less light than when connected individually.
In simple words: In a parallel connection, the potential difference is constant across all resistors, the inverse of total resistance is the sum of inverse individual resistances, and the total current is divided. This setup lowers overall resistance and allows other components to function even if one fails.

🎯 Exam Tip: Understand that in parallel connections, voltage is constant, current divides, and the equivalent resistance is always less than the smallest individual resistance. This is crucial for domestic wiring explanations.

Read The Paragraph And Answer The Questions.

Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket. Before cleaning an electrical appliance it should be switched off and its plug removed from the socket. One's hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles. As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. If a person gets an electric shock, you should not touch that person. You should switch off the main switch and if the switch is too far or you do not know where it is located, then you should remove the plug from the socket if possible. If not, then you should use a wooden pole to push the person away from the electric wire.

(i) Why should the electrical sockets be fitted at a certain height?
Answer:
Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside.
In simple words: Electrical sockets should be fitted high enough to prevent young children from reaching them and inserting objects, ensuring their safety.

🎯 Exam Tip: Questions about safety precautions often test practical understanding; clearly state the reason behind each safety measure.

(ii) Why plug wires should not be pulled out while removing any electrical device?
Answer:
Plug wires should not be pulled out while removing any electrical device as it may cause the wire to break causing short circuit which can lead to fire or death.
In simple words: Pulling plug wires instead of the plug can damage the wire, potentially causing short circuits, electrical fires, or severe electric shocks.

🎯 Exam Tip: Emphasize the potential hazards like short circuits, fire, and electric shock as the consequences of improper handling of electrical plugs and wires.

(iii) Why should a person wear footwear with rubber soles while handling electrical appliances..
Answer:
As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it. Hence a person should wear footwear with rubber soles while handling electrical appliances.
In simple words: Wearing rubber-soled footwear provides insulation, preventing electric current from passing through your body to the ground, thus protecting you from shocks.

🎯 Exam Tip: Highlight the insulating property of rubber and how it breaks the circuit for current flow through the body, which is a common safety principle.

(iv) Saee is touching an electrical button socket with wet hands what will you advise her and why?
Answer:
We will advise her to dry her hands before touching any electrical sockets or devices as water on the hands can cause an electrical short circuit producing shock to the person touching it.
In simple words: Advise Saee to dry her hands immediately because water conducts electricity, increasing the risk of an electric shock if she touches a live socket with wet hands.

🎯 Exam Tip: Focus on water's conductivity and the direct risk of shock when explaining why wet hands are dangerous around electricity.

(v) Sneha is getting an electrical shock what will you do the save her life?
Answer:
We should switch off the main switch and if the switch is too far or we do not know where it is located, then we should remove the plug from the socket if possible. If not, then we should use a wooden pole to push the person away from the electric wire.
In simple words: To save Sneha, immediately cut off the power by switching off the main supply or removing the plug. If direct access to power cut-off is not possible, use a non-conductive object like a wooden pole to push her away from the live wire.

🎯 Exam Tip: Prioritize immediate actions: first, cut power supply (main switch/plug); second, if power cannot be cut, use an insulated object to separate the person from the source.

(vi) Give a title to the above passage.
Answer:
Precautions to be taken while using electricity
In simple words: A suitable title for the passage is "Electrical Safety Precautions" or "Safety Measures While Using Electricity," as it primarily discusses ways to prevent electrical accidents.

🎯 Exam Tip: When titling a passage, choose a concise phrase that accurately reflects the main theme or central idea discussed in the text.

Answer The Questions In Details:

Question 1. Find the expression (i.e., derive the expression) for the resistors connected in series.
Answer:
Expression for the resistance connected in series:
(i) Let \(R_1, R_2\) and \(R_3\) be three resistances connected in series between C and D.
(ii) Let \(R_S\) be the effective resistance in circuit and \(V_1, V_2\) and \(V_3\) be the potential difference across \(R_1, R_2\) and \(R_3\) respectively.
(iii) Let the potential difference across CD be \(V\).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन प्रतिरोधकों \(R_1\), \(R_2\), और \(R_3\) को दर्शाता है जो एक विद्युत परिपथ में श्रृंखला में जुड़े हुए हैं। इसमें एक बैटरी, एक एमीटर, एक वोल्टमीटर और एक कुंजी भी शामिल है, जो श्रृंखला संयोजन में कुल प्रतिरोध और व्यक्तिगत वोल्टेज ड्रॉप को समझने में मदद करता है।
(iv) In series combination,
The total potential difference \(V\) is the sum of the potential differences across individual resistors:
\(V = V_1 + V_2 + V_3\) ...(i)
By using Ohm's law, the potential difference across each resistor is:
\(V_1 = IR_1\)
\(V_2 = IR_2\)
\(V_3 = IR_3\)
Also, for the effective resistance \(R_S\), the total potential difference is:
\(V = IR_S\)
Substituting these values in equation (i) we get:
\(IR_S = IR_1 + IR_2 + IR_3\)
Dividing by \(I\) (since current is same in series):
\(R_S = R_1 + R_2 + R_3\)
For 'n' number of resistors connected in series we get:
\(R_S = R_1 + R_2 + R_3 + \ldots + R_n\)
In simple words: When resistors are connected in series, the total potential difference is the sum of individual voltage drops. Applying Ohm's law to each resistor and the total circuit shows that the effective resistance is simply the sum of all individual resistances.

🎯 Exam Tip: Clearly state Ohm's law and the principle of voltage division in series. Ensure each step of the derivation is logical and mathematically sound, leading to the final formula for effective resistance.

Question 2. Find the expression (i.e., derive the expression) for the resistors connected in parallel.
Answer:
Expression for the resistance connected in parallel.
(i) Let \(R_1, R_2\) and \(R_3\) be three resistances connected in parallel combination between points C and D and let \(R_P\) be their effective resistance.
(ii) Let \(I_1, I_2\) and \(I_3\) be the currents flowing through resistances \(R_1, R_2\) and \(R_3\) respectively. Let \(I\) be the current flowing through the circuit and \(V\) be the potential difference of the cell.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन प्रतिरोधकों \(R_1\), \(R_2\), और \(R_3\) को दर्शाता है जो एक विद्युत परिपथ में समानांतर में जुड़े हुए हैं। इसमें एक बैटरी, एक एमीटर और एक वोल्टमीटर भी शामिल है, जो समानांतर संयोजन में कुल धारा और व्यक्तिगत धारा विभाजन को समझने में मदद करता है।
(iii) For parallel combination of resistances,
The total current \(I\) is the sum of the currents flowing through individual resistors:
\(I = I_1 + I_2 + I_3\) ...(i)
According to Ohm's law, and since the potential difference \(V\) is the same across all parallel resistors:
\(I_1 = \frac{V}{R_1}\)
\(I_2 = \frac{V}{R_2}\)
\(I_3 = \frac{V}{R_3}\)
Also, for the effective resistance \(R_P\), the total current is:
\(I = \frac{V}{R_P}\)
Substituting these values of \(I, I_1, I_2\) and \(I_3\) in equation (i) we get:
\(\frac{V}{R_P} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}\)
Dividing by \(V\) (since voltage is same in parallel):
\(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)
For 'n' number of resistances connected in parallel:
\(\frac{1}{R_P} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots + \frac{1}{R_n}\)
In simple words: In a parallel connection, the total current is the sum of currents through each branch, while the voltage remains constant. Using Ohm's law for each branch and the entire circuit, we derive that the reciprocal of the total resistance equals the sum of the reciprocals of individual resistances.

🎯 Exam Tip: Focus on the principle that voltage is constant across parallel branches, and current divides. Ensure the derivation correctly uses Ohm's law for each branch and for the overall equivalent resistance.

Question 3. Find the expression for resistivity of a material.
Answer:
(i) At a given temperature, the resistance \((R)\) of a conductor depends on its length \((L)\), area of cross-section \((A)\) and the material it is made of. If the resistance of a conductor is \(R\), then
\(R \propto L\) (Resistance is directly proportional to length)
Also, \(R \propto \frac{1}{A}\) (Resistance is inversely proportional to cross-sectional area)
Combining these two proportionalities:
\(R \propto \frac{L}{A}\)
To convert this proportionality into an equation, we introduce a constant of proportionality, \(\rho\) (rho):
\(R = \rho \frac{L}{A}\)
(ii) \(\rho\) is the constant of proportionality and is called the resistivity of the material. Resistivity is a fundamental property of the material itself, independent of its dimensions.
(iii) The unit of resistivity in SI units is Ohm metre (\(\Omega \text{ m}\)).
(iv) Resistivity is a specific property of a material and different materials have different resistivity. For example, conductors have low resistivity, while insulators have high resistivity.
In simple words: The resistance of a material is directly proportional to its length and inversely proportional to its cross-sectional area. The constant of proportionality in this relationship is called resistivity (\(\rho\)), which is a material-specific property with SI unit Ohm-meter.

🎯 Exam Tip: Clearly state the factors affecting resistance (length, area, material) and how they lead to the resistivity formula. Remember the definition and SI unit of resistivity, as these are frequently tested.

Make The Concept Diagram And Explain:

Question 1. Make the concept diagram of an electrical circuit and explain the working of a fuse.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक बुनियादी विद्युत परिपथ का आरेख है जिसमें एक वोल्टमीटर, एक एमीटर, एक प्रतिरोधक और एक बैटरी को उनकी उचित श्रृंखला और समानांतर कनेक्शन में दर्शाया गया है, साथ ही एक कुंजी भी है।

• Fuse wire is used to protect domestic appliances.
• It is made of a mixture of substances and has a specific melting point.
• It is connected in series to the electric appliances. If for some reason, the current in the circuit increases excessively, the fuse wire gets heated and melts. The circuit gets broken and the flow of current stops, thus protecting the appliance.
• This wire is fitted in a groove in a body of porcelain-like non-conducting material. For domestic use, fuse wires with upper limits of 1A, 2A, 3A, 4A, 5A, and 10 A are used.

In simple words: A fuse is a safety device connected in series in an electrical circuit, designed to protect appliances from overcurrent. It melts and breaks the circuit if the current exceeds a safe limit, stopping the flow of electricity and preventing damage.

🎯 Exam Tip: Be able to draw a basic circuit diagram and explain the fuse's role (series connection, low melting point) in simple terms. Mentioning common fuse ratings adds value.

(2) Show motion of electrons in an circuit and explain precautions while using an electrical device.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु के तार में इलेक्ट्रॉनों की गति को दर्शाता है। A में, इलेक्ट्रॉन बिना किसी बाहरी विद्युत क्षेत्र के यादृच्छिक रूप से चलते हैं। B में, जब एक विद्युत क्षेत्र (संभावित अंतर) लागू होता है, तो इलेक्ट्रॉन एक निश्चित दिशा में बहना शुरू करते हैं (पारंपरिक धारा के विपरीत), जो विद्युत प्रवाह का कारण बनता है।

• Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
• Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
• One's hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles.
• As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
• If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
• If not, then you should use a wooden pole to push the person away from the electric wire.

In simple words: Electrons move randomly in a conductor until a potential difference is applied, causing them to drift in a specific direction (forming current). To use electrical devices safely, ensure dry hands, wear rubber-soled shoes, keep switches out of children's reach, avoid pulling wires, and always disconnect power before cleaning or in case of shock.

🎯 Exam Tip: Combine the theoretical concept of electron drift with practical safety rules. Illustrate the difference between random and directed electron motion and list at least 3-4 key electrical safety precautions.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धातु के तार में इलेक्ट्रॉनों की गति को दर्शाता है। चित्र A में, बिना किसी बाहरी विद्युत क्षेत्र के, इलेक्ट्रॉन यादृच्छिक रूप से गति करते हैं। चित्र B में, जब तार के सिरों पर एक विभवांतर लगाया जाता है, तो इलेक्ट्रॉन एक निश्चित दिशा में (परंपरागत धारा के विपरीत) गति करने लगते हैं, जिससे एक विद्युत धारा प्रवाहित होती है।
Answer:

• Electric switches and sockets should be fitted at a height at which small children cannot reach and put pins or nails inside. Plug wires should not be pulled while removing a plug from its socket.
• Before cleaning an electrical appliance it should be switched off and its plug removed from the socket.
• One's hands should be dry while handling an electrical appliance, and, as far as possible, one should use footwear with rubber soles.
• As rubber is an insulator, it prevents the current from flowing through our body, thereby protecting it.
• If a person gets an electric shock, you should not touch that person. You should switch off the main switch or remove the plug from the socket if possible.
• If not, then you should use a wooden pole to push the person away from the electric wire.

In simple words: Electrons move randomly in a wire without a potential difference, but flow in a directed manner when one is applied. To use electricity safely, install switches high, avoid pulling plugs by wire, clean appliances only when unplugged, use dry hands and rubber footwear, and know how to safely disconnect power if someone gets a shock.

🎯 Exam Tip: Focus on explaining both the electron motion (random vs. directed) and the practical safety precautions clearly. Diagrams are for understanding, but the textual explanation and safety rules are key for scoring.

Question 4.4. Complete the incomplete figure and give an explanation:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र ओम के नियम को प्रदर्शित करने वाले एक विद्युत परिपथ को दर्शाता है। इसमें एक बैटरी, एमीटर, वोल्टमीटर, प्रतिरोधक (R), और एक कुंजी को सही ढंग से जोड़ा गया है, जहाँ एमीटर श्रृंखला में और वोल्टमीटर समानांतर में जुड़ा है, जिससे परिपथ पूर्ण होकर ओम के नियम का सत्यापन किया जा सकता है।
Answer: If the physical state of a conductor remains constant, the current (I) flowing through it is directly proportional to the potential difference (V) between its two ends.
\(I \propto V\)
\(I = kV\) (k = constant of proportionality)
\(I \frac{1}{k} = V\)
\(\frac{1}{k} = R = \text{Resistance of the conductor}\)
\(IR = V\)
Hence \(V = IR\) or \(R = \frac{V}{I}\) This is known as Ohm's law. We can obtain the SI unit of resistance from the above formula, Potential difference and current are measured in Volts and Amperes respectively. The unit of resistance is called Ohm. It is indicated by the symbol \(Ω\).
\( \implies \frac{1 \text{ Volt}}{1 \text{ Ampere}} = 1 \text{ Ohm} (Ω)\) The resistance of one Ohm : If one Ampere current flows through a conductor when one Volt potential difference is applied between its ends, then the resistance of the conductor is one Ohm. In simple words: Ohm's Law states that if a conductor's physical conditions don't change, the current flowing through it is directly proportional to the voltage across its ends. The resistance, measured in Ohms (Ω), is calculated by dividing voltage by current.

🎯 Exam Tip: Remember to clearly state Ohm's law, provide its mathematical expression, and define the unit of resistance (Ohm). Understanding how the circuit components are connected to verify the law is also important.