CBSE Class 10 Science Electricity Assignment Set 02

Question. What is the current through a 5.0 ohm resistor if the voltage across it is 10 V
(a) zero
(b) 0.5 A
(c) 2.0 A
(d) 5.0 A
Answer : C

Question. 100 J of heat is produced each second in a 4 W resistance. The potential difference across the resistor is
(a) 20 V
(b) 10 V
(c) 5 V
(d) 15 V
Answer : A

Question. A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A
Answer : C

Question : An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be:
a. 100 W
b. 75 W
c. 50 W
d. 25 W

Answer : D

Very Short Answers :

Question : Write S.I. unit of resistivity.
Answer :  Ohm-metre (Ωm).

Question : Define resistance. Give its S.I. unit.
Answer :  Resistance is the property of a conductor to ppose the flow of charges through it. SI unit of resistance is Ohm (Ω).

Question. State Joule’s law of heating.
Answer: Joule’s law of heating states that the amount of heat produced in a conductor is directly proportional to:
(a) Square of current (I2)
(b) Resistance of wire (R)
(b) Time (t), for which current is passed.

Question. Define electric power.
Answer: The electrical work done per unit time is called electric power.

Question. What does an electric circuit mean?
Answer: A continuous conducting path consisting of wires and other electrical components (like resistance or electric bulb, switch etc.) between the two terminals of a cell or battery, along which an electric current flows, is called an electric circuit.

Question. Distinguish between an open and a closed circuit.
Answer: An electric circuit is said to be an open circuit when the switch is in ‘off’ mode (or key is unplugged) and no current flows in the circuit.
The circuit is said to be a closed circuit when the switch is in ‘on’ mode (or key is plugged) and a current flows in the circuit.

Question. How many 220 Ω resistors (in parallel) are required to carry 5 A on a 220 V line ?
Answer: 5.

Question. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ? 
Answer: 110.

Question. Which uses more energy :
(a) a 250 W T.V. set in 1 hour
(b) a 1200 W toaster in 10 minutes ? 
Answer: (a) 9 × 105 J, (b) 7.2 × 105 J (T.V. set uses more energy)

Question. Study the circuit diagram given in figure carefully and calculate :
(a) Current in main circuit.
(b) Current in each of the resistors in the parallel circuit.
Answer: (a) 1.80 A, (b) 1.2 A and 0.6 A, 1.2 A and 0.6 A.

Question. A wire of resistance 1.5 Ω is stretched to double its length. What will be its new resistance ?
Answer: 6 W.

Question. A wire of resistance 36 Ω and length 60 cm is tripled on itself. What is the new resistance ?
Answer: 4 W.

Question. How many electrons are flowing per second past a point in a circuit in which there is a current of 1 A ?
Answer: 6.25 × 1018.

Question. The graph between V and I for a conductor is a straight line passing through the origin.
(a) Name the law illustrated by such a graph ?
(b) What should remain constant in a statement of this law ?
Answer: (a) Ohm's law, (b) Temperature.

Question : The potential difference across the wire having fixed resistance is tripled. By how much does the electric power increase?
Answer :  The electric power will increase by nine times when the potential difference across the wire having fixed resistance is tripled. According to Ohm’s law, potential difference V is proportional to current, I. Therefore, when V is made 3 times, I will increase 3 times. As Power P = VI, therefore, Power will increase by 9 times.

Question : find the minimum rating of fuse that can be safely used on a line on which two 1.1 KW electric geysers are to run simultaneously.
The supply voltage is 220 V.
Answer :  Power P = VI. As the two geysers have power rating 1.1 kW or 1100 W and are connected in parallel, each geyser draws a current I = P/V = 1100/220 A = 5 A.

Question : Should the resistance of an ammeter be low or high? Give reason. 
Answer :  An ideal ammeter is one which has zero resistance. But that is not possible. Therefore, the resistance of an ammeter should be as close to zero as possible. If it is non-zero and
substantial, it will affect the current flowing through the circuit. This is because an ammeter is connected in series in the circuit for the measurement of electric current.

Question : The electric power consumed by a device may be calculated by either of the two expressions P = I2R or P = V2/R. The frist expression indicates that it is directly proportional to R whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained?
Answer :  The expression P = I2R is used for calculating electric power when only current I and resistance R are known, whereas P = V2/R is used for calculating power when voltage V and
resistance R are known.

Short Answers :

Question : Define the unit of electric current. 

Answer :  The SI unit of electric current is ampere (A). When 1 coulomb of electric charge flows through any cross-section of a conductor in 1 second, the electric current flowing through it is said to be 1 ampere.

 

 

 

Question : How many joule are there in 1 kWh? 

Answer : 

1kWh = 1,000 Ω × 1h = 1,000 J/S × 3600S

1kWh = 3,600,000 J = 3.6 ×106 J 

 

Question. The given figure represents V-I graph for a series combination and for a parallel combination of two resistors. Which of the two, A or B, represents the series combination. Give a reason for your answer. 

Answer: Since, the straight line A is steeper than B, so the straight line A represents a greater resistance. The equivalent resistance in a series combination is greater than in parallel combination. Hence, A represents series combination.

Question. You are given three resistors of resistance 2 W, 4 W and 6 W. Show by a diagram, how you can get a 3 W resistance with the help of these resistors.
Answer: Connect 2 W and 4 W in series then this combination is connected with the parallel 6 W resistance. By this arrangement we can get resistance 3 W as equivalent resistance.

Question. Derive the relation 1/R = 1/R1 + 1/R2 + 1/R3 when resistors are joined in parallel.
Answer: In parallel combination of three resistance R1, R2 and R3, the current in each of the resistance is different. If I is the current drawn from the cell then it is divided into branches as I1, I2 and I3. Thus,
I = I1 + I2 + I3
The potential difference across each of these resistances is the same.
Thus, from Ohm’s law
I1 = V /R1 , I2 = V/R2 , I3 = V/R3
If R is the equivalent resistance then, 

V/R = V/R1 + V/R2 + V/R3 
and 1/R = 1/R1 + 1/R2 + 1/R3 .

Question. How many 10 W resistors are required to get a 25 W resistor?
Answer: Four 10 W resistors are required and they are connected as shown. 

Question. You are given fifty 5 W resistors. What is (a) smallest and (b) largest resistance can be obtained by using these?
Answer: (a) To get the smallest resistance, all the 5 W resistors must be connected in parallel.
Smallest resistance = 5/50 = 0.1 W
(b) To get the largest resistance, the 5 W resistors must be
connected in series.
Largest resistance = 5 × 50 = 250 W.

Question. Why are fairy decorative lights always connected in parallel?
Answer: When the fairy lights are connected in series the resistance offered will be greater and brightness of the bulbs will be affected. But in parallel connection all the bulbs will glow with same intensity and if any bulb gets fused the other bulbs will continue to glow.

Question. Resistance of a conductor of length 80 cm is 4.0 W. Calculate the resistance of a similar conductor of length 400 cm.
Answer: Here, l1 = 80 cm, R1 = 4.0 W, l2 = 400 cm, R2 = ?
∴ R1/R2 = 1₁/1₂
or R2 = R1. 1/2 , 1/1
= 4.0 × 400/80 = 20.0 W.

Question. Out of 60 W and 40 W lamps, which one has a higher electrical resistance when in use.
Answer: Power (P) = V2/R .
From the above formula, P is inversely proportional to R (resistance) as voltage remaining the same. Hence, 40 We lamp has high resistance.

Question. Write the formula for current ‘I’ flowing through a conductor if ‘n’ electrons flow through the cross-section of a conductor in time ‘t’.
Answer: If ‘n’ electrons pass through the cross-section of a conductor in time ‘t’, the total charge ‘Q’ passing through the conductor is :
Q = ne (e is the charge on an electron = 1.6 ×10–19 C) The current ‘I’ in the conductor is :
l = Q/t = ne/t

Question. What is commercial unit of electrical energy ? Convert it into joules.
Answer: The commercial unit of electrical energy is kWh.
1 kWh = 1000 W × 1 hour
= 1000 j/s x 60 x 60s
= 3.6 × 106 J

Question : Heat is generated continuously in an electric heater, but the temperature of its element becomes constant after some time. Why? 

Answer :   When the temperature of the heater becomes greater than the temperature of the surroundings, some of the heat is lost to the surroundings in the form of thermal radiations and through convection. After some time,the rate at which heat is being produced becomes equal to the rate of which heat is lost. Hence, the temperature of the element becomes constant. This happens because an equilibrium or steady state is reached.

 

Question : Define 1 volt. Express it in terms of SI unit of work and charge calculate the amount of energy consumed in carrying a charge of 1 coulomb through a battery of 3 V.
Answer :  When 1 joule of work is done in carrying 1 coulomb of charge, from infinity to a point in the electric field, then potential at that point is called 1 volt. Potential difference between two points is

 

Question :  A heater coil is rated 100 W, 200 V. It is cut into two identical parts. Both parts are connected together in parallel to the same source of 200 V. Calculate the energy liberated per second in the new combination. 

Answer : Resistance of heater coil,

As the coil is cut into two identical parts, resistance of each part = 400/2  = 200 Ω

When connected in parallel, the net resistance is given by

 

Energy liberated per second in the new combination is 400J 

Question : Why is parallel arrangement used in domestic wiring? 
Answer :  Parallel arrangement is used in domestic wiring due to the following reasons:
• Each device will have the same voltage which is equal to the voltage of the supply.
• If two or more devices are used at the same time, then each appliance will be able to draw the required current.
• If one of the devices fails, then the other keeps working.

Answer :

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
(ii) What happens to the reading of A₁, A₂, A₃ and A when the bulb gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?
(i) Since B₁, B₂ and B₃ are in parallel, the potential difference across each of them will remain same. So when the bulb B₁ gets fused,B₁ ,B₂ and B₃ have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.
(ii) Resistance of the parallel combination when all the three bulbs are glowing

Since resistance of each arm is same and p.d. is also same, current divides them equally. So 1A current will pass through each bulb Bl and By 
Therefore, ammeter A₁ and A₃ reads l A current while A₂ will read zero and ammeter A read 2 A current.
(iii) In parallel, total power consumed
Peq=P₁ +P₂ +P₃
So, when all the three bulbs glow together

Question : Find the current drawn from the battery by the network of four resistors Shown in the figure.

Answer :

Question : A 9Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.
Answer :

Question : a.Two resistors R₁ and R₂ may form (i) a series combination or (ii) a parallel combination, and the combination may be connected to a battery of 6 volts. In which combination, will the potential difference across R₁ and across R₂ be the same and in which combination, will the current through R₁ and through R₂ be the same?
b. For the circuit shown in this diagram, calculate

(i) the resultant resistance.
(ii) the total current.
(iii) the voltage across 7 Ω resistor.
Answer. a. Potential difference across R₁ and R₂ is same in parallel combination of R₁ and R₂ and the current through R₁ and R₂ will be same when they are connected in series.

HOTS Questions and Answers

Q.1 What is represented by joule/coulomb?
Answer: Joule per coulomb (\(\text{J/C}\)) represents potential difference or volt. Since potential difference \(V = \frac{W}{Q}\), where work \(W\) is measured in joules and charge \(Q\) is in coulombs, \(1\text{ V} = 1\text{ J/C}\).
In simple words: It is the unit of electric potential difference, also known as volts.

Exam Tip: Always remember the formula \(V = \frac{W}{Q}\) when relating work, charge, and potential difference.

Q.2 A charge of \(2\text{ C}\) moves between two plates, maintained at a potential difference of \(1\text{ V}\). What is the energy acquired by the charge?
Answer:
Given, charge \(Q = 2\text{ C}\)
Potential difference \(V = 1\text{ V}\)
Energy acquired (Work done) \(W = Q \times V\)
\(W = 2\text{ C} \times 1\text{ V} = 2\text{ J}\)
In simple words: The energy gained by the charge is equal to the product of the charge and the potential difference, which is \(2\) joules.

Exam Tip: Energy acquired equals work done by the electric field: \(W = QV\).

Q.3 Why are copper wires used as connecting wires?
Answer: Copper wires are used as connecting wires because copper has a very low electrical resistivity and high electrical conductivity, which ensures minimum loss of electrical energy as heat during transmission.
In simple words: Copper is a very good conductor of electricity, meaning it allows current to flow easily without wasting much energy.

Exam Tip: Mention both low resistivity and high conductivity to score full marks.

Q.4 A wire of resistivity \(\rho\) is stretched to double its length. What is its new resistivity?
Answer: The new resistivity remains the same (\(\rho\)). Resistivity is a characteristic property of a material and depends only on the nature of the material and temperature, not on its dimensions (length or cross-sectional area).
In simple words: Stretching a wire changes its resistance, but not its material, so its resistivity stays constant.

Exam Tip: A classic conceptual trap - remember that resistivity is independent of length and area.

Q.5 What is the resistance of connecting wire?
Answer: The resistance of a connecting wire, which is made of a good conductor (like copper), is extremely small and considered to be nearly zero in practical circuit calculations.
In simple words: Ideal connecting wires offer negligible resistance to the flow of current.

Examist Tip: In numerical problems, connecting wires are assumed to have zero resistance unless specified otherwise.

Q.6 What is the resistance of an ammeter?
Answer: The resistance of an ammeter is very small. For an ideal ammeter, the resistance is exactly zero so that it does not alter the current flowing through the circuit it measures.
In simple words: An ammeter has almost no resistance so it does not block the current it is trying to measure.

Exam Tip: Differentiate clearly between ammeter (very low resistance, connected in series) and voltmeter (very high resistance, connected in parallel).

Q.7 What is the resistance of a Voltmeter?
Answer: The resistance of a voltmeter is very high. For an ideal voltmeter, the resistance is infinity so that it draws no current from the circuit.
In simple words: A voltmeter has extremely high resistance to prevent current from passing through it.

Exam Tip: Infinite resistance ensures that the voltmeter measures the true potential difference across a component.

Q.8 Which has more resistance: \(100\text{ W}\) bulb or \(60\text{ W}\) bulb?
Answer:
Power is given by \(P = \frac{V^2}{R}\), which means \(R = \frac{V^2}{P}\).
Since resistance is inversely proportional to power (\(R \propto \frac{1}{P}\)) for a given voltage, the bulb with lesser power has greater resistance.
Therefore, the \(60\text{ W}\) bulb has more resistance than the \(100\text{ W}\) bulb.
In simple words: Lower wattage appliances have higher resistance because they consume less power at the same voltage.

Exam Tip: Always derive using \(P = V^2 / R\) rather than memorizing directly to avoid confusion.

Q.9 How will you join three resistances, each of \(2\,\Omega\) so that the effective resistance is \(3\,\Omega\)?
Answer: To get an effective resistance of \(3\,\Omega\) using three \(2\,\Omega\) resistors, we connect two resistors in parallel and the third resistor in series with this parallel combination.
Calculation for parallel combination of two \(2\,\Omega\) resistors:
\(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \implies R_p = 1\,\Omega\)
Now, connecting the third resistor (\(2\,\Omega\)) in series with \(R_p\):
\(R_{\text{total}} = R_p + 2 = 1 + 2 = 3\,\Omega\)
In simple words: Put two resistors in parallel to make \(1\,\Omega\), then add the third resistor in series to make a total of \(3\,\Omega\).

Exam Tip: Test combinations systematically: all series (\(6\,\Omega\)), all parallel (\(0.67\,\Omega\)), two parallel + one series (\(3\,\Omega\)).

Q.10 What happens to the current in a circuit if its resistance is doubled?
Answer: According to Ohm's law, \(I = \frac{V}{R}\). If the resistance \(R\) is doubled (\(2R\)) while keeping the potential difference \(V\) constant, the new current becomes half of its original value (\(I' = \frac{V}{2R} = \frac{I}{2}\)).
In simple words: If you double the resistance, the current gets cut in half.

Exam Tip: State Ohm's law clearly before showing the inverse proportionality relationship.

Q.11 What happens to the resistance of a circuit if the current through it is doubled?
Answer: The resistance of the circuit does not change if the current is doubled. Resistance is an intrinsic property of the conductor (depending on material, length, and area) and is independent of the current flowing through it.
In simple words: Changing the current does not change the resistance of the circuit.

Exam Tip: Do not confuse cause and effect - current depends on resistance and voltage, not the other way around.

Q.12 How does the resistance of a wire depend upon its radius?
Answer: Resistance is given by \(R = \rho \frac{l}{A}\). Since the cross-sectional area \(A = \pi r^2\) (where \(r\) is the radius), we have \(R = \rho \frac{l}{\pi r^2}\).
Therefore, resistance is inversely proportional to the square of the radius (\(R \propto \frac{1}{r^2}\)).
In simple words: If you make a wire thicker (larger radius), its resistance decreases rapidly.

Exam Tip: Highlight that area depends on \(r^2\), so doubling the radius reduces resistance to one-fourth.

Q.13 Two wires are of the same length, same radius, but one of them is of copper and the other is of iron. Which will have more resistance?
Answer: Iron will have more resistance. Since both wires have the same length and radius (hence same cross-sectional area), resistance depends directly on resistivity (\(R = \rho \frac{l}{A}\)). The resistivity of iron is higher than that of copper, so iron offers more resistance.
In simple words: Iron is a poorer conductor than copper, so an iron wire has higher resistance than a copper wire of the same size.

Exam Tip: Connect the conclusion directly to the higher resistivity value of iron compared to copper.

Q.14 Two wires of same material and same length have radii \(r_1\) and \(r_2\). Compare their resistances.
Answer:
Resistance of the first wire: \(R_1 = \rho \frac{l}{\pi r_1^2}\)
Resistance of the second wire: \(R_2 = \rho \frac{l}{\pi r_2^2}\)
Taking the ratio of their resistances:
\(\frac{R_1}{R_2} = \frac{\rho \frac{l}{\pi r_1^2}}{\rho \frac{l}{\pi r_2^2}} = \frac{r_2^2}{r_1^2}\)
Therefore, \(\frac{R_1}{R_2} = \left(\frac{r_2}{r_1}\right)^2\).
In simple words: The ratio of their resistances is inversely proportional to the square of the ratio of their radii.

Exam Tip: Cancel out common terms (\(\rho, l, \pi\)) carefully before writing the final ratio.

Q.15 Given \(n\) resistors each of resistors \(R\). How will you combine them to get the (i) maximum and (ii) minimum effective resistance? What is the ratio of the maximum to minimum resistance?
Answer:
(i) For maximum effective resistance, all \(n\) resistors must be connected in series:
\(R_{\text{max}} = nR\)
(ii) For minimum effective resistance, all \(n\) resistors must be connected in parallel:
\(\frac{1}{R_{\text{min}}} = n \left(\frac{1}{R}\right) \implies R_{\text{min}} = \frac{R}{n}\)
Ratio of maximum to minimum resistance:
\(\frac{R_{\text{max}}}{R_{\text{min}}} = \frac{nR}{\frac{R}{n}} = n^2\)
In simple words: Series connection gives maximum resistance, parallel gives minimum, and their ratio is \(n^2\).

Exam Tip: Memorize this standard result: \(R_{\text{max}} / R_{\text{min}} = n^2\) for quick checks in objective questions.

Q.16 A wire of length \(L\) and resistance \(R\) is stretched so that its length is doubled. How will its (a) Resistance change (b) Resistivity change?
Answer:
(a) When a wire is stretched to double its length (\(L' = 2L\)), its volume remains constant. Since volume \(V = A \times L\) is constant, when length becomes 2 times, the cross-sectional area becomes half (\(A' = \frac{A}{2}\)).
New resistance \(R' = \rho \frac{L'}{A'} = \rho \frac{2L}{\frac{A}{2}} = 4 \left(\rho \frac{L}{A}\right) = 4R\).
Thus, the resistance becomes 4 times the original value.
(b) Resistivity remains unchanged because it depends only on the material of the wire, not on its dimensions.
In simple words: Stretching the wire makes it thinner and longer, increasing its resistance 4 times, but the material's resistivity stays the same.

Exam Tip: Always remember that volume conservation (\(AL = \text{constant}\)) is the key step when a wire is stretched.

Q.17 Two students perform the experiments on series and parallel combinations of two given resistors \(R_1\) and \(R_2\) and plot the following V-I graphs. Which of the graphs is (are) correctly labelled in terms of the words 'series' and parallel' Justify your answer.
Answer:
Both graphs are correct. In a \(V-I\) graph (with \(V\) on the Y-axis and \(I\) on the X-axis), the slope of the graph represents resistance (\(\text{Slope} = \frac{V}{I} = R\)).
- Series combination has higher effective resistance, so it will have a steeper slope (\(V\) increases more for a given \(I\)).
- Parallel combination has lower effective resistance, so it will have a smaller slope.
In the first graph, Series has a steeper slope than Parallel (correct). In the second graph (with axes interchanged or correctly interpreted), the labeling is also consistent with the respective slopes.
In simple words: A steeper slope means higher resistance (series), and a flatter slope means lower resistance (parallel).

Exam Tip: Always check which variable is plotted on the X-axis and Y-axis before determining slope = resistance or conductance.

Q.18 A household uses the following electric appliances:
(i) Refrigerator of rating \(400\text{ W}\) for ten hours each day.
(ii) Two electric fans of rating \(80\text{ W}\) each for twelve hours each day.
(iii) Six electric tubes of rating \(18\text{ W}\) each for \(6\) hours each day.
Calculate the electricity bill of the household for the month of June if the cost per unit of electric energy is Rs. 3.00.
Answer:
Energy consumed by refrigerator per day = \(400\text{ W} \times 10\text{ h} = 4000\text{ Wh}\)
Energy consumed by 2 fans per day = \(2 \times 80\text{ W} \times 12\text{ h} = 1920\text{ Wh}\)
Energy consumed by 6 tubes per day = \(6 \times 18\text{ W} \times 6\text{ h} = 648\text{ Wh}\)
Total energy consumed per day = \(4000 + 1920 + 648 = 6568\text{ Wh} = 6.568\text{ kWh}\)
Number of days in June = \(30\)
Total energy consumed in 30 days = \(6.568\text{ kWh} \times 30 = 197.04\text{ kWh (units)}\)
Cost per unit = Rs. \(3.00\)
Total Electricity Bill = \(197 \times 3 = \text{Rs. } 591\)
In simple words: Calculate total watt-hours per day, multiply by 30 days to get kilowatt-hours (units), and multiply by the rate per unit.

Exam Tip: Remember that \(1\text{ kWh} = 1\text{ unit}\) and June has 30 days.

Q.19 Ammeter burns out when connected in parallel. Give reasons.
Answer: An ammeter has very low resistance. When it is wrongly connected in parallel across a voltage source, a very large amount of current passes through it (\(I = \frac{V}{R}\) where \(R\) is extremely small). Due to the excessive heating effect of current (\(H = I^2Rt\)), the ammeter burns out.
In simple words: Connecting an ammeter in parallel allows too much current to rush through it, destroying the instrument.

Exam Tip: Always emphasize that low resistance + parallel connection = massive current flow.

More Questions for Practice

Q.1 Name a substance whose resistance almost remains unchanged by increase of temperature.
Answer: Manganin (or Constantan) is an alloy whose resistance remains almost unchanged with an increase in temperature because it has a very low temperature coefficient of resistance.
In simple words: Alloys like manganin do not change their resistance when heated, making them useful for standard resistors.

Exam Tip: Mention alloys like Manganin or Constantan as standard examples.

Q.2 Name two special characteristics of heater coil.
Answer:
1. High resistivity so that it produces a large amount of heat.
2. High melting point so that it does not melt or burn out at high temperatures.
In simple words: Heater wires need to resist electricity strongly to make heat and must not melt when they get red hot.

Exam Tip: High resistivity and high melting point are the two mandatory keywords expected by examiners.

Q.3 A wire of resistance \(4\,\Omega\) is bent to form a circle. What is the resistance between two diametrically opposite ends?
Answer:
When a wire of total resistance \(4\,\Omega\) is bent into a circle and connected across two diametrically opposite ends, the wire is divided into two equal halves, each of resistance \(\frac{4}{2} = 2\,\Omega\).
These two halves are in parallel combination:
\(\frac{1}{R_p} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} = 1 \implies R_p = 1\,\Omega\)
In simple words: Bending the wire splits it into two \(2\,\Omega\) halves in parallel, giving an equivalent resistance of \(1\,\Omega\).

Exam Tip: Remember that dividing a wire into two halves in parallel quarters its total resistance (\(R/4\)).

Q.4 How does the resistance of a conductor change if its temperature is increased?
Answer: The resistance of a metallic conductor increases with an increase in temperature. Higher temperature causes increased thermal agitation and frequent collisions of free electrons with lattice ions, hindering their smooth flow.
In simple words: Heating a metal wire makes it harder for electrons to pass through, so its resistance increases.

Exam Tip: Mention increased frequency of collisions between electrons and positive ions for complete explanation.

Q.5 A current of \(4\text{ A}\) flows in a wire of resistance \(60\,\Omega\). Calculate electrical energy consumed in \(2\) minutes.
Answer:
Given, current \(I = 4\text{ A}\)
Resistance \(R = 60\,\Omega\)
Time \(t = 2\text{ minutes} = 2 \times 60 = 120\text{ seconds}\)
Electrical energy consumed \(E = I^2Rt\)
\(E = (4)^2 \times 60 \times 120\)
\(E = 16 \times 60 \times 120 = 115,200\text{ J} = 1.152 \times 10^5\text{ J}\)
In simple words: Use the formula \(E = I^2Rt\) with time converted into seconds to find the total energy in joules.

Exam Tip: Always convert minutes into seconds before calculating energy or heat.

Q.6 V-I graph for two resistors is given. Which of the two has minimum resistance?
Answer: In a \(V-I\) graph, the slope represents resistance (\(R = \frac{V}{I}\)). The graph with the smaller slope (lesser angle with the current axis) has minimum resistance. Therefore, resistor B (having a smaller slope compared to A) has minimum resistance.
In simple words: The flatter line represents the smaller resistance.

Exam Tip: Slope = \(V/I = R\). Steeper = higher resistance, flatter = lower resistance.

Q.7 Alloys are used in electrical heating devices rather than pure metals. Give one reason.
Answer: Alloys have much higher electrical resistivity than pure metals and do not oxidize (burn) readily at high temperatures.
In simple words: Alloys produce more heat and do not burn out easily when red hot.

Exam Tip: High resistivity and resistance to oxidation are key points.

Q.8 An electric geyser has the ratings \(2000\text{ W}, 220\text{ V}\) marked on it. What should be the minimum rating, in whole number of fuse wire that may be required for safe use with this geyser?
Answer:
Given, Power \(P = 2000\text{ W}\), Voltage \(V = 220\text{ V}\)
Current \(I = \frac{P}{V} = \frac{2000}{220} = \frac{100}{11} \approx 9.09\text{ A}\)
The minimum rating of the fuse wire in whole numbers for safe use must be slightly higher than the operating current, so a \(10\text{ A}\) fuse is required.
In simple words: Calculate the operating current (~9.09 A) and choose the next higher standard whole-number fuse rating (10 A).

Exam Tip: Always round up to the next available whole-number standard fuse rating.

Q.9 The electrical resistivity of few materials is given below in ohm-meter. Which of these materials can be used for making element of a heating device?
A \(6.84 \times 10^{-8}\)
B \(1.60 \times 10^{-8}\)
C \(1.00 \times 10^{-4}\)
D \(2.50 \times 10^{12}\)
E \(4.40 \times 10^{-5}\)
F \(2.30 \times 10^{17}\)
Answer: (C) \(1.00 \times 10^{-4}\,\Omega\text{m}\)
Materials used for heating elements (alloys) require moderately high resistivity compared to pure metals but lower than insulators. Material C has the appropriate high resistivity range suitable for heating elements.
In simple words: Heating elements need high resistivity (like alloys), which corresponds to option C.

Exam Tip: Conductors have low resistivity (\(10^{-8}\)), insulators have extremely high resistivity (\(10^{12}\) to \(10^{17}\)), and alloys/semiconductors lie in between.

Q.10 Where do we connect a fuse: with live wire or with neutral wire?
Answer: A fuse is always connected in series with the live wire. If a fault occurs and excessive current flows, the fuse melts and disconnects the live wire, cutting off power to the appliance so it remains safe to touch.
In simple words: Always put the fuse on the live wire to cut off incoming high voltage during a short circuit.

Q.11 What is the resistance of an air gap?
Answer: The resistance of an air gap is infinite (or extremely high), as air is an excellent insulator under normal conditions.
In simple words: Air does not conduct electricity, so its resistance is infinite.

Exam Tip: Air breaks down only at very high voltages (causing sparks/lightning), otherwise it acts as a perfect insulator.

Q.12 Name two safety measures commonly used in electric circuits and appliances.
Answer:
1. Use of electric fuses.
2. Earthing of metallic appliances.
In simple words: Fuses protect against overloading, and earthing prevents electric shocks.

Exam Tip: Mentioning fuses and earthing covers both overcurrent protection and shock prevention.

Q.13 Two metallic wires A and B are connected in parallel. Wire A has length \(l\) and radius \(r\), wire B has a length \(2l\) and radius \(2r\). Compute the ratio of the total resistance of parallel combination and the resistance of wire A.
Answer:
Resistance of wire A: \(R_A = \rho \frac{l}{\pi r^2}\)
Resistance of wire B: \(R_B = \rho \frac{2l}{\pi (2r)^2} = \rho \frac{2l}{4\pi r^2} = \frac{1}{2} \left(\rho \frac{l}{\pi r^2}\right) = \frac{R_A}{2}\)
When connected in parallel, equivalent resistance \(R_p\):
\(\frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} = \frac{1}{R_A} + \frac{1}{\frac{R_A}{2}} = \frac{1}{R_A} + \frac{2}{R_A} = \frac{3}{R_A}\)
\(R_p = \frac{R_A}{3}\)
Ratio of total parallel resistance to resistance of wire A: \(\frac{R_p}{R_A} = \frac{1}{3}\).
In simple words: Wire B has half the resistance of wire A, and their parallel combination gives \(1/3\) of wire A's resistance.

Exam Tip: Simplify \(R_B\) first in terms of \(R_A\) before applying the parallel formula.

Q.14 What is the meaning of the term 'frequency' of an alternating current? What is its value in India? Why is an alternating current considered to be advantageous over direct current for long-range transmission of electric energy?
Answer:
- Frequency of AC is the number of complete cycles completed by the alternating current in one second.
- Its value in India is \(50\text{ Hz}\).
- AC is advantageous over DC for long-range transmission because AC can be easily stepped up or stepped down using transformers with very little loss of electrical energy over long distances.
In simple words: Frequency is how fast AC alternates per second (50 Hz in India), and AC is preferred because it travels long distances without major power loss using transformers.

Exam Tip: Mention transformers and low energy loss to explain why AC is preferred for transmission.

Q.15 A TV set picture tube shoots out a beam of electrons. The current due to this beam is \(10\text{ mA}\). How many electrons will strike the TV screen every second?
Answer:
Given, current \(I = 10\text{ mA} = 10 \times 10^{-3}\text{ A} = 10^{-2}\text{ A}\)
Time \(t = 1\text{ s}\)
Total charge \(Q = I \times t = 10^{-2}\text{ A} \times 1\text{ s} = 10^{-2}\text{ C}\)
We know \(Q = ne\), where charge on one electron \(e = 1.6 \times 10^{-19}\text{ C}\)
Number of electrons \(n = \frac{Q}{e} = \frac{10^{-2}}{1.6 \times 10^{-19}} = 6.25 \times 10^{16}\)
In simple words: Use \(Q = It\) to find total charge, then divide by the charge of a single electron to find the total number of electrons.

Exam Tip: Remember standard electronic charge \(e = 1.6 \times 10^{-19}\text{ C}\).

Q.16 An electric wire is stretched to increase its length by \(25\%\). By what \(\%\) will the resistance be increased and what will be increase in its resistivity?
Answer:
Let original length be \(l_1 = l\) and new length \(l_2 = l + 0.25l = 1.25l\).
Since volume remains constant (\(A_1l_1 = A_2l_2\)):
\(A_1l = A_2(1.25l) \implies A_2 = \frac{A_1}{1.25} = 0.8 A_1\)
New resistance \(R_2 = \rho \frac{l_2}{A_2} = \rho \frac{1.25l}{0.8 A_1} = 1.5625 \left(\rho \frac{l}{A_1}\right) = 1.5625 R_1\)
Percentage increase in resistance = \(\frac{R_2 - R_1}{R_1} \times 100 = (1.5625 - 1) \times 100 = 56.25\%\)
The increase in resistivity is \(0\%\) because resistivity depends only on the material.
In simple words: A 25% increase in length increases the resistance by 56.25%, while resistivity remains completely unchanged.

Exam Tip: Handle percentage increases carefully by setting up proper ratio equations.

Q.17 An electric iron of resistance \(20\,\Omega\) takes a current of \(5\text{ A}\). Calculate the heat developed in \(30\text{ sec}\).
Answer:
Given, resistance \(R = 20\,\Omega\)
Current \(I = 5\text{ A}\)
Time \(t = 30\text{ s}\)
Heat developed \(H = I^2Rt\)
\(H = (5)^2 \times 20 \times 30\)
\(H = 25 \times 20 \times 30 = 15,000\text{ J} = 15\text{ kJ}\)
In simple words: Substitute the given values into Joule's heating formula \(H = I^2Rt\).

Exam Tip: Double-check units and write the final answer in Joules or kilojoules clearly.

Q.18 A \(60\text{ W}\) electric lamp gives off energy in the form of light at the rate of \(7.5\text{ J/s}\). What percentage of energy does the lamp transform into light?
Answer:
Total electrical power consumed = \(60\text{ W} = 60\text{ J/s}\)
Power emitted as light = \(7.5\text{ J/s}\)
Percentage of energy transformed into light = \(\frac{\text{Light Energy}}{\text{Total Energy}} \times 100\)
\(= \frac{7.5}{60} \times 100 = \frac{750}{60} = 12.5\%\)
In simple words: Divide the light energy output by the total electrical power input and multiply by 100.

Q.19 The voltage-current variation of two metallic wires A and B at constant temperature are shown in fig. Assuming that the wires have the same length and same diameter, explain which of the two wires will have larger resistivity?
Answer:
In the given graph (with current \(I\) on the Y-axis and voltage \(V\) on the X-axis or vice-versa), resistance is related to the slope. For a \(V-I\) graph, \(R = \frac{V}{I}\).
Wire A has a steeper slope than wire B, meaning wire A has higher resistance (\(R_A > R_B\)).
Since \(R = \rho \frac{l}{A}\) and both wires have the same length and diameter (same area), resistance is directly proportional to resistivity (\(R \propto \rho\)).
Therefore, wire A will have larger resistivity.
In simple words: The wire with the higher resistance (steeper slope) has the higher resistivity.

Exam Tip: Always identify the axes correctly before determining resistance from graph slopes.

Q.20 You are given following current-time graphs from two different sources:
(i) Name the type of current in two cases.
(ii) Identify any one source for each type of current.
(iii) What is the frequency of current in case II in India?
Answer:
(i) Case I: Direct Current (DC); Case II: Alternating Current (AC).
(ii) Source for DC: Cell or Battery; Source for AC: AC Generator (or Mains supply).
(iii) The frequency of alternating current in India is \(50\text{ Hz}\).
In simple words: Constant unidirectional current is DC (from batteries), while reversing current is AC (from AC generators with 50 Hz frequency in India).

Exam Tip: Straight line parallel to time axis = DC; wave graph crossing zero = AC.

Q.21 The electric power consumed by a device may be calculated by using either of the two expressions \(P = I^2R\) or \(P = V^2/R\). The first expression indicates that it is directly proportional to \(R\) whereas the second expression indicates inverse proportionality. How can the seemingly different dependence of P on R in these expressions be explained?
Answer: The difference arises from the operating conditions:
- In \(P = I^2R\), the expression is used when appliances are connected in series, where current \(I\) remains constant through each resistor. Thus, \(P \propto R\).
- In \(P = \frac{V^2}{R}\), the expression is used when appliances are connected in parallel, where potential difference \(V\) remains constant across each resistor. Thus, \(P \propto \frac{1}{R}\).
In simple words: Power increases with resistance in series circuits (constant current), but decreases with resistance in parallel circuits (constant voltage).

Exam Tip: This is a crucial conceptual question - always mention whether current or voltage is held constant.

Q.22 Draw a schematic diagram of a circuit containing the following electrical components: (a) a resistance (b) a voltmeter (c) an electric bulb (d) a cell (e) an ammeter and (f) plug key
Answer:
Description of the circuit schematic diagram:
1. A cell is connected in series with a plug key, an ammeter, a resistor, and an electric bulb.
2. A voltmeter is connected in parallel across the resistor.
In simple words: Connect the cell, key, ammeter, bulb, and resistor in a single loop (series), and place the voltmeter across the resistor in parallel.

Exam Tip: Ensure the ammeter is in series and the voltmeter is in parallel with correct polarity markings (+ and -).