Q: Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 V and the average resistance per km is 0.5 Ω. The power loss in the wire is

19.2 kW. Total potential drop = 150 \times 8 = 1200 V; total resistance = 150 \times 0.5 = 75 Ω. P = \frac{V^2}{R} = \frac{1200^2}{75} = 19200 W = 19.2 kW

Q: An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? (specific heat of water = 4200 J/kg·°C)

6.3 min. Vit = ms\Delta t \Rightarrow t = \frac{1 \times 4200 \times 80}{220 \times 4} = 381.8 s \approx 6.3 min

Q: A current of 3 A flows through the 2 Ω resistor in a circuit where the 2 Ω resistor is in series with a parallel combination of a 4 Ω resistor and a (1 Ω + 5 Ω) series branch. The power dissipated in the 5 Ω resistor is

5 W. Voltage across the 2 Ω = 3 \times 2 = 6 V, which also appears across the 4 Ω branch and the (1+5) Ω branch. Current in the (1+5) branch = \frac{6}{6} = 1 A. P_{5\Omega} = i^2R = 1^2 \times 5 = 5 W

Question 1

The charge flowing through a resistance R varies with time t as Q = at - bt^2, where a and b are positive constants. The total heat produced in R is

Accepted Answer

\frac{a^3R}{6b}. Current I = \frac{dQ}{dt} = a - 2bt, which is zero at t = \frac{a}{2b}.

H = \int_0^{a/2b} I^2R\,dt = R\int_0^{a/2b}(a - 2bt)^2 dt = \frac{a^3R}{6b}

Question 2

A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R is

Accepted Answer

26 Ω. Bulb current i = \frac{500}{100} = 5 A; bulb resistance = \frac{100}{5} = 20 Ω.

Total needed \frac{230}{5} = 46 Ω, so R = 46 - 20 = 26 Ω

Question 3

Two cities are 150 km apart. Electric power is sent from one city to another city through copper wires. The fall of potential per km is 8 V and the average resistance per km is 0.5 Ω. The power loss in the wire is

Accepted Answer

19.2 kW. Total potential drop = 150 	imes 8 = 1200 V; total resistance = 150 	imes 0.5 = 75 Ω.

P = \frac{V^2}{R} = \frac{1200^2}{75} = 19200 W = 19.2 kW

Question 4

If voltage across a bulb rated 220 V-100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is

Accepted Answer

5%. P = \frac{V^2}{R}, so \frac{\Delta P}{P} = 2\frac{\Delta V}{V} = 2 	imes 2.5\% = 5\%

Question 5

In a circuit, a 9 Ω resistor is in parallel with a 6 Ω resistor, and this parallel combination is in series with a 2 Ω resistor. If the power dissipated in the 9 Ω resistor is 36 W, the potential difference across the 2 Ω resistor is

Accepted Answer

10 V. Current in 9 Ω: i_1 = \sqrt{\frac{36}{9}} = 2 A; voltage across the parallel section = 2 	imes 9 = 18 V, so current in 6 Ω = \frac{18}{6} = 3 A. Total I = 2 + 3 = 5 A.

V_{2\Omega} = 5 	imes 2 = 10 V

Question 6

An electric kettle takes 4 A current at 220 V. How much time will it take to boil 1 kg of water from temperature 20°C? (specific heat of water = 4200 J/kg·°C)

Accepted Answer

6.3 min. Vit = ms\Delta t \Rightarrow t = \frac{1 	imes 4200 	imes 80}{220 	imes 4} = 381.8 s \approx 6.3 min

Question 7

A current of 3 A flows through the 2 Ω resistor in a circuit where the 2 Ω resistor is in series with a parallel combination of a 4 Ω resistor and a (1 Ω + 5 Ω) series branch. The power dissipated in the 5 Ω resistor is

Accepted Answer

5 W. Voltage across the 2 Ω = 3 	imes 2 = 6 V, which also appears across the 4 Ω branch and the (1+5) Ω branch. Current in the (1+5) branch = \frac{6}{6} = 1 A.

P_{5\Omega} = i^2R = 1^2 	imes 5 = 5 W

Question 8

In a circuit a 6 Ω and a 3 Ω resistor are in parallel, and this combination is in series with a 4 Ω resistor across an 18 V supply. The total power dissipated in the circuit is (in watts)

Accepted Answer

54. 6 Ω ∥ 3 Ω = 2 Ω; total R = 2 + 4 = 6 Ω.

P = \frac{V^2}{R} = \frac{18^2}{6} = 54 W

Question 9

In a circuit, a 1 Ω and a 3 Ω resistor are in series (total 4 Ω), and this branch is in parallel with an 8 Ω resistor. Power dissipated across the 8 Ω resistor is 2 W. The power dissipated across the 3 Ω resistor is (in watts)

Accepted Answer

3.0. Same voltage across the 8 Ω and the (1+3) = 4 Ω branch: 8i_2 = 4i_1, so i_1 = 2i_2. From P_8 = i_2^2 	imes 8 = 2 W, i_2^2 = 0.25.

P_{3\Omega} = i_1^2 	imes 3 = (2i_2)^2 	imes 3 = 12i_2^2 = 12 	imes 0.25 = 3 W

Question 10

A 5 A fuse wire can withstand a maximum power of 1 W in a circuit. The resistance of the fuse wire is

Accepted Answer

0.04 Ω. R = \frac{P}{i^2} = \frac{1}{5^2} = 0.04 Ω

Question 11

In India, electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance of a 60 W bulb for use in India is R, the resistance of a 60 W bulb for use in USA will be

Accepted Answer

\frac{R}{4}. R = \frac{V^2}{P} at the same power: \frac{R_{USA}}{R} = \frac{110^2}{220^2} = \frac{1}{4}, so R_{USA} = \frac{R}{4}

Question 12

When three identical bulbs of 60 W, 200 V rating are connected in series to a 200 V supply, the power drawn by them will be

Accepted Answer

20 W. In series the total resistance triples, so the power drawn is one-third of a single bulb's rated power at that voltage: P = \frac{60}{3} = 20 W

Question 13

Two 220 V, 100 W bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 V AC supply line. The power drawn by the combination in each case respectively will be

Accepted Answer

50 W, 200 W. Series: P_{eq} = \frac{P_1P_2}{P_1 + P_2} = \frac{100 	imes 100}{200} = 50 W. Parallel: P_{eq} = P_1 + P_2 = 200 W

Question 14

An electric kettle has two heating coils. When one of the coils is connected to an AC source, the water in the kettle boils in 10 min. When the other coil is used the water boils in 40 min. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be

Accepted Answer

8 min. For parallel coils the times combine as \frac{1}{t} = \frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{10} + \frac{1}{40}, giving t = 8 min

Question 15

Fuse wire is a wire of

Accepted Answer

high resistance and low melting point. A fuse wire is made of a material of high resistance and low melting point so that it heats up and melts quickly when excessive current flows, breaking the circuit.

Question 16

Two bulbs 25 W, 220 V and 100 W, 220 V are given. Which has higher resistance?

Accepted Answer

25 W bulb. R = \frac{V^2}{P}; for the same voltage, R \propto \frac{1}{P}. The lower-power (25 W) bulb has the higher resistance.

Question 17

A 5°C rise in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately

Accepted Answer

20°C. Heat \propto i^2, so \Delta	heta \propto i^2. Doubling the current gives \Delta	heta_2 = 2^2 	imes 5 = 20°C

Question 18

Three equal resistors connected in series across a source of emf together dissipate 10 W of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of emf?

Accepted Answer

90. Series R_s = 3R; parallel R_p = \frac{R}{3}. Since P = \frac{V^2}{R}, \frac{P_p}{P_s} = \frac{R_s}{R_p} = 9, so P_p = 9 	imes 10 = 90 W

Question 19

A 100 W, 200 V bulb is connected to a 160 V power supply. The power consumption would be

Accepted Answer

64 W. P \propto V^2 (R constant): P' = \left(\frac{160}{200}ight)^2 	imes 100 = 0.64 	imes 100 = 64 W

Question 20

A heating coil is labelled 100 W, 220 V. The coil is cut into two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is

Accepted Answer

400 J. Cutting in half then joining in parallel reduces the resistance to \frac{1}{4}. Since P \propto \frac{1}{R} at constant voltage, power becomes 4×: 4 	imes 100 = 400 W (400 J/s).

Heating Effect of Current — NEET Physics PYQ MCQs with Solutions

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