In India electricity is supplied for domestic use at 220 V. It is supplied at 110 V in USA. If the resistance

I. Multiple Choice Questions · Q8

Q. 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

(a) R

(b) 2R

(c)

\dfrac{R}{4}

(d)

\dfrac{R}{2}

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Step 1. The India bulb has resistance $R=(220)^2/60$ (from $R=V^2/P$ ).

Step 2. The USA bulb, rated for the same power 60 W but at 110 V, has resistance $R_{USA}=(110)^2/60$ .

Step 3. Taking the ratio, $\dfrac{R_{USA}}{R}=\dfrac{(110)^2}{(220)^2}=\left(\dfrac{110}{220}\right)^2=\left(\dfrac12\right)^2=\dfrac14$ .

Step 4. So $R_{USA}=R/4$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set A1 markMCQ
$Q. Which of the following expressions represents electrical energy? (A) V.I.t (B) V.I (C) I.R^2.t (D) V^2/R$
›Reveal solutionSolution

Electrical energy W = VIt (power VI multiplied by time t).

Electrical power dissipated is $P=VI$ . Energy is power multiplied by the time for which it flows:

$W=P\,t=V\,I\,t$

…

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CBSE 2025Set D1 markMCQ
$Q. Kilowatt-hour (kWh) is the unit of (A) energy (B) power (C) torque (D) force$
›Reveal solutionSolution

kWh = power × time = energy; 1 kWh = 3.6 × 10⁶ J.

The kilowatt-hour is the product of a unit of power and a unit of time:

$1\,kWh = 1\,kW \times 1\,hour = 1000\,W \times 3600\,s = 3.6\times10^{6}\,J$

…

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CBSE 2025Set D1 markMCQ
$Q. Power of electric circuit is (A) V.R (B) V^2.R (C) V^2/R (D) V^2.R.I$
›Reveal solutionSolution

Power dissipated in a resistor is P = V²/R.

Electrical power is the rate of doing work, P = VI. Using Ohm's law V = IR we can write it three equivalent ways:

$P = VI = I^{2}R = \frac{V^{2}}{R}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A 100 watt electric bulb is connected to a 220 volt electric source. The resistance of the bulb's filament is$
(a) 484 ohm
(b) 100 ohm
(c) 22000 ohm
(d) 242 ohm
›Reveal solutionSolution

For a resistive load, power P = V^2/R, so the filament resistance is R = V^2/P.

Given V = 220 V and P = 100 W:

R = V^2 / P = (220)^2 / 100 = 48400 / 100 = 484 ohm

…

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CBSE 2025Set ANNUAL1 mark
$Q. A current of 5.0 A flows through a press of resistance 11Ω. Calculate the energy consumed by the press in 5 minutes.$
›Reveal solutionSolution

Using W = I²Rt with t converted to seconds gives the energy dissipated as heat in the press: 82.5 kJ.

Given: I = 5.0 A, R = 11 Ω, t = 5 min = 5 × 60 = 300 s

Energy consumed (heat dissipated):

$W = I^2 R t = (5.0)^2 \times 11 \times 300$

$W = 25 \times 11 \times 300 = 275 \times 300 = 82500\ \text{J}$

…

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CBSE 2024Set A1 markMCQ
$Q. Two bulbs of 40 W and 60 W are connected to 220 V source. The ratio of their resistances will be (A) 4 : 3 (B) 3 : 4 (C) 2 : 3 (D) 3 : 2$
›Reveal solutionSolution

At the same voltage R = V²/P, so R ∝ 1/P → R₄₀:R₆₀ = 60:40 = 3:2.

The power rating is defined at the operating voltage, so $R = \dfrac{V^2}{P}$ . Since both bulbs use the same 220 V, resistance is inversely proportional to power.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. 1 mA current is flowing through a conductor of 2 kΩ resistance. How much power is lost in it ?$
(a) 0.2 W
(b) 2 mW
(c) 2 W
(d) 2 kW
›Reveal solutionSolution

Use P = I²R with I in amperes.

Given: I = 1 mA = 1 × 10⁻³ A, R = 2 kΩ = 2 × 10³ Ω.

Power dissipated (Joule heating) in a resistor carrying current I is:

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Calculate the number of kWh in one joule :$
(a) 6.3 × 10⁶ kWh
(b) 2.8 × 10⁻⁷ kWh
(c) 3.6 × 10⁶ kWh
(d) 3.6 × 10³ kWh
›Reveal solutionSolution

Convert using 1 kWh = 3.6 × 10⁶ J, then invert.

One kilowatt-hour is the energy consumed by a 1 kW appliance running for 1 hour:

$1\ \text{kWh} = 1000\ \text{W} \times 3600\ \text{s} = 3.6\times 10^{6}\ \text{J}$

So to express 1 joule in kWh, invert this relation: …

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CBSE 2024Set ANNUAL1 markMCQ
$\Omega$
(a) 240 W
(b) 400 W
(c) 480 W
(d) 2 W
›Reveal solutionSolution

Using $P=V^2/R$ with $V=240$ V and $R=120\,\Omega$ gives $P=480$ W.

Working

For a resistive appliance operating at a fixed voltage $V$ with resistance $R$ , the power dissipated is

$P = \dfrac{V^2}{R}$

Substituting $V=240$ V, $R=120\,\Omega$ : …

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CBSE 2023Set F1 markMCQ
$Q. Two resistors R and 2R are connected in series in an electric circuit. The thermal energy developed in R and 2R will be in ratio (A) 1:2 (B) 1:4 (C) 2:1 (D) 4:1$
›Reveal solutionSolution

Series → same current → heat $\propto R$ → ratio 1:2.

In a series circuit the same current $I$ flows through both resistors. The thermal (Joule) power dissipated is $P = I^2 R$ , so with equal $I$ the heat produced is directly proportional to the res …

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CBSE 2023Set ANNUAL1 mark
$Q. Power of 60 W is being supplied to an electrical appliance, under a potential difference of 240 V. What is the current flowing through the appliance? OR A heating element connected to a 12 V battery draws a current of 5 A. How much electric power is supplied?$
›Reveal solutionSolution

Using $P=VI$ , the current drawn by the appliance is 0.25 A; in the alternative part, the power supplied by a 12 V battery driving 5 A is 60 W.

Solution:

Electrical power: $P = VI$

Given $P = 60\text{ W}$ , $V = 240\text{ V}$ :

$I = \frac{P}{V} = \frac{60}{240} = 0.25\text{ A}$

Alternative (Or):

…

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CBSE 2022Set GC1 mark
$Q. In a house 200 bulbs each of 60 W are connected in parallel with the mains of 200 V. Calculate the consumption of electrical energy per day (24 hrs).$
›Reveal solutionSolution

$200$ bulbs $\times\,60$ W $=12$ kW; over $24$ h that is $12\times24=288$ kWh $=288$ units.

Bulbs in parallel each receive the full mains voltage, so each dissipates its rated $60$ W. Total power:

$P=200\times60=12000\text{ W}=12\text{ kW}.$

Energy consumed in one day ( $t=24$ h): …

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Q.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

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Q.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

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