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 :

Showing the 12 most recent of 22 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.A toaster operating at 240 V has resistance of 120 $\Omega$. Its power is :

   (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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10. 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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11. 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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12. 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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