Q.A toaster operating at 240 V has a resistance of 120 $\Omega$. Its power is

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:

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

   Using Ohm's law ($V=IR$) this can also be written $W=I^2Rt$ or $W=\dfrac{V^2}{R}t$ — note these other forms all carry a factor of time $t$. Among the options, only V·I·t correctly has the dimensions of energy (the others V·I, V²/R are powers, and I·R²·t is dimensionally wrong).

   ✓Final answer

   (A) V·I·t.

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:

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

   Since energy = power × time, the kWh is a (commercial) unit of energy, commonly called one 'unit' on an electricity bill. It is not a unit of power, torque or force.

   ✓Final answer

   (A) energy.

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:

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

   Of the options, only V²/R is correct (it is dimensionally power: V²/Ω = W). Options V·R, V²·R and V²·R·I are dimensionally wrong for power.

   ✓Final answer

   (C) V²/R.

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

   This is the resistance of the glowing filament at its rated operating temperature; note that a cold filament has a much lower resistance, which is why bulbs draw a large inrush current the instant they are switched on.

   ✓Final answer

   (a) 484 ohm.

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):

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

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

   $W = 8.25 \times 10^4\ \text{J} = 82.5\ \text{kJ}$

   ✓Final answer

   The energy consumed by the press in 5 minutes is $8.25 \times 10^4\ \text{J}$ (82.5 kJ).

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.

   $\frac{R_{40}}{R_{60}} = \frac{P_{60}}{P_{40}} = \frac{60}{40} = \frac{3}{2}$

   So the lower-wattage (40 W) bulb has the higher resistance.

   ✓Final answer

   (D) 3 : 2.

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:

   P = I²R = (1 × 10⁻³)² × (2 × 10³) = (1 × 10⁻⁶) × (2 × 10³) = 2 × 10⁻³ W

   ✓Final answer

   P = 2 × 10⁻³ W = 2 mW — option (b).

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:

   $1\ \text{J} = \dfrac{1}{3.6\times 10^{6}}\ \text{kWh} = 2.78\times 10^{-7}\ \text{kWh} \approx 2.8\times 10^{-7}\ \text{kWh}$

   ✓Final answer

   1 J = 2.8 × 10⁻⁷ kWh — option (b).

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, $R=120\,\Omega$:

   $P = \dfrac{240^2}{120} = \dfrac{57600}{120} = 480\ \text{W}$

   ✓Final answer

   The correct option is (c): $P = 480$ W

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 resistance:

    $\frac{H_R}{H_{2R}} = \frac{I^2 R}{I^2 (2R)} = \frac{R}{2R} = \frac{1}{2}.$

    ✓Final answer

    (A) 1:2.

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):

    Given $V = 12\text{ V}$, $I = 5\text{ A}$:

    $P = VI = 12 \times 5 = 60\text{ W}$

    ✓Final answer

    Main: $I = 0.25\text{ A}$. Alternative: $P = 60\text{ W}$.

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):

    $E=P\,t=12\text{ kW}\times24\text{ h}=288\text{ kWh}.$

    That is 288 units of electrical energy per day.

    ✓Final answer

    $E=288$ kWh (288 units) per day.