An electric bulb is rated 220 V, 100 W. Assuming that its resistance…

The correct answer is Option (4) → (A), (B) and (D) only

The rating of a bulb gives the power it consumes when it works at its rated voltage. With the resistance taken as constant, the power on any supply follows from \( P=V^{2}/R \), and the current from Ohm’s law \( I=V/R \).

(A) The resistance of the bulb is \( 484\,\Omega \). — True. From the rated values:

    \( \begin{aligned} R &= \frac{V^{2}}{P} \\ &= \frac{(220\,\mathrm{V})^{2}}{100\,\mathrm{W}} \\ &= 484\,\Omega \end{aligned} \)

(B) At its rated voltage the bulb draws a current of about 0.45 A. — True. Using \( P=IV \):

    \( \begin{aligned} I &= \frac{P}{V} \\ &= \frac{100\,\mathrm{W}}{220\,\mathrm{V}} \\ &\approx 0.45\,\mathrm{A} \end{aligned} \)

(C) On a 110 V supply the bulb consumes 50 W. — False. The power varies as the square of the voltage. So half the voltage gives one-fourth the power:

    \( \begin{aligned} P' &= \frac{(110\,\mathrm{V})^{2}}{484\,\Omega} \\ &= 25\,\mathrm{W} \end{aligned} \)

(D) On a 110 V supply the current through the bulb is half of its value on a 220 V supply. — True. For a fixed resistance the current is proportional to the voltage. On 110 V it is \( 110/484\approx0.23\,\mathrm{A} \), half of the 0.45 A drawn on 220 V.

NCERT (2026-27):
Electrical Energy, Power – “… The energy dissipated per unit time is the power dissipated P = ΔW/Δt and we have,”

    \( P=IV \)\( \qquad (3.32) \)

“Using Ohm’s law V = IR, we get”

    \( P=I^{2}R= \)\( V^{2}/R \)\( \qquad (3.33) \)

“as the power loss (‘ohmic loss’) in a conductor of resistance R carrying a current I. It is this power which heats up, for example, the coil of an electric bulb to incandescence, radiating out heat and light.”