The electric power consumed by a \( 220\,\mathrm{V} \), \( 60\,\mathrm{W} \) bulb, when operated at \( 110\,\mathrm{V} \), is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 15\,\mathrm{W} \)
The rating fixes the resistance of the filament, which is taken as constant; at a different voltage the power follows from \( P = V^{2}/R \).
\( \displaystyle R = \frac{V_{\mathrm{rated}}^{2}}{P_{\mathrm{rated}}} = \frac{(220)^{2}}{60} = 806.7\,\Omega \)
\( \displaystyle P = \frac{V^{2}}{R} = \frac{(110)^{2}\times60}{(220)^{2}} = 60\times\frac{1}{4} = 15\,\mathrm{W} \)
Shortcut: With R fixed, \( P \propto V^{2} \): halving the voltage makes the power one-fourth, \( 60/4 = 15\,\mathrm{W} \).