An electric bulb rated as 220 W, 110 V is used in a circuit having a 220 V supply. For the bulb to deliver 220 W, the resistance to be added in the series is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 55\,\Omega \)
The bulb delivers its rated power of 220 W only when it carries its rated current. The extra resistance must drop the surplus voltage, so it is connected in series with the bulb.
\( \displaystyle I = \frac{P}{V} = \frac{220\,\mathrm{W}}{110\,\mathrm{V}} = 2\,\mathrm{A} \)
The voltage across the bulb stays 110 V, so the series resistor must take the remaining \( 220 - 110 = 110\,\mathrm{V} \) at the same current of \( 2\,\mathrm{A} \).
\( \displaystyle R = \frac{110\,\mathrm{V}}{2\,\mathrm{A}} = 55\,\Omega \)