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 Ω
Given:
Bulb rating: $P = 220 \, W$, $V_{rated} = 110 \, V$
Supply voltage: $V_s = 220 \, V$
Bulb resistance: $R_b = \frac{V_{rated}^2}{P} = \frac{110^2}{220} = 55 \, \Omega$
Let $R_s$ be the series resistance to limit voltage across bulb to 110 V.
Total current in series: $I = \frac{V_{rated}}{R_b} = \frac{110}{55} = 2 \, A$
Total series resistance for supply voltage: $R_{total} = \frac{V_s}{I} = \frac{220}{2} = 110 \, \Omega$
Series resistance needed: $R_s = R_{total} - R_b = 110 - 55 = 55 \, \Omega$
Final Answer: $R_s = 55 \, \Omega$