Q88 / 90 · JEE Main 2021 (25 Jul Shift 1) · Physics

An electric bulb rated as $200 W$ at $100 V$ is used in a circuit having $200 V$ supply. The resistance $R$ that must be put in series with the bulb so that the bulb delivers the same power is ______ $Ω$ .

Correct answer: 50

Solution

Power, $P = \frac{V^{2}}{R_{B}}$

$R_{B} = \frac{V^{2}}{P} = \frac{100 \times 100}{200}$

$R_{B} = 50 Ω$

To produce same power, same voltage (i.e., $100 V$ ) should be across the bulb.

Hence, $R = R_{B}$

$R = 50 Ω$

Hence, the correct answer is 50.