Q82 / 90 · JEE Main 2022 (27 Jun Shift 1) · Physics

A $220 V, 50 Hz$ AC source is connected to a $25 V, 5 W$ lamp and an additional resistance $R$ in series (as shown in figure) to run the lamp at its peak brightness, then the value of $R$ (in ohm) will be

Correct answer: 975

Solution

Resistance of the bulb can be calculated as,

$P = \frac{V^{2}}{R_{B}} \Rightarrow R_{B} = \frac{V^{2}}{P} = \frac{25^{2}}{5} = 125 Ω$

The current through the bulb for peak brightness should be,

$i = \frac{25}{125} = \frac{1}{5} A$

Now, $i_{rms} = \frac{1}{5} = \frac{220}{R_{B} + R}$

$R B + R = 1100$

$R = 1100 - 125 = 975$

Hence, the correct answer is 975.