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

Question

TestHub · Accepted Answer

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

$R B + R = 1100$

$R = 1100 - 125 = 975$

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