Question

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 _____ .

Answer 1

$R_b= \frac {(25)^2}{5} = 125 Ω$
$R_b= 125 Ω$
rms value of current.
$I_{rms} = \sqrt {\frac {5}{125}}= \frac 15A$
$I_{rms} = \frac 15A$

$⇒\frac {220}{R+125}=\frac 15$
$⇒ R = 1100 -125$
$⇒ R = 975 Ω$

So, the answer is $975 Ω$.