A 220V, 50Hz AC source is connected to a 25V, 5W lamp and an...

Resistance of the bulb can be calculated as,

$P = \dfrac{{V}^{2}}{{R}_{B}} ⇒ {R}_{B} = \dfrac{{V}^{2}}{P} = \dfrac{{25}^{2}}{5} = 125Ω$

The current through the bulb for peak brightness should be,

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

Now, ${i}_{rms} = \dfrac{1}{5} = \dfrac{220}{{R}_{B} + R}$

$RB + R = 1100$

$R = 1100 - 125 = 975$