a
Resistance of bulb, $R_{B}=\frac{V^{2}}{P}=\frac{(100)^{2}}{500}=20\, \Omega$
Power of the bulb in the circuit,
$P =V I $
$I =\frac{P}{V_{B}} $
$=\frac{500}{100}=5\, \mathrm{A}$
$V_{R} =I R \Rightarrow(230-100)=5 \times R $
$\therefore \quad R =26 \,\Omega$
