A 20 watt, 50 V lamp is to be connected to AC mains of 260 V, 50 Hz.…

The correct answer is Option (4) → 4.9 µF

The Power P in an AC circuit is related to Voltage V by-

$P=V_{lamp}×I$

$I=\frac{P}{V_{lamp}}=\frac{20}{50}=0.4A$

$V_{AC}=I.Z_{total}$

$⇒Z_{total}=\frac{V_{AC}}{I}=\frac{210}{0.4}=650Ω$

The lamp's impedance $Z_{lamp}$ is -

$Z_{lamp}=\frac{V_{lamp}}{I}=\frac{50}{0.4}=125Ω$

$∴Z_{total}=\sqrt{{Z_{lamp}}^2+{X_C}^2}$

$650=\sqrt{125^2+{X_C}^2}$

$⇒X_C=\sqrt{406875}=638.4Ω$

and,

$X_C=\frac{1}{2πfc}$

$⇒c=\frac{1}{2πfX_C}=\frac{1}{2π×50×638.4}$

$≃4.9µF$