A 20 watt, 50 V lamp is to be connected to AC mains of 260 V, 50 Hz. Calculate the value of series capacitor to run the lamp.

4.9 µF

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,

$X_C=\frac{1}{2πfc}$

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

$≃4.9µF$