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.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.99 µF
The lamp must carry its rated current. The voltage across the lamp and the voltage across the capacitor are 90° out of phase, so they add as \( V^{2} = V_{R}^{2} + V_{C}^{2} \), not arithmetically. The capacitor takes up the remaining voltage without dissipating any power.
Rated current of the lamp
\( \displaystyle I = \frac{P}{V_{\mathrm{lamp}}} = \frac{20}{50} = 0.4\,\mathrm{A} \)
Voltage across the capacitor (supply 260 V, rms)
\( V_C = \sqrt{260^{2} - 50^{2}} = \sqrt{65100}\,\mathrm{V} \approx 255\,\mathrm{V} \)
Capacitive reactance
\( \displaystyle X_C = \frac{V_C}{I} = \frac{255}{0.4}\,\Omega \approx 638\,\Omega \)
Capacitance
\( \displaystyle X_C = \frac{1}{2\pi fC} \Rightarrow C = \frac{1}{2\pi fX_C} = \frac{1}{2\pi\times50\times638}\,\mathrm{F} \)
\( C \approx 4.99\times10^{-6}\,\mathrm{F} = 4.99\,\mu\mathrm{F} \)