Question

A $50$ W, $100$ V lamp is to be connected to an AC mains of $200$V, $50$Hz. What capacitance is essential to be put in series with the lamp?

Answer 1

$V_C=$ Voltage across capacitor

$V_L=$ Voltage across lamp

Current $I$ is

$Power=V_{L}I$

$\Rightarrow 50=100I$

$I=0.5$ $A$

$V_C=\sqrt{V^2-V_L^2}=\sqrt{{200}^2-{100}^2}=173.2$

$V_C=173.2$ $V$

$i{X}_C=173.2$

$\Rightarrow X_C=\frac{173.2}{0.5}=2\times 173.2$

$\Rightarrow \frac{1}{2\pi fC}=2\times 173.2$

$\Rightarrow C=\frac{1}{2\pi \times 50\times 2\times 173.2}$

$C=9.1\times {10}^{-6}$ $F=9.1$ $\mu F$