Question

To light, a W, 100 V lamp is connected, in series with a capacitor of capacitance
$\frac{50}{π\sqrt{x}}\mu F$ with $200 V$, 50 Hz AC source.
The value of x will be ___.

Answer 1

The correct answer is 3
$X_C=\frac{1}{wc}=\frac{π\sqrt{x}}{2π×50×50}×10^6$
$v^{2}_{R}+v^{2}_{C}=(200)^2$
$v^{2}_{C}=200^2–100^2$
$v_C=100\sqrt3V$
$v_R=100V$
$P=\frac{V^2}{R}$
$R=\frac{100×100}{50}=200 Ω$
$i_m=\frac{1}{2}A$
$\frac{1}{2}×x_C=100\sqrt3$
$⇒10^{−6}×\frac{\sqrt{x}}{5000}×\frac{1}{2}=100\sqrt3$
$\frac{10^{−6}\sqrt{x}}{10000×100}=\sqrt3$
$\sqrt{x} = \sqrt3$
x = 3
$\therefore$ value of x will be 3