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

Correct Answer: 3

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