Question

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

Answer 1

We have, $X_c=\frac{1}{\omega c}=\frac{\pi \sqrt{x}}{2\pi \times 50\times 50}\times {10}^6...\left(i\right)$
Also,
$v_R^2+v_c^2=(200{)}^2$
$\Rightarrow v_c^2={200}^2-{100}^2$
$\Rightarrow v_c=100\sqrt{3}\text{V}$
and, $v_R=100\text{V}$
According to the question
$P=50=\frac{V^2}{R}\Rightarrow R=\frac{100\times 100}{50}=200\Omega$
Now we have,
$i_{rms}=\frac{50}{100}=\frac{1}{2}\text{A}$
$\Rightarrow v_c=\frac{1}{2}\times X_c=100\sqrt{3}$ $\Rightarrow {10}^6\times \frac{\sqrt{x}}{5000}\times \frac{1}{2}=100\sqrt{3}$
$\Rightarrow \frac{{10}^6\sqrt{x}}{10000\times 100}=\sqrt{3}$
$\Rightarrow \sqrt{x}=\sqrt{3}$
$\Rightarrow x=3$