Question

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

Answer 1

Answer: 3

$P=\frac{V^2}{R}\Rightarrow R=\frac{V^2}{P}$

$R=\frac{100\times 10^2}{50}=R=200\Omega$
$V_R^2+V_C^2=V^2$
$(100{)}^2+V_C^2=(200{)}^2$
$i=\frac{100}{200}=\frac{1}{2};V^2=40000$
$V=I\times X_C;V_C^2=30000$
$V_C=100\sqrt{3}$
$X_C=200\sqrt{3}$
$200\sqrt{3}=\frac{1}{\omega C}$
$C=\frac{1}{20\times 50\times 20\sqrt{3}}=\frac{50\times 10^{-6}}{\sqrt{x}}$
$\sqrt{x}=50\times 10^{-6}\times 100\times 200\sqrt{3}$
$X=3$