Question

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

Answer 1

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

$ 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} $
$X =3$