Question

In the above circuit, $C=\frac{\sqrt{3}}{2}\mu F.R_2=20\Omega ,L=\frac{\sqrt{3}}{10}H$ and $R_1=10\Omega$ Current in $L-R_1$, path is $I_1$ and in C-R${}_2$path it is $I_2$ The voltage of A.C source is given by $V=200\sqrt{2}$ sin (100t) volts. the phase difference between $I_1$ and $I_2$is :

• A. $30^{\circ }$
• B. $0^{\circ }$
• C. $90^{\circ }$
• D. $60^{\circ }$

Answer 1

Answer: (C)

$x_e=\frac{1}{\omega }=\frac{4}{10^{-6}\times \sqrt{3}\times 100}=\frac{2\times 10^4}{\sqrt{3}}$
$tan\theta /2\frac{x_e}{R_e}=\frac{10^3}{\sqrt{3}}$
${\theta }_1$is close to $90$
For L-R circuit
$x_L=w_L=100\times \frac{\sqrt{3}}{10}=\sqrt{3}$
tan${\theta }_2=\frac{x_e}{R}$
tan${\theta }_2=\sqrt{3}$
${\theta }_2=60$
So phase difference comes out $90+60=150$.
Therefore Ans. is Bonus
If $R_2$ is 20 K$\Omega$
then phase difference comes out to be $60+30$
= $90^{\circ }$