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 $V=200\sqrt{2}\sin \left(100t\right)$ volts. The phase difference between $I_1$ and $I_2$ is:

• A. ${30}^o$
• B. $0^o$
• C. ${150}^o$
• D. ${60}^o$

Answer 1

The correct option is C ${150}^o$

$x_e=\frac{1}{{\omega }_c}=\frac{4}{{10}^{-6}\times \sqrt{3}\times 100}=\frac{2\times {10}^4}{\sqrt{3}}$
$\tan \frac{\theta }{2}\frac{x_e}{R_e}=\frac{{10}^3}{\sqrt{3}}$
${\theta }_1$ is close to $90$
For $L-R$ circuit
$x_L=w_L=100\times \frac{\sqrt{3}}{10}=\sqrt{3}$
$R_1=10$
$\tan {\theta }_2=\frac{x_e}{R}$
$\tan {\theta }_2=\sqrt{3}$
${\theta }_2=60$
So phase difference comes out $90+60=150$