Question

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

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

Answer 1

The correct option is D ${90}^{\circ }$

As the two branches are in parallel, the potential difference will be the same.

So, let us solve the question using phasor diagram, keeping voltage common for both the branches.
In $C-R_2$ path, Capacitance reactance,

$X_c=\frac{1}{\omega C}=\frac{2}{100\times {10}^{-6}\times \sqrt{3}}$

$X_c=\frac{2\times {10}^4}{\sqrt{3}}$

$\therefore \tan {\theta }_1=\frac{X_c}{R_2}=\frac{2\times {10}^4}{\sqrt{3}\times 2\times {10}^4}$

$\tan {\theta }_1=\frac{1}{\sqrt{3}}$

${\theta }_1={30}^{\circ }$


For $L-R_1$ path, Inductive reactance,
$X_L=\omega L=100\times \frac{\sqrt{3}}{10}=10\sqrt{3}\Omega$

Here, the currents lags behind the voltage by a phase difference of ${\theta }_2$

$R_1=10\Omega$
$\tan {\theta }_2=\frac{X_L}{R_1}$
$\tan {\theta }_2=\sqrt{3}\Rightarrow {\theta }_2={\tan }^{-1}\left(\sqrt{3}\right)$
$\Rightarrow {\theta }_2={60}^{\circ }$


Therefore, the phase difference between $I_1\text{and}{I}_2$ is given by,

$\theta ={\theta }_1+{\theta }_2=60+30={90}^{\circ }$
Hence, option $\left(C\right)$ is correct.