Question

In a series RC circuit shown in the figure, at an instant, the applied voltage is $10\text{V}$ and the voltage across the capacitor is found to be $8\text{V}$. The phase difference between the current and the applied voltage will be -

• A. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{3}{4}\right)$
• C. ${\tan }^{-1}\left(\frac{5}{3}\right)$
• D. ${\tan }^{-1}\left(\frac{3}{5}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{4}{3}\right)$

For an AC RC circuit,

$V^2=V_R^2+V_C^2$

$\Rightarrow {10}^2=V_R^2+8^2$

$\Rightarrow V_R=6\text{V}$



The phase difference between the current and the applied voltage,

$\varphi ={\tan }^{-1}\left(\frac{V_C}{V_R}\right)={\tan }^{-1}\left(\frac{8}{6}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$

Hence, option $\left(A\right)$ is the correct answer.