Question

In a series $CR$ circuit shown in figure, the applied voltage is $10\text{V}$ and the voltage across capacitor is found to be $8\text{V}$. Then the voltage across $R$ and the phase difference between current and the applied voltage will respectively be

• A. $6\text{V},{\tan }^{-1}\left(\frac{4}{3}\right)$
• B. $3\text{V},{\tan }^{-1}\left(\frac{3}{4}\right)$
• C. $6\text{V},{\tan }^{-1}\left(\frac{5}{3}\right)$
• D. None

Answer 1

The correct option is A $6\text{V},{\tan }^{-1}\left(\frac{4}{3}\right)$

From given data,
Applied voltage $V_{\text{applied}}=10\text{V}$
and Voltage across capacitor $V_C=8\text{V}$
Let $V_R$ be voltage across resistor.


Then, $V_{applied}=\sqrt{V_c^2+V_R^2}$
i.e $8^2+V_R^2={10}^2$
or $V_R=6\text{V}$

Current is in phase with $V_R$.
Therefore, phase difference between current and applied voltage is
$\theta ={\tan }^{-1}\left(\frac{8}{6}\right)={\tan }^{-1}\left(\frac{4}{3}\right)$