Question

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

Answer 1

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


From phasor diagram,

$\Rightarrow V_{net}^2=V_R^2+V_C^2$

Given data are the $rms$ values. So,
$V_{net}=10\text{V}$ and $V_C=8\text{V}$

$\Rightarrow V_R^2={10}^2-8^2=36$

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


Phase difference between current and voltage,

$\tan \theta =\left|\frac{V_C}{V_R}\right|=\frac{8}{6}=\frac{4}{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{4}{3}\right)$

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