Question

In a series $\text{CR}$ 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. $6\text{V};{\tan }^{-1}\left(\frac{1}{2}\right)$
• D. None

Answer 1

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

Given, $E=10\text{V};V_C=8\text{V}$


For, $\text{CR}$ circuit,

$E=\sqrt{V_R^2+V_C^2}$

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

$\Rightarrow (V_R{)}^2=36$

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

From phasor digram,

$\tan \varphi =\frac{V_C}{V_R}$

$\Rightarrow \tan \varphi =\frac{4}{3}$

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

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.