Question

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

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

Answer 1

Answer: (A)

$6V,{\tan }^{-1}\left(\frac{4}{3}\right)$

$V_R=\sqrt{V^2-V_C^2}=\sqrt{(10{)}^2-(8{)}^2}=6V$
$\tan \varphi =\frac{X_C}{X_R}=\frac{V_C}{V_R}=\frac{8}{6}=\frac{4}{3}$