Question

In a series $CR$ circuit shown in figure, the applied voltage is $10V$ and the voltage across capacitor is found to be $8V$. If the voltage across $R$ and the phase difference between current and the applied voltage respectively be $x$ and ${\tan }^{-1}\left(y\right)$, then the value of $\frac{2x}{y}$ is

Answer 1

We have.
$V_C^2+V_R^2={10}^2{8}^2+V_R^2={10}^2{V}_R=6V.\Rightarrow x=6\tan \varphi =\frac{V_c}{V_R}=\frac{8}{6}\tan \varphi =\frac{4}{3}\Rightarrow y=\frac{4}{3}\therefore \frac{x}{y}=\frac{2\times 6}{\left(\frac{4}{3}\right)}=9$