Question

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

• A. $6V,ta{n}^{-1}\left(\frac{4}{3}\right)$
• B. $3V,ta{n}^{-1}\left(\frac{3}{4}\right)$
• C. $6V,ta{n}^{-1}\left(\frac{5}{3}\right)$
• D. Zero, Zero

Answer 1

The correct option is A

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

Hence $V_Q=\sqrt{{10}^2-8^2}=6V$ and $\tan \varphi =\frac{8}{C}=\frac{4}{3}$