Q. In a circuit L, C and R are connected in series with an alternating voltage source of frequency $f.$ The current leads the voltage by $45{^\circ}$. The value of C is
A) $\frac{1}{2\pi\ f\ (2\pi fL + R)}$
B) $\frac{1}{\pi f\ (2\pi fL + R)}$
C) $\frac{1}{2\pi f\ (2\pi fL - R)}$
D) $\frac{1}{\pi f\ (2\pi fL - R)}$
Explanation
(c) $\tan{\phi = \frac{\omega L - \frac{1}{\omega C}}{R}}$
$\phi$ being the angle by which the current leads the voltage.
Given, $\phi = 45{^\circ}$
$\therefore\ \ \ \ \ \ \ \ \ \tan{45{^\circ} = \frac{\omega L - \frac{1}{\omega C}}{R}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ 1 = \frac{\omega L - \frac{1}{\omega C}}{R}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ R = \omega L - \frac{1}{\omega C}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ C = \frac{1}{\omega(\omega L - R)}$
$= \frac{1}{2\pi f(2\pi fL - R)}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.