Q. In an L-C-R series AC circuit the voltage across L, C and R is 10 V each. If the inductor is short circuited, the voltage across the capacitor would become
A) 10 V
B) $\frac{20}{\sqrt{2}\ }\ \ V$
C) $20\sqrt{2}\ V$
D) $\frac{10}{\sqrt{2}\ }\ \ V$
Explanation
(d) Circuit is resonant.
Hence supply voltage equals
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{R} = 10\ V$
Also, $X_{C} = R$
As the voltage drops are equals across them when L is shortened
$Z = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{2R}$
$V_{C} = iX_{C}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \because i = \frac{V}{2} \right)$
or $V_{C} = \frac{V}{Z}X_{C} = \frac{10}{\sqrt{2R}}R$
$V_{C} = \frac{10}{\sqrt{2}}V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.