In an LCR 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 on

Question

In an LCR 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 on (A)  10 V  (B)  (20/√2) V  (C)  20√2 V  (D)  (10/√2) V

Tardigrade Admin · Accepted Answer

Circuit is resonant, hence supply voltage equals to

V_{R}

.

V_{R} = 10

volt Also,

X_{C} = R

As the voltage drops are equals across L, C and R then when L is short

Z = R^{2} + X_{C}^{2}

= 2 R

V_{C} = i X_{C}

[∵ i = \frac{V}{Z}]

Or

V_{C} = \frac{V}{Z} X_{C}

Or

= \frac{10}{2 R} R

Or

V_{C} = \frac{10}{2} v o lt

Q. In an $L CR$ series $A C$ 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 on

$10 V$

$\frac{20}{2} V$

$202 V$

$\frac{10}{2} V$

$20 V$

Q. In an LCR series AC circuit the voltage across L,C and R is 10V each. If the inductor is short circuited, the voltage across the capacitor would become on

10V

2​20​V

202​V

2​10​V

20V