A 50 Hz AC source of 20 volts is connected across R and C as shown in figure. The voltage across R is 12 volt. The voltage across C is

Question

A 50 Hz AC source of 20 volts is connected across R and C as shown in figure. The voltage across R is 12 volt. The voltage across C is (A) 8 V (B) 16 V (C) 10 V (D) not possible to determine unless values of R and C are given

Tardigrade Admin · Accepted Answer

Voltage across capacitor lags behind current by

9 0^{\circ}

.
The given circuit is a

C - R

series circuit,

V_{R}

is in phase with

i

, while

V_{C}

lags behind

i

by

9 0^{\circ}

. Hence, resultant potential is

V = V_{R}^{2} + V_{C}^{2}

Given,

V_{R} = 12 V, V = 20 V

∴ (20)^{2} = (12)^{2} + V_{C}^{2}

\Rightarrow V_{C}^{2} = (20)^{2} - (12)^{2}

= 400 - 144 = 256

\Rightarrow V_{C} = 256 = 16 V

Q. A 50Hz AC source of 20 volts is connected across R and C as shown in figure. The voltage across R is 12 volt. The voltage across C is

8 V

16 V

10 V

not possible to determine unless values of R and C are given

Q. A $50 Hz$ AC source of $20$ volts is connected across $R$ and $C$ as shown in figure. The voltage across $R$ is $12$ volt. The voltage across $C$ is