An alternating e.m.f. having voltage $V = V_0 \sin \omega t$ is applied to a series $L-C-R$ circuit. Given: $|X_L - X_C| = R$. The r.m.s. value of potential difference across the capacitor will be:

  • A
    $V_0 R \omega C$
  • B
    $\frac{V_0}{R \omega C}$
  • C
    $\frac{V_0}{2 R \omega C}$
  • D
    $\frac{V_0}{\sqrt{2} R \omega C}$

Explore More

Similar Questions

In an $LCR$ series $a.c.$ circuit,the voltage across each of the components $L, C$ and $R$ is $60 \,V$. The voltage across the $LC$ combination is

In an $L-C-R$ circuit,the potential difference across the inductor is $60\,V$,across the capacitor is $30\,V$,and across the resistor is $40\,V$. The supply voltage will be: (in $,V$)

Difficult
View Solution

What will be the reading in the voltmeter and ammeter of the circuit shown?

The current $I$,potential difference $V_L$ across the inductor,and potential difference $V_C$ across the capacitor in the circuit as shown in the figure are best represented vectorially as:

In an $ac$ circuit,the reactance of a coil is $\sqrt{3}$ times its resistance. The phase difference between the voltage across the coil and the current through the coil will be:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo