$A$ charged capacitor discharges through a resistance $R$ with time constant $\tau$. The two are now placed in series across an $AC$ source of angular frequency $\omega = \frac{1}{\tau}$. The impedance of the circuit will be

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

Explore More

Similar Questions

In a given series $LCR$ circuit,$R = 4\,\Omega$,$X_L = 5\,\Omega$,and $X_C = 8\,\Omega$. The current:

Difficult
View Solution

When an $AC$ source of $e.m.f.$ $e = E_0 \sin(100t)$ is connected across a circuit,the phase difference between the $e.m.f.$ $e$ and the current $i$ in the circuit is observed to be $\pi/4$,as shown in the diagram. If the circuit consists possibly only of $RC$ or $LC$ in series,find the relationship between the two elements.

Two resistors are connected in series across a $5\,V\,rms$ source of alternating potential. The potential difference across the $6\,\Omega$ resistor is $3\,V$. If $R$ is replaced by a pure inductor $L$ of such magnitude that the current remains the same,then the potential difference across $L$ is.......$V$.

For the given $RL$ circuit,the maximum power loss across the circuit can be $W$.

In an $AC$ circuit,the potential differences across an inductance and a resistance joined in series are $16 \, V$ and $20 \, V$ respectively. The total potential difference of the source is .......... $V$.

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