In the circuit shown below, the switch is kept in position $a$ for a long time and is then thrown to position $b$. The amplitude of the resulting oscillating current is given by

  • A
    $E \sqrt{L/C}$
  • B
    $E / R$
  • C
    infinity
  • D
    $E \sqrt{C/L}$

Explore More

Similar Questions

The square root of the product of inductance $(L)$ and capacitance $(C)$ has the dimension of:

If maximum energy is stored in a capacitor at $t=0$,then the time after which the current in the circuit will be maximum is:

In an oscillating $LC$ circuit,the total stored energy is $U$ and the maximum charge on the capacitor is $Q$. When the charge on the capacitor is $\frac{Q}{2}$,the energy stored in the inductor is:

Initially,the key was placed on $(1)$ until the capacitor got fully charged. Now,the key is placed on $(2)$ at $t = 0$. Find the minimum time when the energy in both the capacitor and the inductor will be the same.

Difficult
View Solution

In the given circuit,when $S_1$ is closed,the capacitor $C$ gets fully charged. Then $S_1$ is kept open and $S_2$ is closed. Hence

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