Question

An LC circuit with negligible resistance containing a 20 mH inductor and a 50 \(\mu F\) capacitor with an initial charge of 10 mC is closed at \(t=0\). Then the minimum time taken (in \(\mu s\)) for the total energy to be shared equally between the inductor and capacitor is:

• A. \(375\pi\)
• B. \(125\pi\)
• C. \(500\pi\)
• D. \(250\pi\)

Hint:

In an LC circuit, equal energy sharing first occurs at \[ t=\frac{T}{8} = \frac{\pi}{4\omega}. \]

Answer 1

The Correct Option is D

Concept: In an LC circuit, \[ q=q_0\cos\omega t \] where \[ \omega=\frac1{\sqrt{LC}}. \] Energy in capacitor: \[ U_C=U_0\cos^2\omega t. \] Energy in inductor: \[ U_L=U_0\sin^2\omega t. \] Equal sharing occurs when \[ U_C=U_L. \]

Step 1: Apply equal energy condition.
\[ \cos^2\omega t=\sin^2\omega t \] \[ \tan^2\omega t=1 \] Minimum value: \[ \omega t=\frac{\pi}{4} \] \[ t=\frac{\pi}{4\omega} \]

Step 2: Find angular frequency.
\[ L=20\times10^{-3}H \] \[ C=50\times10^{-6}F \] \[ LC=10^{-6} \] \[ \sqrt{LC}=10^{-3} \] \[ \omega=10^3\;rad/s \]

Step 3: Calculate time.
\[ t = \frac{\pi}{4\times10^3} \] \[ = 250\pi\times10^{-6}s \] \[ = 250\pi\;\mu s \] \[ \boxed{250\pi} \]