A circuit contains 50 `muF ` capacitor and 20 mH inductor connected together with the help of a key which is open initially. Charge stored in the capacitor is 50 mC. Key is closed at t=0 . Calculate the minimum time interval in which energy stored in the inductor becomes equal to the energy stored in capacitor . Neglect any resistance in the circuit .

Explanation

To solve the problem, we need to determine the minimum time interval in which the energy stored in the inductor becomes equal to the energy stored in the capacitor after the key is closed at \( t = 0 \).

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance \( C = 50 \, \mu F = 50 \times 10^{-6} \, F \)
- Inductance \( L = 20 \, mH = 20 \times 10^{-3} \, H \)
- Initial charge on the capacitor \( Q = 50 \, mC = 50 \times 10^{-3} \, C \)
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