Question

A $4\mu F$ capacitor is charged to 10 V. The battery is then disconnected and a pure 10 mH coil is connected across the capacitor so that LC oscillations are set up. The maximum current in the coil is

• A. 0.2 A
• B. 0.1 A
• C. 0.4 A
• D. 0.25 A

Hint:

Energy oscillates between electric (capacitor) and magnetic (inductor) forms in an LC circuit.

Answer 1

The Correct Option is A

Step 1: Energy Conservation
Max energy in capacitor = Max energy in inductor.
$\frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2$
Step 2: Substitution
$4 \times 10^{-6} \times (10)^2 = 10 \times 10^{-3} \times I^2$
$4 \times 10^{-4} = 10^{-2} \times I^2$
Step 3: Calculation
$I^2 = \frac{4 \times 10^{-4}}{10^{-2}} = 4 \times 10^{-2} = 0.04$
$I = \sqrt{0.04} = 0.2 \text{ A}$
Step 4: Conclusion
The maximum current is 0.2 A.
Final Answer:(A)