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: Understanding the Question:
In an ideal LC circuit, energy oscillates between the electric field of the capacitor and the magnetic field of the inductor. Energy is conserved.
Step 2: Key Formula or Approach:
Max energy in Capacitor = Max energy in Inductor
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I_{max}^2 \]
Step 3: Detailed Explanation:
Given:
$C = 4 \mu F = 4 \times 10^{-6} F$
$V = 10 V$
$L = 10 mH = 10 \times 10^{-3} H = 10^{-2} H$
Using the energy conservation equation:
\[ C V^2 = L I_{max}^2 \]
\[ (4 \times 10^{-6}) \times (10)^2 = (10^{-2}) \times I_{max}^2 \]
\[ 4 \times 10^{-4} = 10^{-2} \times I_{max}^2 \]
\[ I_{max}^2 = \frac{4 \times 10^{-4}}{10^{-2}} = 4 \times 10^{-2} = 0.04 \]
Taking square root:
\[ I_{max} = \sqrt{0.04} = 0.2 A \]
Step 4: Final Answer:
The maximum current in the coil is 0.2 A.