A capacitor of capacity `2muF` is changed to a potential different of `12V` . It is then connected across an inductor of inductance `0.6mH` What is the current in the circuit at a time when the potential difference across the capacitor is `6.0V` ?

Options

A. 3.6A
B. 2.4A
C. 1.2A
D. 0.6A

Answer

To solve the problem, we will use the principle of conservation of energy in an LC circuit. The energy stored in the capacitor will be transferred to the inductor as the circuit oscillates.

### Step-by-Step Solution:

1. **Calculate the initial energy stored in the capacitor:**
The energy \( U_C \) stored in a capacitor is given by the formula:
\[ U_C = \frac{1}{2} C V^2 \]
where:
- \( C = 2 \mu F = 2 \times 10^{-6} F \)
- \( V = 12 V \)

Substituting the values:
\[ U_C = \frac{1}{2} \times 2 \times 10^{-6} \times (12)^2 \]
\[ U_C = \frac{1}{2} \times 2 \times 10^{-6} \times 144 = 1.44 \times 10^{-4} J \]

2. **Calculate the energy in the capacitor when the voltage is 6V:**
When the potential difference across the capacitor is \( 6V \), the energy \( U_{C2} \) stored in the capacitor is:
\[ U_{C2} = \frac{1}{2} C V^2 \]
where \( V = 6 V \):
\[ U_{C2} = \frac{1}{2} \times 2 \times 10^{-6} \times (6)^2 \]
\[ U_{C2} = \frac{1}{2} \times 2 \times 10^{-6} \times 36 = 3.6 \times 10^{-5} J \]

3. **Use conservation of energy to find the energy in the inductor:**
The initial energy in the capacitor will be equal to the sum of the energy in the capacitor and the energy in the inductor:
\[ U_C = U_{C2} + U_L \]
Rearranging gives:
\[ U_L = U_C - U_{C2} \]
Substituting the values:
\[ U_L = 1.44 \times 10^{-4} - 3.6 \times 10^{-5} = 1.08 \times 10^{-4} J \]

4. **Relate the energy in the inductor to the current:**
The energy \( U_L \) stored in the inductor is given by:
\[ U_L = \frac{1}{2} L I^2 \]
where \( L = 0.6 mH = 0.6 \times 10^{-3} H \). Setting the two expressions for \( U_L \) equal gives:
\[ 1.08 \times 10^{-4} = \frac{1}{2} \times 0.6 \times 10^{-3} \times I^2 \]
Rearranging for \( I^2 \):
\[ I^2 = \frac{2 \times 1.08 \times 10^{-4}}{0.6 \times 10^{-3}} = \frac{2.16 \times 10^{-4}}{0.6 \times 10^{-3}} = 0.36 \]

5. **Calculate the current \( I \):**
Taking the square root:
\[ I = \sqrt{0.36} = 0.6 A \]

### Final Answer:
The current in the circuit when the potential difference across the capacitor is \( 6.0 V \) is \( 0.6 A \).

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