A capacitor of capacitance `100 mu F` is charged to a potential of "12V" and connected to a `6.4mH` inductor to produce oscillations.The maximum current in the circuit would be

Options

A. 1.5 A
B. 2.0 A
C. 1.2 A
D. 3.2 A

Answer

To find the maximum current in the LC oscillating circuit, we can use the principle of conservation of energy. The energy stored in the capacitor will be equal to the energy stored in the inductor at maximum current.

### Step-by-step Solution:

1. **Identify the given values**:
- Capacitance, \( C = 100 \, \mu F = 100 \times 10^{-6} \, F \)
- Inductance, \( L = 6.4 \, mH = 6.4 \times 10^{-3} \, H \)
- Maximum voltage across the capacitor, \( V = 12 \, V \)

2. **Write the energy formulas**:
- Energy stored in the capacitor:
\[ U_C = \frac{1}{2} C V^2 \]
- Energy stored in the inductor:
\[ U_L = \frac{1}{2} L I^2 \]

3. **Set the energies equal** (conservation of energy):
\[ U_C = U_L \]
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I^2 \]

4. **Cancel the \( \frac{1}{2} \)** from both sides:
\[ C V^2 = L I^2 \]

5. **Rearrange to find \( I^2 \)**:
\[ I^2 = \frac{C V^2}{L} \]

6. **Substitute the values into the equation**:
\[ I^2 = \frac{(100 \times 10^{-6} \, F)(12^2 \, V^2)}{6.4 \times 10^{-3} \, H} \]
\[ I^2 = \frac{(100 \times 10^{-6})(144)}{6.4 \times 10^{-3}} \]

7. **Calculate the numerator**:
\[ 100 \times 10^{-6} \times 144 = 0.0144 \, J \]

8. **Calculate \( I^2 \)**:
\[ I^2 = \frac{0.0144}{6.4 \times 10^{-3}} = \frac{0.0144}{0.0064} = 2.25 \]

9. **Take the square root to find \( I \)**:
\[ I = \sqrt{2.25} = 1.5 \, A \]

### Final Answer:
The maximum current in the circuit is \( I = 1.5 \, A \).

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