A `8 muH `inductor is shorted across a `2 muF` capacitor. The capacitor is initially charged to 20 V. The maximum value of the current in the circuit is

Options

A. `10.0 A `
B. `7.5 A`
C. `12.0 A `
D. `8.2 A `

Explanation

To solve the problem, we need to find the maximum current in a circuit consisting of an inductor and a capacitor. The inductor has an inductance of \( L = 8 \, \mu H \) and the capacitor has a capacitance of \( C = 2 \, \mu F \). The capacitor is initially charged to \( V_0 = 20 \, V \).

### Step-by-Step Solution:

1. **Identify the maximum charge on the capacitor:**
The maximum charge \( Q_0 \) on the capacitor can be calculated using the formula:
\[ Q_0 = C \cdot V_0 \]
Substituting the values:
\[ Q_0 = 2 \times 10^{-6} \, F \cdot 20 \, V = 4 \times 10^{-5} \, C \]

2. **Calculate the angular frequency \( \omega \):**
The angular frequency \( \omega \) for the LC circuit can be found using the formula:
\[ \omega = \frac{1}{\sqrt{L \cdot C}} \]
Substituting the values:
\[ \omega = \frac{1}{\sqrt{8 \times 10^{-6} \, H \cdot 2 \times 10^{-6} \, F}} = \frac{1}{\sqrt{16 \times 10^{-12}}} = \frac{1}{4 \times 10^{-6}} = 2.5 \times 10^{5} \, rad/s \]

3. **Calculate the maximum current \( I_m \):**
The maximum current \( I_m \) in the circuit can be calculated using the formula:
\[ I_m = \omega \cdot Q_0 \]
Substituting the values:
\[ I_m = 2.5 \times 10^{5} \, rad/s \cdot 4 \times 10^{-5} \, C = 10 \, A \]

### Final Answer:
The maximum value of the current in the circuit is \( \boxed{10 \, A} \).