An LC circuit contains a 20 mH inductor and `25 mu F` capacitor with an initial charge of 5 mC. The total energy stored in the circuit initially is

Options

A. 5 J
B. `0.5 J`
C. 50 J
D. 500 J

Explanation

To find the total energy stored in the LC circuit initially, we will focus on the energy stored in the capacitor since at time \( t = 0 \), all the energy is stored in the capacitor.

### Step-by-Step Solution:

1. **Identify the given values:**
- Inductance \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 25 \, \mu\text{F} = 25 \times 10^{-6} \, \text{F} \)
- Initial charge \( Q = 5 \, \text{mC} = 5 \times 10^{-3} \, \text{C} \)

2. **Formula for energy stored in a capacitor:**
The energy \( E \) stored in a capacitor is given by the formula:
\[ E = \frac{Q^2}{2C} \]

3. **Substitute the values into the formula:**
\[ E = \frac{(5 \times 10^{-3})^2}{2 \times (25 \times 10^{-6})} \]

4. **Calculate \( Q^2 \):**
\[ Q^2 = (5 \times 10^{-3})^2 = 25 \times 10^{-6} \, \text{C}^2 \]

5. **Calculate the denominator:**
\[ 2C = 2 \times (25 \times 10^{-6}) = 50 \times 10^{-6} \, \text{F} \]

6. **Substitute \( Q^2 \) and \( 2C \) into the energy formula:**
\[ E = \frac{25 \times 10^{-6}}{50 \times 10^{-6}} = \frac{25}{50} = 0.5 \, \text{J} \]

7. **Conclusion:**
The total energy stored in the circuit initially is \( 0.5 \, \text{J} \).

### Final Answer:
The total energy stored in the circuit initially is \( 0.5 \, \text{J} \).

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