A capacitor of capacitance `1 muF` is charged to a potential of `1 V`, it is connected in parallel to an inductor of inductance `10^(-3)H`. The maximum current that will flow in the circuit has the value

Options

A. `sqrt 1000` mA
B. 1A.
C. 1 mA
D. 1000 mA

Explanation

To find the maximum current that will flow in the circuit when a capacitor is connected in parallel to an inductor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Energy Conservation**:
The energy stored in the capacitor will be transferred to the inductor. The total energy in the circuit remains constant and oscillates between the capacitor and the inductor.

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

Plugging in the values:
\[ U = \frac{1}{2} \times 1 \times 10^{-6} \times (1)^2 = \frac{1}{2} \times 10^{-6} = 0.5 \times 10^{-6} \, J = 5 \times 10^{-7} \, J \]

3. **Relate Energy to Maximum Current in the Inductor**:
The energy stored in the inductor when the current is at its maximum (\( I_{max} \)) is given by:
\[ U = \frac{1}{2} L I_{max}^2 \]
where:
- \( L = 10^{-3} H \)

Setting the energies equal (since energy is conserved):
\[ \frac{1}{2} L I_{max}^2 = 5 \times 10^{-7} \]

4. **Solve for Maximum Current**:
Rearranging the equation to solve for \( I_{max} \):
\[ I_{max}^2 = \frac{2 \times 5 \times 10^{-7}}{L} \]
Substituting \( L \):
\[ I_{max}^2 = \frac{2 \times 5 \times 10^{-7}}{10^{-3}} = 10^{-4} \]
Taking the square root:
\[ I_{max} = \sqrt{10^{-4}} = 10^{-2} \, A = 0.1 \, A \]

5. **Convert to Milliamperes**:
To express the current in milliamperes:
\[ I_{max} = 0.1 \, A = 100 \, mA \]

### Final Answer:
The maximum current that will flow in the circuit is \( 100 \, mA \).

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