A `100 mu F` capacitor is charged with a 50 V source supply. Then source supply is removed and the capacitor is connected across an inductor, as a result of which 5 A current flows through the inductance. Calculate the value of inductance.

Options

A. 0.01
B. 0.02
C. 0.03
D. 0.04

Explanation

To find the value of inductance when a charged capacitor is connected across an inductor, we can follow these steps:

### Step 1: Calculate the 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:
- \(C\) is the capacitance in farads (F)
- \(V\) is the voltage in volts (V)

Given:
- \(C = 100 \, \mu F = 100 \times 10^{-6} \, F\)
- \(V = 50 \, V\)

Substituting the values:
\[ U = \frac{1}{2} \times (100 \times 10^{-6}) \times (50)^2 \]
\[ U = \frac{1}{2} \times (100 \times 10^{-6}) \times 2500 \]
\[ U = \frac{1}{2} \times 0.25 = 0.125 \, J \]

### Step 2: Relate the energy stored in the inductor to the current
When the capacitor discharges through the inductor, the energy stored in the inductor (also \(U\)) is given by:
\[ U = \frac{1}{2} L I^2 \]
where:
- \(L\) is the inductance in henries (H)
- \(I\) is the current in amperes (A)

Given:
- \(I = 5 \, A\)

### Step 3: Set the energies equal to each other
Since the energy stored in the capacitor is equal to the energy stored in the inductor:
\[ \frac{1}{2} C V^2 = \frac{1}{2} L I^2 \]

We can cancel \(\frac{1}{2}\) from both sides:
\[ C V^2 = L I^2 \]

### Step 4: Solve for inductance \(L\)
Rearranging the equation to find \(L\):
\[ L = \frac{C V^2}{I^2} \]

### Step 5: Substitute the known values
Substituting \(C\), \(V\), and \(I\):
\[ L = \frac{(100 \times 10^{-6}) \times (50)^2}{(5)^2} \]
\[ L = \frac{(100 \times 10^{-6}) \times 2500}{25} \]
\[ L = \frac{(100 \times 10^{-6}) \times 100}{1} \]
\[ L = 100 \times 10^{-4} = 0.01 \, H \]

### Final Answer
The value of the inductance \(L\) is \(0.01 \, H\) or \(10 \, mH\).

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