If a current of 10 A changes in one second through a coil, and the induced emf is 20 V, then the self-inductance of the coil is

Options

A. `(2)/(5)H`
B. `(4)/(5)H`
C. `(5)/(4)H`
D. 2 H

Explanation

To find the self-inductance of the coil, we can use the formula for induced electromotive force (emf) in terms of self-inductance:

\[ \text{emf} = -L \frac{dI}{dt} \]

Where:
- \( \text{emf} \) is the induced electromotive force (in volts),
- \( L \) is the self-inductance (in henries),
- \( \frac{dI}{dt} \) is the rate of change of current (in amperes per second).

### Step-by-step Solution:

1. **Identify the given values:**
- Current change (\( \Delta I \)) = 10 A
- Time interval (\( \Delta t \)) = 1 s
- Induced emf = 20 V

2. **Calculate the rate of change of current:**
\[ \frac{dI}{dt} = \frac{\Delta I}{\Delta t} = \frac{10 \, \text{A}}{1 \, \text{s}} = 10 \, \text{A/s} \]

3. **Substitute the values into the emf formula:**
\[ 20 \, \text{V} = L \cdot 10 \, \text{A/s} \]

4. **Solve for self-inductance \( L \):**
\[ L = \frac{20 \, \text{V}}{10 \, \text{A/s}} = 2 \, \text{H} \]

5. **Conclusion:**
The self-inductance of the coil is \( L = 2 \, \text{H} \).

### Final Answer:
The self-inductance of the coil is 2 Henry.

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