When current in a coil changes from 5 A to 2 A in 0.1 s, average voltage of 50 V is produced. The self - inductance of the coil is :

Options

A. 1.67 H
B. 6 H
C. 3 H
D. 0.67 H

Answer

To find the self-inductance of the coil, we can use the formula for the average voltage induced in an inductor, which is given by:

\[ V = -L \frac{di}{dt} \]

Where:
- \( V \) is the average voltage (50 V in this case),
- \( L \) is the self-inductance,
- \( di \) is the change in current,
- \( dt \) is the change in time.

### Step 1: Determine the change in current (\( di \))
The current changes from 5 A to 2 A. Thus, the change in current (\( di \)) is:

\[ di = I_{final} - I_{initial} = 2\, \text{A} - 5\, \text{A} = -3\, \text{A} \]

### Step 2: Determine the change in time (\( dt \))
The time interval over which this change occurs is given as 0.1 s:

\[ dt = 0.1\, \text{s} \]

### Step 3: Substitute values into the formula
We can now substitute the values into the formula:

\[ 50 = -L \frac{-3}{0.1} \]

### Step 4: Simplify the equation
This simplifies to:

\[ 50 = L \cdot 30 \]

### Step 5: Solve for \( L \)
Now, we can solve for \( L \):

\[ L = \frac{50}{30} = \frac{5}{3} \approx 1.67\, \text{H} \]

Thus, the self-inductance of the coil is approximately **1.67 Henry**.

### Final Answer:
The self-inductance of the coil is **1.67 H**.

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