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

We are given that the current in a coil changes from 5 A to 2 A in 0.1 seconds, and an average voltage of 50 V is produced. We need to find the self-inductance of the coil.

The induced electromotive force (EMF) in a coil due to self-inductance is given by Faraday's law. The formula for the magnitude of the average induced EMF is:

$$ \varepsilon = L \left| \frac{\Delta I}{\Delta t} \right| $$

where:

• $$\varepsilon$$ is the average voltage (50 V),
• $$L$$ is the self-inductance (which we need to find),
• $$\Delta I$$ is the change in current,
• $$\Delta t$$ is the time interval (0.1 s).

First, calculate the change in current ($$\Delta I$$). The initial current is 5 A, and the final current is 2 A. Since the current decreases, the change is:

$$ \Delta I = I_{\text{final}} - I_{\text{initial}} = 2 \, \text{A} - 5 \, \text{A} = -3 \, \text{A} $$

The magnitude of the change in current is $$ |\Delta I| = |-3| = 3 \, \text{A} $$.

Now, the time interval $$\Delta t = 0.1 \, \text{s}$$. So, the magnitude of the rate of change of current is:

$$ \left| \frac{\Delta I}{\Delta t} \right| = \frac{3 \, \text{A}}{0.1 \, \text{s}} = 30 \, \text{A/s} $$

Substitute the known values into the formula:

$$ 50 = L \times 30 $$

Solve for $$L$$:

$$ L = \frac{50}{30} = \frac{5}{3} \approx 1.6667 \, \text{H} $$

Rounding 1.6667 H to two decimal places gives 1.67 H.

Now, comparing with the options:

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

Hence, the correct answer is Option A.