Question

In a coil, an increase in current from 5 A to 10 A in 100 ms induces an emf of 100 V. The self-inductance of the coil is:

• A. 2 H
• B. 10 H
• C. 20 H
• D. 2000 H

Answer 1

The Correct Option is A

Faraday's Law of Induction describes the induced electromotive force (EMF) in a coil due to a changing current. This EMF is related to the coil's self-inductance (L) by the equation:

\( EMF = -L \frac{\Delta I}{\Delta t} \)

In this formula:

• \( EMF \) represents the induced electromotive force, measured in Volts.
• \( L \) is the self-inductance of the coil, measured in Henries.
• \( \Delta I \) denotes the change in current, measured in Amperes.
• \( \Delta t \) signifies the change in time, measured in seconds.

The provided values are:

• \( EMF = 100 \, \text{V} \)
• The change in current, \( \Delta I \), is \( 10 \, \text{A} - 5 \, \text{A} = 5 \, \text{A} \).
• The change in time, \( \Delta t \), is \( 100 \, \text{ms} \), which is equivalent to \( 0.1 \, \text{s} \).

The objective is to determine the self-inductance \( L \). By rearranging the formula to solve for \( L \), we get:

\( L = -\frac{EMF}{\frac{\Delta I}{\Delta t}} = -\frac{EMF \cdot \Delta t}{\Delta I} \)

Substituting the given values into the rearranged formula yields:

\( L = -\frac{100 \, \text{V} \cdot 0.1 \, \text{s}}{5 \, \text{A}} = -\frac{10}{5} \, \text{H} = -2 \, \text{H} \)

As self-inductance \( L \) is inherently a positive value, we take its absolute value:

\( L = |-2 \, \text{H}| = 2 \, \text{H} \)

Consequently, the self-inductance of the coil is determined to be 2 H.