Question

In a coil, the current changes form –2 A to +2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :

• A. 5 mH
• B. 1 mH
• C. 2.5 mH
• D. 4 mH

Answer 1

The Correct Option is A

The self-inductance of the coil can be determined using the formula relating induced electromotive force (emf), self-inductance, and the rate of change of current:

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

Where:

• \(E\) is the induced emf (0.1 V).
• \(L\) is the self-inductance of the coil.
• \(\Delta I\) is the change in current (from -2 A to +2 A, thus \(\Delta I = 4 \, \text{A}\)).
• \(\Delta t\) is the time interval for the current change (0.2 s).

Rearranging to solve for \(L\):

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

Substituting the given values:

\(L = -\frac{0.1 \, \text{V} \cdot 0.2 \, \text{s}}{4 \, \text{A}}\)

\(L = -\frac{0.02}{4}\)

\(L = -0.005 \, \text{H}\)

Inductance is conventionally positive, so we use the absolute value:

\(L = 0.005 \, \text{H} = 5 \, \text{mH}\)

The self-inductance of the coil is 5 mH.

Analyzing the provided options:

• 5 mH: Matches the calculated value; Correct.
• 1 mH: Incorrect.
• 2.5 mH: Incorrect.
• 4 mH: Incorrect.

The correct option is 5 mH.