Question:medium

An e.m.f of 5 volts is produced by a self-inductance when the current changes at a steady rate from 3A to 2A in 1 millisecond. The value of self-inductance is:

Show Hint

The self-inductance of a coil determines the opposition to the change in current and can be calculated using the induced e.m.f and rate of change of current.
Updated On: Jan 13, 2026
Show Solution

Solution and Explanation

Step 1: Formula for induced electromotive force (e.m.f.) due to self-inductance. The induced e.m.f. (\( \mathcal{E} \)) in a coil is defined as: \[ \mathcal{E} = -L \frac{\Delta I}{\Delta t}, \] where \( \mathcal{E} = 5 \, \text{V} \) represents the induced e.m.f., \( \Delta I = 3 \, \text{A} - 2 \, \text{A} = -1 \, \text{A} \) is the change in current, and \( \Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \) is the time interval.

Step 2: Calculation of self-inductance. The formula is rearranged to solve for \( L \): \[ L = \frac{\mathcal{E} \cdot \Delta t}{\Delta I}. \] Substituting the provided values yields: \[ L = \frac{5 \times 1 \times 10^{-3}}{-(-1)} = 5 \times 10^{-3} \, \text{H}. \] \[ \therefore \text{The calculated self-inductance is: } L = 5 \times 10^{-3} \, \text{H}. \]

Was this answer helpful?
0