Question

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:

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.

Answer 1

Step 1: Formula for e.m.f induced by self-inductance.
The induced e.m.f (\( \mathcal{E} \)) in a coil is given by: \[ \mathcal{E} = -L \frac{\Delta I}{\Delta t}, \] where: \( \mathcal{E} = 5 \, \text{V} \) (induced e.m.f), \( \Delta I = 3 \, \text{A} - 2 \, \text{A} = -1 \, \text{A} \) (change in current), \( \Delta t = 1 \, \text{ms} = 1 \times 10^{-3} \, \text{s} \). Step 2: Calculating self-inductance.
Rearranging the formula for \( L \): \[ L = \frac{\mathcal{E} \cdot \Delta t}{\Delta I}. \] Substitute the given values: \[ L = \frac{5 \times 1 \times 10^{-3}}{-(-1)} = 5 \times 10^{-3} \, \text{H}. \] \[ \therefore \text{The value of self-inductance is: } L = 5 \times 10^{-3} \, \text{H}. \]