Question

The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self inductance of the inductor _______ mH.

Answer 1

The emf ($\varepsilon$) induced in an inductor is related to the rate of change of current and self-inductance ($L$) by the formula:

$|\varepsilon| = L \frac{dI}{dt}.$

Step 1: Determine the rate of change of current
The given current is:

$I = 3t + 8.$

Differentiate $I$ with respect to $t$:

$\frac{dI}{dt} = 3 \, \text{A/s}.$

Step 2: Substitute the given values
The magnitude of induced emf is $|\varepsilon| = 12 \, \text{mV} = 12 \times 10^{-3} \, \text{V}$. Substituting into the formula:

$12 \times 10^{-3} = L \cdot 3.$

Step 3: Solve for $L$
Rearrange to find $L$:

$L = \frac{12 \times 10^{-3}}{3} = 4 \times 10^{-3} \, \text{H}.$

Convert to millihenries:

$L = 4 \, \text{mH}.$

Thus, the self-inductance of the inductor is $L = 4 \, \text{mH}$.