Question

When a current of $4\text{A}$ changes to $12\text{A}$ in $0.5\text{s}$ in the primary coil, it induces an emf of $50\text{mV}$ in the secondary coil. The mutual inductance is -

• A. $5.125\text{mH}$
• B. $2.125\text{mH}$
• C. $3.125\text{mH}$
• D. $4.125\text{mH}$

Answer 1

The correct option is C $3.125\text{mH}$

Given:

Change in current in primary coil,

$d{i}_1=12-4=8\text{A}$

Time taken, $dt=0.5\text{s}$

Induced emf in secondary coil,

$E_2=50\times {10}^{-3}\text{V}$

So, mutual inductance is,

$M=\frac{E_2}{d{i}_1/dt}=\frac{E_{2}dt}{d{i}_1}$

$=\frac{50\times {10}^{-3}\times 0.5}{8}$

$=3.125\times {10}^{-3}\text{H}=3.125\text{mH}$

Hence, option $\left(B\right)$ is the correct answer.