Question

The mutual inductance between a primary and secondary coil is $0.5\text{H}$. The resistance of the primary and the secondary coils are $20\Omega$ and $5\Omega$ respectively. To generate a current of $0.4\text{A}$ in the secondary coil, current in the primary must be changed at a rate of-

• A. $4.0{\text{As}}^{-1}$
• B. $1.6{\text{As}}^{-1}$
• C. $16.0{\text{As}}^{-1}$
• D. $8.0{\text{As}}^{-1}$

Answer 1

The correct option is A $4.0{\text{As}}^{-1}$

Given,
$M=0.5\text{H};R_p=20\Omega R_s=5\Omega ;i_s=0.4\text{A}$

From, the principle of mutual inductance,

${\varphi }_s=M{i}_p$

On differentiating both sides w.r.t time we get, induced emf as,

$\left|\begin{array}{c}{E}_s\end{array}\right|=\frac{d{\varphi }_s}{dt}=M\left(\frac{d{i}_p}{dt}\right)$

$\Rightarrow E_s=M\left(\frac{d{i}_p}{dt}\right)$

$\Rightarrow i_s{R}_s=M\left(\frac{d{i}_p}{dt}\right)$

$\Rightarrow \left(\frac{d{i}_p}{dt}\right)=\frac{i_s{R}_s}{M}=\frac{0.4\times 5}{0.5}=4{\text{As}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.