Question

Two coils $\text{X}$ and $\text{Y}$ are placed in a circuit such that when the current changes by $2\text{A}$ in coil $\text{X}$ the magnetic flux changes by $0.4\text{Wb}$ in $\text{Y}$. The value of mutual inductance of the coils is-

• A. $0.2\text{H}$
• B. $5\text{H}$
• C. $0.8\text{H}$
• D. $4\text{H}$

Answer 1

The correct option is A $0.2\text{H}$

From the principle of mutual inductance,

${\varphi }_2=M{i}_1\text{(or)}$

$\Delta {\varphi }_2=M\left(\Delta i_1\right)$

$\Rightarrow M=\frac{\Delta {\varphi }_2}{\Delta i_1}....\left(1\right)$

Where, ${\varphi }_2=\text{Flux through the second coil}{i}_1=\text{Current in the first coil}M=\text{Co-efficient of mutual inductance}$

Here, $\Delta i_1=2\text{A};\Delta {\varphi }_2=0.4\text{Wb}$

Putting this values in (1) we get,

$M=\frac{\Delta {\varphi }_2}{\Delta i_1}=\frac{0.4}{2}=0.2\text{H}$

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