Question

Two inductances connected in parallel are equivalent to a single inductance of $1.5\text{H}$ and when connected in series are equivalent to a single inductance of $8\text{H}.$ The difference in their inductance is-

• A. $3\text{H}$
• B. $7.5\text{H}$
• C. $2\text{H}$
• D. $4\text{H}$

Answer 1

The correct option is D $4\text{H}$

Let, $L_1$ and $L_2$ be the individual inductors,

When, $L_1$ and $L_2$ are connected in parallel, their equivalent inductance is,

$\frac{L_1{L}_2}{L_1+L_2}=1.5........\left(1\right)$

When, $L_1$ and $L_2$ are connected in series, their equivalent inductance is,

$L_1+L_2=8.........\left(2\right)$

From (1) and (2) we get,

$L_1{L}_2=12$

And, we can write,

$(L_1-L_2{)}^2=(L_1+L_2{)}^2-4L_1{L}_2$

$=(8{)}^2-4\times 12=64-48=16$

$\therefore L_1-L_2=4\text{H}$

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