Question

Two inductors, each of inductance $L$ are connected in parallel. One more inductor of inductance $5\text{mH}$ is connected in series with this combination. Now, the effective inductance is $15\text{mH}$. The value of $L$ is

• A. $10\text{mH}$
• B. $50\text{mH}$
• C. $15\text{mH}$
• D. $20\text{mH}$

Answer 1

The correct option is D $20\text{mH}$

The given configuration is shown in the figure.

For inductors between points $A$ and $B$, net inductance is,

$L_{\text{eq}}=\frac{L_1{L}_2}{L_1+L_2}=\frac{L\times L}{L+L}=\frac{L}{2}$

The configuration reduced into the below shown circuit part.


For inductors between points $A$ and $C$, net inductance,

$L_{\text{eq}}=L_1+L_2$

$\Rightarrow 15=\frac{L}{2}+5$

$\Rightarrow L=20\text{mH}$

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