Question

The inductor arrangement of Figure, with

${\mathbf{L}}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{mH}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{mH}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{mHandL}}_{\mathbf{4}}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{mH}$is to be connected to a varying current source. What is the equivalent inductance of the arrangement?

Short answer

$L_{eq}=59.3mH$

Step-by-step solution

Given

$L_1=30.0mH;L_2=50.0mH;L_3=20.0mH;L_4=15.0mH$

Understanding the concept

Equivalent inductance forinductors connected in series isthesum of all inductances. And reciprocal of equivalent inductance for parallel connection of inductances isthesum of reciprocals of inductances.

Formulae:

Equivalent inductance forinductors connected inparallel is

$\frac{1}{L_{eq}}=\sum _{n=1}^N\frac{1}{L_n}$

Equivalent inductance forinductors connected inseries is

$L_{eq}=\sum _{n=1}^N{L}_n$

Calculate the equivalent inductance of the arrangement

$L_2$and $L_3$are connected in parallel, so their equivalent is given by

$$\begin{gathered} \frac{1}{L_{23}}=\frac{1}{L_2}+\frac{1}{L_3} \\ \frac{1}{L_{23}}=\frac{1}{50}+\frac{1}{20} \end{gathered}$$

Hence,

$L_{23}=14.3mH$

Now, $L_{23},L_{1}and{L}_4$are in series.

$$\begin{gathered} L_{eq}=L_1+L_{23}+L_4 \\ {L}_{eq}=30+14.3+15 \\ {L}_{eq}=59.3m{H}_z \end{gathered}$$

Therefore the equivalent inductance = 59.6mH