Chapter 32, Problem 83CP

To determine

The equivalent inductance for the system.

Answer to Problem 83CP

The equivalent inductance for the system is $\frac{L_1{L}_2-M^2}{L_1+L_2-2M}$.

Explanation of Solution

The flow of current in the circuit is as shown in the figure below.

Figure-(1)

Here, $i$, $i_1$ and $i_2$ are the flow of current as shown in the figure below.

Write the expression based on junction rule.

    ${\displaystyle \sum _{\text{junction}}I}=0$

Here, ${\displaystyle \sum _{\text{junction}}I}$ is the sum current at the junction.

Write the expression to obtain the loop rule.

    ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}=0$

Here, ${\displaystyle \sum _{\text{closed}\text{loop}}\Delta V}$ is the potential drop across each element in a closed circuit.

Write the expression based on junction rule to obtain the current division in the circuit.

    $i=i_1+i_2$                                                                                                         (I)

Here, $i$ is the total current through the battery, $i_1$ is the current in $L_1$ inductor and $i_2$ is the current in $L_2$ inductor.

Differentiate the above equation with respect to time $t$ to find the current with respect to time.

    $\frac{di}{dt}=\frac{d{i}_1}{dt}+\frac{d{i}_2}{dt}$                                                                                               (II)

Write the expression based on loop rule to obtain the potential drop in the left loop.

    $V=L_1\frac{d{i}_1}{dt}+M\frac{d{i}_2}{dt}$                                                                                          (III)

Write the expression based on loop rule to obtain the potential drop in the right loop.

    $V=L_2\frac{d{i}_2}{dt}+M\frac{d{i}_1}{dt}$                                                                                           (IV)

As the inductor $L_1$ and $L_2$ are connected in parallel, thus the voltage across the $L_1$ and $L_2$ is same.

Compare equation (III) and (IV).

    $L_1\frac{d{i}_1}{dt}+M\frac{d{i}_2}{dt}=L_2\frac{d{i}_2}{dt}+M\frac{d{i}_1}{dt}$

Further solve the above equation.

    $\begin{array}{c}{L}_1\frac{d{i}_1}{dt}-M\frac{d{i}_1}{dt}=L_2\frac{d{i}_2}{dt}-M\frac{d{i}_2}{dt}\\ \left(L_1-M\right)\frac{d{i}_1}{dt}=\left(L_2-M\right)\frac{d{i}_2}{dt}\\ \frac{d{i}_1}{dt}=\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}\end{array}$

Substitute $\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}$ for $\frac{d{i}_1}{dt}$ in equation (II).

    $\begin{array}{c}\frac{di}{dt}=\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}+\frac{d{i}_2}{dt}\\ =\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}+1\right)\frac{d{i}_2}{dt}\end{array}$                                                                                  (V)

Write the expression to obtain the voltage across the circuit.

    $V=L_{\text{eq}}\frac{di}{dt}$

Here, $V$ is the voltage across the circuit and $L_{\text{eq}}$ is the equivalent inductance in the circuit.

Substitute $L_1\frac{d{i}_1}{dt}+M\frac{d{i}_2}{dt}$ for $V$ in the above equation.

    $L_{\text{eq}}\frac{di}{dt}=L_1\frac{d{i}_1}{dt}+M\frac{d{i}_2}{dt}$

Further substitute $\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}$ for $\frac{d{i}_1}{dt}$ in the above equation.

    $\begin{array}{c}{L}_{\text{eq}}\frac{di}{dt}=L_1\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}\right)+M\frac{d{i}_2}{dt}\\ \frac{di}{dt}=\frac{1}{L_{\text{eq}}}\left[L_1\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}\right)+M\frac{d{i}_2}{dt}\right]\end{array}$                                                             (VI)

Compare equation (V) and (VI).

    $\begin{array}{c}\frac{1}{L_{\text{eq}}}\left[L_1\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\frac{d{i}_2}{dt}\right)+M\frac{d{i}_2}{dt}\right]=\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}+1\right)\frac{d{i}_2}{dt}\\ \frac{1}{L_{\text{eq}}}\left[L_1\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\right)+M\right]=\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}+1\right)\\ L_{\text{eq}}=\frac{L_1\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}\right)+M}{\left(\frac{\left(L_2-M\right)}{\left(L_1-M\right)}+1\right)}\\ =\frac{\frac{L_1{L}_2-M{L}_1+M{L}_1-M^2}{L_1-M}}{\frac{L_2-M+L_1-M}{L_1-M}}\end{array}$

Further solve the above equation.

    $L_{\text{eq}}=\frac{L_1{L}_2-M^2}{L_1+L_2-2M}$

Therefore, the equivalent inductance for the system is $\frac{L_1{L}_2-M^2}{L_1+L_2-2M}$.