Question 5.23: Determine the equivalent inductance of the series-parallel-c......

Let current through the parallel branches be \overline{I}_{ a } \text { and } \overline{ I }_{ b }    as shown in Fig. 2.

The coupling between the series-connected coils 2 H and 3 H is additive because the current \overline{I}_{ a } leaves at the dotted end in both the coils. Hence, the magnetic coupling can be eliminated by adding the mutual inductance to the self-inductances as shown in Fig. b. Then the series-connected inductances 4 H and 5 H are combined to form a single equivalent as shown in Fig. c.
The coupling between parallel-connected coils 2 H and 9 H is opposive because the current enters at the dotted end in one coil and leaves at the dotted end in the other coil. This parallel-connected coupled coil is combined to a single equivalent as shown in Fig. d.

The equivalent inductance in parallel opposing, L_{e q}=\frac{L_1 L_2-M^2}{L_1+L_2+2 M}

Alternate Method

Alternatively, we can estimate the looking back impedance (inductive reactance) from the two terminals of the inductive network by connecting a source of value E at the two terminals of the network.
The looking back inductive reactance is the equivalent inductive reactance at the two terminals, which is given by the ratio of voltage to current at the two terminals.

Let us connect a sinusoidal voltage source of value \overline{E} as shown in Fig. 3. Let \overline{ I }_1 \text { and } \overline{ I }_2 be the mesh currents.

Now the equivalent inductive reactance, j \omega L _{ eq }=\frac{\overline{ E }}{\overline{ I }_1}

Let us name the coils as coil-A, coil-B and coil-C as shown in Fig. 3. Now, \overline{ I }_1-\overline{ I }_2 be the current through coil-A and \overline{ I }_2 be the current through coil-B and coil-C. The current flowing in each coupled coil will induce an emf in the other coil. Therefore, in the circuit of Fig. 3, there will be four mutual induced emfs as explained below :

The self- and mutual induced emfs in the coils are shown in Fig. 4.

By KVL in mesh-1,

\begin{aligned} & j 2 \omega\left(\overline{ I }_1-\overline{ I }_2\right)= j \omega \overline{ I }_2+\overline{ E } \\ & \therefore j 2 \omega \overline{ I }_1- j 3 \omega \overline{ I }_2=\overline{ E } \end{aligned}                                   …….(1)

By KVL in mesh-2,

\begin{aligned}& j \omega \bar{I}_2+ j 2 \omega \bar{I}_2+ j 2 \omega \bar{I}_2+ j 2 \omega \bar{I}_2+ j 3 \omega \bar{I}_2= j 2 \omega\left(\overline{ I }_1-\overline{ I }_2\right)+ j \omega\left(\overline{ I }_1-\overline{I }_2\right) \\ & \therefore j 3 \omega \overline{ I }_1+ j 13 \omega \overline{ I }_2=0 \end{aligned}                       …….(2)

On arranging equations (1) and (2) in matrix form, we get,

\left[\begin{array}{cc} j 2 \omega & -j 3 \omega \\ -j 3 \omega & j 13 \omega \end{array}\right]\left[\begin{array}{l} \bar{I}_1 \\\bar{I}_2\end{array}\right]=\left[\begin{array}{l} \bar{E} \\0\end{array}\right]

\text { Now, } \Delta_1=\left|\begin{array}{cc} \bar{E} & -j 3 \omega \\ 0 & j 13 \omega \end{array}\right|=\bar{E} \times j 13 \omega-0=j 13 \omega \bar{E}

\begin{aligned} & \overline{ I }_1=\frac{\Delta_1}{\Delta}=\frac{ j 13 \omega \overline{ E }}{-17 \omega^2}=\frac{13 \overline{ E }}{ j 17 \omega} \\ & \therefore \frac{\overline{ E }}{\overline{ I }_1}=\frac{ j 17 \omega}{13} \end{aligned}

RESULT

Equivalent inductance, L _{ eq }=\frac{17}{13} H=1.3077 H