Question 5.24: Determine the equivalent inductance of the inductive networ......

Let us connect a sinusoidal voltage source of value, $\overline{E}$ as shown in Fig. 2. Let $\bar{I}_1 \text { and } \bar{I}_2$ be the mesh currents.

Now, Equivalent inductive reactance, $j \omega L_{e q}=\frac{\overline{E}}{\overline{I}_1}$

Let us name the coils as shown in Fig. 2. Now, $\overline{I}_1$ is the current through coil-A, $\overline{ I }_1-\overline{ I }_2$ is the current through coil-B and $\overline{I}_2$ is the current through coils C and D.

The current flowing in each coupled coil will induce an emf in the other coil. Therefore, in the circuit of Fig. 2, there will be four mutual induced emfs as explained below :

The self- and mutual induced emfs in the coils are shown in Fig. 3.

By KVL in mesh-1,

$\begin{aligned} & j 5 \omega \overline{ I }_1+ j 6 \omega\left(\overline{ I }_1-\overline{ I }_2\right)+ j 3 \omega \overline{ I }_2= j 2 \omega \overline{ I }_2+\overline{ E } \\ & \therefore j 11 \omega \overline{ I }_1- j 5 \omega \overline{ I }_2=\overline{ E } \end{aligned}$                  …..(1)

By KVL in mesh-2,

$\begin{aligned} & j 8 \omega \overline{ I }_2+ j 9 \omega \overline{ I }_2+ j 3 \omega\left(\overline{ I }_1-\overline{ I }_2\right)= j 6 \omega\left(\overline{ I }_1-\overline{ I }_2\right)+ j 3 \omega \overline{ I }_2+ j 2 \omega \overline{ I }_1 \\ & \therefore- j 5 \omega \overline{ I }_1+ j 17 \omega \overline{ I }_2=0 \end{aligned}$      …..(2)

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

$\left[\begin{array}{cc} j 11 \omega & – j 5 \omega \\ – j 5 \omega & j 17 \omega \end{array}\right]\left[\begin{array}{l} \overline{ I }_1 \\ \overline{ I }_2 \end{array}\right]=\left[\begin{array}{l} \overline{ E } \\ 0 \end{array}\right]$

$\text { Now, } \Delta_1=\left|\begin{array}{cc} \bar{E} & -j 5 \omega \\ 0 & j 17 \omega \end{array}\right|=\bar{E} \times j 17 \omega-0=j 17 \omega \bar{E}$

$\begin{aligned}\Delta=\left|\begin{array}{rr} j 11 \omega & -j 5 \omega \\ -j 5 \omega & j 17 \omega \end{array}\right| & =j 11 \omega \times j 17 \omega-(-j 5 \omega)^2 \\ & =-187 \omega^2+25 \omega^2=-162 \omega^2 \end{aligned}$

RESULT

Equivalent inductance, $L _{ eq }=\frac{162}{17} H=9.5294 H$