Question 5.24: Determine the equivalent inductance of the inductive networ......
Determine the equivalent inductance of the inductive network with coupled coils shown in Fig. 1.
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 :
\text{Emf-1: The current $\bar{I}_1$ entering at the undotted end in coil-A will induce an emf $j 2 \omega \bar{I}_1$ in coil-C such that the undotted end is positive.} \text{Emf-2 : The current $\bar{I}_1-\bar{I}_2$ entering at the dotted end in coil-B will induce an emf $j 3 \omega\left(\overline{ I }_1-\overline{ I }_2\right)$ in coil-D such that the dotted end is positive.} \text { Emf-3 : The current } \bar{I}_2 \text { entering at the dotted end in coil-C will induce an emf } j 2 \omega \bar{I}_2 \text { in coil-A such that the dotted end is positive. } \text { Emf-4: The current } \overline{ I }_2 \text { entering at the dotted end in coil-D will induce an emf } j 3 \omega \overline{ I }_2 \text { in coil-B such that the dotted end is positive. }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}
\begin{aligned} & \overline{ I }_1=\frac{\Delta_1}{\Delta}=\frac{ j 17 \omega \overline{ E }}{-162 \omega^2}=\frac{17}{ j 162 \omega} \overline{ E } \\ & \therefore \frac{\overline{ E }}{\overline{ I}_1}=\frac{ j 162}{17} \omega \end{aligned} \text { Here, } \frac{\bar{E}}{\bar{I}_1}=j \omega L_{\text {eq }}, \quad \therefore j \omega L_{\text {eq }}=\frac{j 162}{17} \omega \Rightarrow \quad L_{\text {eq }}=\frac{162}{17} HRESULT
Equivalent inductance, L _{ eq }=\frac{162}{17} H=9.5294 H
