Question 5.28: Determine the equivalent conductively coupled circuit for th......
Determine the equivalent conductively coupled circuit for the magnetically coupled circuit as shown Fig. 1 and solve the mesh currents.
Let us name the coupled coils as coil-A and coil-B as shown in Fig. 2. The current \overline{I}_1 entering at the dotted end in coil-A will induce an emf j6\overline{I}_1 in coil-B such that the dotted end is positive. The current entering at the undotted end in coil-B will induce an emf j6\overline{I}_2 in coil-A such that the undotted end is positive.
The self- and mutual induced emfs in the coupled coils are shown in Fig.2.
By KVL in mesh-1,
j 5 \bar{I}_1+3\left(\bar{I}_1-\bar{I}_2\right)+\left[- j 4\left(\bar{I}_1-\bar{I}_2\right)\right]=50+ j 6 \bar{I}_2
\begin{aligned} & j 5 \overline{ I }_1+3 \overline{ I }_1-3 \overline{ I }_2- j 4 \overline{ I }_1+ j 4 \overline{ I }_2- j 6 \overline{ I }_2=50 \\ & \therefore(3+ j ) \overline{ I }_1-(3+ j 2) \overline{ I }_2=50 \end{aligned} …..(1)
By KVL in mesh-2,
\begin{aligned} & j 10 \overline{ I }_2+5 \overline{ I }_2=- j 4\left(\overline{ I }_1-\overline{ I }_2\right)+3\left(\overline{ I }_1-\overline{ I }_2\right)+ j 6 \overline{ I }_1 \\ & j 10 \overline{ I }_2+5 \overline{ I }_2+ j 4 \overline{ I }_1- j 4 \overline{ I }_2-3 \overline{ I }_1+3 \overline{ I }_2- j 6 \overline{ I }_1=0 \\ & \therefore-(3+ j 2) \overline{ I }_1+(8+ j 6) \overline{ I }_2= \end{aligned} …(2)
Using equations (1) and (2) the conductively coupled equivalent circuit can be drawn as shown in Fig. 3.
On arranging equations (1) and (2) in matrix form we get,
\left[\begin{array}{cc} 3+ j & -(3+ j 2) \\ -(3+ j 2) & 8+ j 6\end{array}\right]\left[\begin{array}{l} \bar{I}_1 \\ \overline{ I }_2 \end{array}\right]=\left[\begin{array}{c} 50 \\0\end{array}\right]
\text { Now, } \Delta=\left|\begin{array}{cc} 3+ j & -(3+ j 2) \\ -(3+ j 2) & 8+ j 6 \end{array}\right|=(3+ j ) \times(8+ j 6)-(3+ j 2)^2=13+ j 14\begin{aligned} & \Delta_1=\left|\begin{array}{rr} 50 & -(3+j 2) \\ 0 & 8+j 6 \end{array}\right|=50 \times(8+j 6)-0=400+j 300 \\ & \Delta_2=\left|\begin{array}{cc} 3+j & 50 \\ -(3+j 2) & 0 \end{array}\right|=0-[-(3+j 2) \times 50]=150+j 100 \end{aligned}
\begin{aligned} \therefore & \overline{ I }_1=\frac{\Delta_1}{\Delta}=\frac{400+ j 300}{13+ j 14}=25.7534- j 4.6575 A =26.1712 \angle-10.3^{\circ} A \\ & \overline{ I }_2=\frac{\Delta_2}{\Delta}=\frac{150+ j 100}{13+ j 14}=9.1781- j 2.1918 A =9.4432 \angle-13.4^{\circ} A \end{aligned}
