Question 5.29: Calculate the primary and secondary current in the coupled ......
Calculate the primary and secondary current in the coupled circuit shown in Fig. 1.
The current \overline{I}_p entering at the dotted end in the primary coil will induce an emf j 10\overline{I}_p in the secondary such that the dotted end is positive. The current \overline{I}_s entering at the dotted end in the secondary
coil will induce an emf 10\overline{I}_sin the primary such that the dotted end is positive. The self- and mutual induced emfs are shown in Fig. 2.
By KVL in primary we get,
\begin{aligned} & 6 \bar{I}_p+ j 12 \overline{ I }_p+ j 10 \overline{ I }_{ s }+\left(- j 3 \overline{ I }_{ p }\right)=220 \\ & \therefore(6+ j 9) \overline{ I }_{ p }+ j 10 \overline{ I }_{ s }=220 \end{aligned} …..(1)
By KVL in secondary we get,
\begin{aligned} & j 10 \overline{ I }_{ p }+ j 40 \overline{ I }_{ s }+8 \overline{ I }_{ s }+\left(- j 20 \overline{ I }_{ s }\right)=0 \\ & \therefore j 10 \overline{ I }_{ p }+(8+ j 20) \overline{ I }_{ s }=0 \end{aligned} …..(2)
On arranging equations (1) and (2) in matrix form we get,
\left[\begin{array}{rr} 6+ j 9 & j 10 \\ j 10 & 8+ j 20 \end{array}\right]\left[\begin{array}{l} \bar{I}_p \\ \bar{I}_{ s } \end{array}\right]=\left[\begin{array}{r} 220 \\ 0 \end{array}\right]
Alternatively, the mesh basis matrix equation can be obtained from the electrical equivalent of the coupled coils shown in Fig. 3. (Here it is group-II coupled coil) With reference to Fig. 3, we can write,
\left[\begin{array}{rr} 6+ j 22- j 10- j 3 & -(- j 10) \\ -(- j 10) & – j 10+ j 50+8- j 20 \end{array}\right]\left[\begin{array}{l} \bar{I}_{ p } \\ \bar{I}_{ s } \end{array}\right]=\left[\begin{array}{r} 220 \angle 0^{\circ} \\ 0 \end{array}\right]
\left[\begin{array}{rr} 6+ j 9 & j 10 \\ j 10 & 8+ j 20 \end{array}\right]\left[\begin{array}{l} \bar{I}_{ p } \\ \bar{I}_{ s } \end{array}\right]=\left[\begin{array}{r} 220 \\ 0 \end{array}\right]
\text { Now, } \quad \Delta=\left|\begin{array}{rr} 6+j 9 & j 10 \\ j 10 & 8+j 20 \end{array}\right|=(6+j 9) \times(8+j 20)-(j 10)^2=-32+j 192
\begin{aligned} & \Delta_p=\left|\begin{array}{rr} 220 & j 10 \\ 0 & 8+j 20 \end{array}\right|=220 \times(8+ j 20)-0=1760+ j 4400 \\ & \Delta_{ S }=\left|\begin{array}{rr} 6+ j 9 & 220 \\ j 10 & 0 \end{array}\right|=0- j 10 \times 220=- j 2200 \end{aligned}
\begin{aligned} &\overline{ I }_{ p }=\frac{\Delta_{ p }}{\Delta}=\frac{1760+ j 4400}{-32+ j 192}=20.8108- j 12.6351 A =24.3462 \angle-31.3^{\circ} A \\ &\overline{ I }_{ s }=\frac{\Delta_{ s }}{\Delta}=\frac{- j 2200}{-32+ j 192}=-11.1486+ j 1.8581 A =11.3024 \angle 170.5^{\circ} A \end{aligned}
