Example 7.3. Replace branches b and c between node-pairs ① (3) and ③ of Fig. 7.5 by the mutually coupled branches of Fig. 7.7. Then, find Ybus and the nodal equations of the new network. Solution. The admittance diagram of the new network including the mutual coupling is shown in Fig. 7.11. From Example 7.2 we know that the mutually coupled branches have the nodal admittance matrix. (3) T -j6.25 j3.75 j2.50 2 j3.75 -j6.25 j2.50 3 j2.50 j2.50 -j5.00 which corresponds to the equivalent circuit shown encircled in Fig. 7.12. The remaining portion of Fig. 7.12 is drawn from Fig. 7.5. Since mutual coupling is not evident in Fig. 7.12, we may apply the standard rules of Yus formation to the 0000 -j8.0 -j4.0 j5.0 0000 0000 Ale-j4.0 00002 -j2.5I/A (1) 1.00/-90° -j0.8 j0.8 0.68135 0 FIGURE 7.5 Per-unit admittance diagram for Fig. 7.4 with current sources replacing voltage sources. Branch names a to g correspond to the subscripts of branch voltages and currents. (3 2 j0.25 000 (2 -6.25 000 j0.15 j3.75 (3) ண j0.25 (a) 1.00/-90° 000 -j6.25 (6) FIGURE 7.7 The two mutually coupled branches of Example 7.2, their (a) primitive impedances and (b) primitive admittances in per unit. 0000 -j6.25 -j8.0 j3.75 -j5.0 0000 00006.25 V 0000 -j2.5 -0.8 140 -j0.8 0.68/-135 00002 FIGURE 7.11 Per-unit admittance diagram for Example 7.3. How exactly are we getting these values for the admittance? 1.00/-90° Neele j3.75 j8.0 -j2.5 -j5.0 00002 -j2.5 0000 -j2.5 0000 -j0.8 ⑩ -0.8 0.68-135° FIGURE 7.12 Nodal admittance network for Example 7.3. The shaded portion represents two mutually coupled branches connected between buses ①. 2. and ③. Θ (③ How are we getting this table? Θα -j16.75 j11.75 j2.50 j11.75 -j19.25 j2.50 j2.50 j5.00 V₂ 0 0 j2.50 j2.50 0 -j5.80 1.00/-90° 4 j2.50 j5.00 0 -j8.30 V. 0.68 -135° Note that the two admittances between nodes ① and 2 combine in parallel to yield Y12 (13.75-j8.00) -j11.75