(a) Nominal π-method will be used to solve this problem. Capacitive susceptance (or admittance) at each end = Y /2 = 3 × 10−4 /2 = 1.5 × 10−4 siemens as shown in Fig.
As seen from Fig.
Line value of sending-end voltage
Angle between VS and IS = 41.6º + 3.2º = 44.8º;
cos φS = cos 44.8º = 0.71 (lag)
Power input = √3 × 122.2 × 103 × 195.7 × 0.7 = 29.41 MW
Power output = 40 × 0.7 = 28 MW
∴ power transmission η = 28/29.41 = 0.952 or 95.2%
(b) vector diagram is similar to that shown in Fig.(b).
(c) As seen from Fig.under no-load condition, current in the conductor is