A one line diagram of a simple electrical system is shown below, together with the per unit equipment reactances. You may assume that all network resistances are negligible. Draw each of the three sequence networks and calculate the rms fault current resulting from: (1) a line-to- earth fault at bus no. 1, (ii) a line-to-line fault at bus no. 3, and (iii) a line-to-line to earth fault at bus no. 2. [25 marks] G Gen. 1 V=1/0 pu X=0.6 pu X-0.2 pu Xo-0.05 pu X=0.1 pu bus 1 LA bus 2 T X=0.08 pu X=0.08 pu X-0.08 pu Answers: 5.017 pu; 2.797 pu; and 2.96 pu. Transmission Line X-0.16 pu X-0.16 pu Xo-0.06 pu bus 3 AL T X=0.1 pu X=0.1 pu Xo-0.1 pu bus 4 Gen. 2 V=1/0 pu X-0.7 pu X=0.3 pu X-0.06 pu X=0.1 pu