MGVCL Exam Paper (30-07-2021 Shift 3)
The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil.
Die-electric test on Circuit breaker: These tests are performed to check power frequency and impulse voltage withstand capacity. Power frequency tests are kept on a new circuit breaker, the test voltage changes with a circuit breaker rated voltage. In impulse tests, impulse voltage of particular value is employed to the breaker.
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
The ratio of full load current to short circuit current = 1/6 Xsc = j/(1/6) External reactance required = j*((1/6) - 0.06)) = j*0.106 pu Full load current = (40*1000)/(√3*15) = 1539.6 A Per unit reactance = j*0.106 = (I*Xr)/V j*0.15 = (1539.6*Xb)/((15/√3)*1000)) = 0.60 ohm
EXAMIANS