GETCO Exam Paper (11-06-2017) I _____ tennis every Sunday morning playing am play play am playing playing am play play am playing ANSWER DOWNLOAD EXAMIANS APP
GETCO Exam Paper (11-06-2017) The efficiency of a nuclear power plant is less than that of a conventional fuel fired thermal plant because of less rejection of heat in the condenser low temperature and pressure conditions higher temperature conditions higher pressure conditions less rejection of heat in the condenser low temperature and pressure conditions higher temperature conditions higher pressure conditions ANSWER DOWNLOAD EXAMIANS APP
GETCO Exam Paper (11-06-2017) In air blast circuit breaker resistance switching is used to damp out the fast transient change the fault current power factor control the CB operating time reduce the magnitude of fault current damp out the fast transient change the fault current power factor control the CB operating time reduce the magnitude of fault current ANSWER DOWNLOAD EXAMIANS APP
GETCO Exam Paper (11-06-2017) By increasing the transmission voltage to double of its original value the same power can be dispatched half the original value equal to original value double the original value one fourth of original value half the original value equal to original value double the original value one fourth of original value ANSWER DOWNLOAD EXAMIANS APP
GETCO Exam Paper (11-06-2017) In a super heater pressure rises and temperature remains the same temperature rises and pressure drops pressure rises and temperature drops temprature rises and pressure remains unchanged pressure rises and temperature remains the same temperature rises and pressure drops pressure rises and temperature drops temprature rises and pressure remains unchanged ANSWER DOWNLOAD EXAMIANS APP
GETCO Exam Paper (11-06-2017) If the reading of two wattmeters are equal and positive in two wattmeter method, the load power factor in abalanced 3-phase 3-wire circuit will be 0.866 zero unity 0.5 0.866 zero unity 0.5 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP W1 = W2 (+Ve) → PF = 1W1, W2 Positive → PF > 0.5W1 +Ve, W2 = 0 → PF = 0.5W1 +Ve, W2 = -Ve → PF < 0.5W1 = - W2 → PF = 0
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