DGVCL Exam Paper (11-12-2011) The efficiency of a transformer at full load, 0.8 p.f. lag is 90%. Its efficiency at full load, 0.8 p.f. lead will be 80 % 90 % More than 90% Less than 90% 80 % 90 % More than 90% Less than 90% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Efficiency= output power/input powerefficiency = (v*i*cosφ)/((v*i*cosφ)+(x² Pc)+(Pi)) x=fractional part of full load
DGVCL Exam Paper (11-12-2011) Which type of back up protection is widely used for the protection of transmission lines? Breaker backup Control backup Remote backup Relay backup Breaker backup Control backup Remote backup Relay backup ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Under operating conditions, the secondary of a current transformer is always short ciruited because It is safe to human beings It avoids core saturation and high voltage induction It protects the secondary ciruits It protects the primary ciruits It is safe to human beings It avoids core saturation and high voltage induction It protects the secondary ciruits It protects the primary ciruits ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Bulk power transmission over long HVDC lines are preferred on account of Minimum line power losses Simple protection Low cost of HVDC terminals No harmonic problems Minimum line power losses Simple protection Low cost of HVDC terminals No harmonic problems ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The best location of the power factor correction equipment to be installed in the transmission line is at The receiving end Any place Middle of the line The sending end The receiving end Any place Middle of the line The sending end ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) A 3 phase, 6 pole induction motor operates on 440 V, 50 Hz, supply. If the actual speed of the motor is 960 rpm the slip will be 4 % 6 % 0.4 % 5 % 4 % 6 % 0.4 % 5 % ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Ns = 120f/PSlip=(Ns-Nr)/NsNs = 1000 and Nr = 960 from equation of slip, S = (1000-960)/1000S = 0.04 or 4%
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