DGVCL Exam Paper (11-12-2011) Buchholz relay causes Alarm for both major and minor faults Tripping for both major and minor faults Tripping for major faults and alarm for minor faults Tripping for minor faults and alarm for major faults Alarm for both major and minor faults Tripping for both major and minor faults Tripping for major faults and alarm for minor faults Tripping for minor faults and alarm for major faults ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) If the voltage at the two ends of a line are 132 kV, and its reactance is 40 Ω, the stability limit of the line is 217.8 MW 435.6 MW 251.5 MW 500 MW 217.8 MW 435.6 MW 251.5 MW 500 MW ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The purpose of providing a choke in a tube light is To limit the current to an appropriate value To eliminate corona effect To improve the power factor To avoid radio interference To limit the current to an appropriate value To eliminate corona effect To improve the power factor To avoid radio interference 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 primary ciruits It protects the secondary ciruits It is safe to human beings It avoids core saturation and high voltage induction It protects the primary ciruits It protects the secondary ciruits ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) What type of insulator is used if the direction of the transmission line is to be changed? Shakle type Strain type Pin type Suspebsion type Shakle type Strain type Pin type Suspebsion type ANSWER DOWNLOAD EXAMIANS APP
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 Less than 90% 80 % More than 90% 90 % Less than 90% 80 % More than 90% 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
EXAMIANS