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 More than 90% 90 % 80 % Less than 90% More than 90% 90 % 80 % 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) An AC series circuit has R = 6 Ω, XL = 20 Ω and Xc = 12 Ω. The ciruit power factor will be 0.8 lagging 0.5 leading 0.6 leading 0.6 lagging 0.8 lagging 0.5 leading 0.6 leading 0.6 lagging ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) If the fault current is 2000 A, the relay plug setting is 50% and the CT ratio is 400/5, the plug setting multiple will be 10 50 15 25 10 50 15 25 ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) DC series motors are best suited for traction work because Its torque is proportional to the square of the armature current and speed is inversely proportional to torque Torque and speed are proportional to the square of the armature current Torque and speed are inversely proportional to the square of the armature current Its torque is proportional to the square of the armature current and speed is directional proporational to torque Its torque is proportional to the square of the armature current and speed is inversely proportional to torque Torque and speed are proportional to the square of the armature current Torque and speed are inversely proportional to the square of the armature current Its torque is proportional to the square of the armature current and speed is directional proporational to torque ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Torque in d.c motor = k* φ * IaIn series motor φ α IaTα I²N α 1/φN α 1/T
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 435.6 MW 500 MW 217.8 MW 251.5 MW 435.6 MW 500 MW 217.8 MW 251.5 MW ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) It is difficult interrupt a capacitive current because It has a leading power factor Stored energy in the capacitor is very high The restriking voltage can be high Current magnitude is very small It has a leading power factor Stored energy in the capacitor is very high The restriking voltage can be high Current magnitude is very small ANSWER DOWNLOAD EXAMIANS APP
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