DGVCL Exam Paper (11-12-2011) First zone distance relay protects the transmission line section upto Its full length 80 % of its full length 50 % of its full length 25 % of its full length Its full length 80 % of its full length 50 % of its full length 25 % of its full length ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Two parallel conductors are carrying the current in the same direction. Which of the following statements is correct? There is an attracting force between the conductors There is a repelling force between the conductors The force increases when the distance between the conductors is increased The force increases when the distance between the conductors is decreased There is an attracting force between the conductors There is a repelling force between the conductors The force increases when the distance between the conductors is increased The force increases when the distance between the conductors is decreased 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 25 50 15 10 25 50 15 10 ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The purpose of providing a choke in a tube light is 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 To limit the current to an appropriate value To eliminate corona effect To improve the power factor ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The transient stability of a power system can be improved by Excitation control Increasing the turbine valve opening Single pole switching of circuit breakers Phase shifting transformer Excitation control Increasing the turbine valve opening Single pole switching of circuit breakers Phase shifting transformer 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 More than 90% Less than 90% 80 % 90 % More than 90% Less than 90% 80 % 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
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