DGVCL Exam Paper (11-12-2011) Which type of back up protection is widely used for the protection of transmission lines? Relay backup Breaker backup Remote backup Control backup Relay backup Breaker backup Remote backup Control backup ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) If the percentage reactance of the system upto the point of fault is 20% and the base kVA is 10,000 the short ciruit kVA is 60 MVA 10 MVA 50 MVA 20 MVA 60 MVA 10 MVA 50 MVA 20 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Short circuit kVA = Base kVA/%X
DGVCL Exam Paper (11-12-2011) It is difficult interrupt a capacitive current because It has a leading power factor Current magnitude is very small The restriking voltage can be high Stored energy in the capacitor is very high It has a leading power factor Current magnitude is very small The restriking voltage can be high Stored energy in the capacitor is very high ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) If a 5 A fuse is blown frequently in a house lighting circuit, it is required to Insert a fuse of higher voltage rating Reduce the load Insert a fuse of higher current rating Check the load Insert a fuse of higher voltage rating Reduce the load Insert a fuse of higher current rating Check the load ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The purpose of interpoles in DC machines is to nullify The demagnetizing effect of armature mmf The cross-magnetizing effect of armature mmf Both the cross-magnetizing mmf and the reactance voltage The reactance voltage The demagnetizing effect of armature mmf The cross-magnetizing effect of armature mmf Both the cross-magnetizing mmf and the reactance voltage The reactance voltage ANSWER EXPLANATION DOWNLOAD EXAMIANS APP interpoles is widely use to improve commutationplace between main poles fitted to yokefunctions:to neutralize armature mmf in commutating zoneto produce some flux in commutating zone
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 90 % 80 % More than 90% Less than 90% 90 % 80 % 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
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