DGVCL Exam Paper (11-12-2011) Ybus used in power flow studies and Zbus used in short circuit studies Are transpose of each other Are the same Are inverse of each other Have no inter relation Are transpose of each other Are the same Are inverse of each other Have no inter relation 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% More than 90% 90 % 80 % Less than 90% More than 90% 90 % 80 % 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) The purpose of providing a choke in a tube light is To limit the current to an appropriate value To eliminate corona effect To avoid radio interference To improve the power factor To limit the current to an appropriate value To eliminate corona effect To avoid radio interference To improve the power factor ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Which motor is used to start heavy loads? Shunt motor Cumulatively compounded motor Series motor Differentially compounded motor Shunt motor Cumulatively compounded motor Series motor Differentially compounded motor ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Torque in d.c motor = k* φ * IaIn series motor φ α IaTα I²starting torque of dc series motor is high
DGVCL Exam Paper (11-12-2011) The main advantage of using fractional pitch winding is to refuce Cost of the machine Amount of copper in the winding Size of the machine Harmonics in the generated emf Cost of the machine Amount of copper in the winding Size of the machine Harmonics in the generated emf ANSWER EXPLANATION DOWNLOAD EXAMIANS APP due to fractional pitch EMF can be approximately made into sine wavelosses due to harmonics reducedsaving in conductor materialmechanical strength of coil is increased.
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 20 MVA 50 MVA 60 MVA 10 MVA 20 MVA 50 MVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Short circuit kVA = Base kVA/%X
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