DGVCL Exam Paper (11-12-2011) The circuit breaker generally used in railway traction is Minimum Oil CB Air break CB Vacuum CB Bulk Oil CB Minimum Oil CB Air break CB Vacuum CB Bulk Oil CB 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? Suspebsion type Pin type Shakle type Strain type Suspebsion type Pin type Shakle type Strain type ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Due to skin effect Portion of the conductor near the surface carries less current and the core of the conductor carries more current Current flows through the half cross section of the conductor Uniform current flows through the conductor Portion of the conductor near the surface carries more current and the core of the conductor carries less current Portion of the conductor near the surface carries less current and the core of the conductor carries more current Current flows through the half cross section of the conductor Uniform current flows through the conductor Portion of the conductor near the surface carries more current and the core of the conductor carries less current ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Skin Effect:tendency of the current to crowd near the conductor surface.Skin effect is proportional to frequency, conductor diameter and permiability. Skin effect increases effective resistance of conductor.In DC, no skin effect.
DGVCL Exam Paper (11-12-2011) Different computers are connected to a LAN by a cable and a Modem Special wires Interface card Telephone line Modem Special wires Interface card Telephone line ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The bus bar zone faults are generally Three phase short ciruits Single line to ground faults Double line to ground faults Phase to phase faults Three phase short ciruits Single line to ground faults Double line to ground faults Phase to phase faults 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% 80 % 90 % Less than 90% More than 90% 80 % 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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