DGVCL Exam Paper (11-12-2011) Permittivity is expressed in terms of N/m Farad/m farad/sq. m. Weber /m N/m Farad/m farad/sq. m. Weber /m ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) The form factor in AC means the ration of Peak value to r.m.s value Peak value to average value R.M.S. value to average value R.M.S. value to peak value Peak value to r.m.s value Peak value to average value R.M.S. value to average value R.M.S. value to peak value ANSWER DOWNLOAD EXAMIANS APP
DGVCL Exam Paper (11-12-2011) Bulk power transmission over long HVDC lines are preferred on account of Minimum line power losses Low cost of HVDC terminals No harmonic problems Simple protection Minimum line power losses Low cost of HVDC terminals No harmonic problems Simple protection 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 Uniform current flows through the conductor Current flows through the half cross section of 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 Uniform current flows through the conductor Current flows through the half cross section of 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) Kilobyte, megabyte and gigabyte are measuring units used in a computer. How many megabytes make one gigabyte? 1204 1042 1000 1024 1204 1042 1000 1024 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% 90 % 80 % More than 90% Less 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
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