MGVCL Exam Paper (30-07-2021 Shift 3)
A DC shunt generator has an induced voltage of 220 V on open circuit. When the machine is on load the terminal voltage is 200 V. Find the load current if the field resistance is 100 Ω and armature resistance is 0.2 Ω.
Voltage equation for Dc shunt generator, Eb = V + Ia*Ra V = Ish*Rsh Ish = 200/100 = 2 A Eb = V + Ia*Ra 20 = Ia*Ra Ia = 20/0.2 = 100 A I = 100 - 2 = 98 A
Here auto transformer change voltage from 600 V to 200 V. Voltage ratio (a) = 600/200 = 3 kVA rating of the auto-transformer = (a/(a - 1))*(kVA rating of auto transformer) = (3/2)*20 = 30 kVA
Points should be consider for new transformer installation: Location Environmental impact Soils condition Survey of land Ventilation Grounding Wiring Fluid check Inspection
Value of power factor is varies from 0 to 1. R→PF = 1 L→PF = 0 lag C→PF = 0 lead R-L→0 lag < PF < 1 R-C→0 lead < PF < 1 R-C-L→depends on value of R, L and C. PF = R/Z
EXAMIANS