MGVCL Exam Paper (30-07-2021 Shift 2) In a synchronous motor with field under excited, the power factor will be leading zero unity lagging leading zero unity lagging ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For synchronous motor,Under excitation - Lagging power factorOver excitation - Leading power factorFor synchronous generatorUnder excitation - Leading power factorOver excitation - lagging power factor
MGVCL Exam Paper (30-07-2021 Shift 2) In transformer, bushing failure occurs due to(i) transformer vibrations(ii) dielectric losses(ii) partial discharge (i) and (ii) only (ii) and (iii) only (i) and (iii) only (i), (ii) and (iii) (i) and (ii) only (ii) and (iii) only (i) and (iii) only (i), (ii) and (iii) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing. 730 mH 1135 mH 1270 mH 932 mH 730 mH 1135 mH 1270 mH 932 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
MGVCL Exam Paper (30-07-2021 Shift 2) Magnetic tape is not practical for applications where data must be quickly recalled because tape is a/an____. Sequential access medium Random access medium None of these expensive storage medium Sequential access medium Random access medium None of these expensive storage medium ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 350 kA 35 kA 25 kA 2500 A 350 kA 35 kA 25 kA 2500 A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) In the RLC series circuit shown in figure, resonance occurs when the value of C is 20 µF. The supply voltages is v(t) = 20 sin (500t) V. Find the value of L. 0.2 H 5 H 0.2 mH 5 mH 0.2 H 5 H 0.2 mH 5 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,ω = 500 rad/sec = 2πf.At Resonance condition,ωL = 1/(ωC)L 0.2 HAlso at series resonace1. fr = 1/[2π√(LC)]2. PF = 13. Maximum current4. Source voltage = voltage across resistor.
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