MGVCL Exam Paper (30-07-2021 Shift 3) What is the value of the characteristic impedance of a transmission line with impedance and admittance of 15 and 5? 1.414 0.577 1.732 0.7.7 1.414 0.577 1.732 0.7.7 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Characteristic impedance = √(impedance/admittance) = 1.73
MGVCL Exam Paper (30-07-2021 Shift 3) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 8 V 6 V 4 V 2 V 8 V 6 V 4 V 2 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
MGVCL Exam Paper (30-07-2021 Shift 3) Which is the function of Operating System? Devices Management Process Management All of these Security Devices Management Process Management All of these Security ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In a power network, 375 kV is recorded at a 400 kV bus. A 45 MVAR, 400 kV shunt reactor is connected to the bus. What is the reactive power absorbed by the shunt reactor? 49.55 MVAR 69.55 MVAR 59.55 MVAR 39.55 MVAR 49.55 MVAR 69.55 MVAR 59.55 MVAR 39.55 MVAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Q = V²/XX = 400²/45= 3555.55 ohmFor 375 kV,Q = 375²/3555.55= 39.55 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 700 kVA 1430 kVA 1050 kVA 715 kVA 700 kVA 1430 kVA 1050 kVA 715 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1500 rpm 1440 rpm 1475 rpm 1525 rpm 1500 rpm 1440 rpm 1475 rpm 1525 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
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