MGVCL Exam Paper (30-07-2021 Shift 3) The elements are specialized computers used to connect two or more transmission lines. Networking Switching Broadcasting Transfering Networking Switching Broadcasting Transfering ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Find the word which is correctly spelt from the given options. Clasification Idantical Profitable Dimnished Clasification Idantical Profitable Dimnished ANSWER DOWNLOAD EXAMIANS APP
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 715 kVA 1050 kVA 1430 kVA 700 kVA 715 kVA 1050 kVA 1430 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) The International boundary 'Radcliffe Line’ lies between India and China Japan Pakistan Afghanistan China Japan Pakistan Afghanistan ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Match the following shown in table A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) A = (iii), B = (i), C = (ii) A = (ii), B = (iii), C = (i) A = (ii), B = (i), C = (iii) A = (iii), B = (ii), C = (i) ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Bus Type - Known Parameter - Unknown ParameterLoad Bus -P, Q - V, phase angleGenerator Bus - P, V (magnitude) - Q, Voltage phase angleSlack Bus Voltage - magnitude and phase angle - P, Q
MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.25 H 0.5 H 0.75 H 1.25 H 0.25 H 0.5 H 0.75 H 1.25 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
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