MGVCL Exam Paper (30-07-2021 Shift 3) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વીસ + અમ વી + સમ વિસ્ + અમ વિ + સમ વીસ + અમ વી + સમ વિસ્ + અમ વિ + સમ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end. 400 m 200 m 100 m 300 m 400 m 200 m 100 m 300 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
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. 104 A 98 A 102 A 96 A 104 A 98 A 102 A 96 A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Voltage equation for Dc shunt generator,Eb = V + Ia*RaV = Ish*RshIsh = 200/100= 2 AEb = V + Ia*Ra20 = Ia*RaIa = 20/0.2= 100 AI = 100 - 2= 98 A
MGVCL Exam Paper (30-07-2021 Shift 3) Two-wattmeter method is used to measure the power taken by a 3-phase induction motor on no load. The wattmeter readings are 400 W and -50 W. Calculate the reactive power taken by the load 350/√3 VAR 350√3 VAR 450√3 VAR 450/√3 VAR 350/√3 VAR 350√3 VAR 450√3 VAR 450/√3 VAR ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Reactive power taken (Q) = √3*(W1 - W2)= √3*(450 + 50)= √3*450 kVAR
MGVCL Exam Paper (30-07-2021 Shift 3) નીચે પૈકી 'ઓસોડ' નો સમાનર્થી શબ્દ કયો છે? અનરવું વૃંદા ઔષધ અમર અનરવું વૃંદા ઔષધ અમર ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A star connected synchronous generator rated at 500 MVA, 50 kV has a reactance of 0.5 pu. Find the ohmic value of the reactance. 2.5 Ω 0.1 Ω 1 Ω 0.25 Ω 2.5 Ω 0.1 Ω 1 Ω 0.25 Ω ANSWER EXPLANATION DOWNLOAD EXAMIANS APP pu = actual/baserated phase consired at base value for alternator.ohmic pu = ohmic actual/ohmic basepu = ohmic actual*current base/voltage baseConsider all above phase valuesMVA (3phase) = 3 Vph IphIph = 500*10⁶/(3*28901.73) = 5766.66 Apu = ohmic actual*current base/voltage base0.5 = ohmic actual*5766.66/28901.73ohmic actual = 2.5 Ω
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