MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Bold Italic Regular Superscript Bold Italic Regular Superscript ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a single phase energy meter, the shunt electromagnet is energized by a ____of ____ current coil, thin wire of more turns pressure coil, thin wire of more turns pressure coil, thick wire of few turns current coil, thick wire of few turns current coil, thin wire of more turns pressure coil, thin wire of more turns pressure coil, thick wire of few turns current coil, thick wire of few turns ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In single phase energy meter,Current coil is made up from thick wire having few no. of turns.Pressure coil is made up from thin wire having more no. of turns.
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 2500 A 25 kA 350 kA 35 kA 2500 A 25 kA 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) A single phase 240 V, 1 kW heater is connected across single phase 240 V, 50 Hz supply through an SCR. For firing angle delay of 90˚, calculate the power absorbed in the heater element. 225 W 275 W 250 W 300 W 225 W 275 W 250 W 300 W ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Resistance of heater, R = V²/PR = 57.6 ohmRMS value of voltage through controlled SCR = (Vm/2π)*(1 + cosα)= 120 VPower absorb by heater = Vrms²/R= 120²/57.6= 250 W
MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Jawaharlal Nehru Subhas Chandra Bose Ram Mohan Roy Dadabhai Naoroji Jawaharlal Nehru Subhas Chandra Bose Ram Mohan Roy Dadabhai Naoroji ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.4π V zero 0.2π V 0.8π V 0.4π V zero 0.2π V 0.8π V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
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