MGVCL Exam Paper (30-07-2021 Shift 2) Who started the daily newspaper 'National Herald' in 1938? Subhas Chandra Bose Ram Mohan Roy Dadabhai Naoroji Jawaharlal Nehru Subhas Chandra Bose Ram Mohan Roy Dadabhai Naoroji Jawaharlal Nehru ANSWER DOWNLOAD EXAMIANS APP
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. 275 W 225 W 250 W 300 W 275 W 225 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) Consider the following statements associated with Buchholz's relays.Statement 1: Buchholz's relay is a current operated device.Statement 2 : Buchholz's relay is placed between the transformer tank and the conservator. Both Statement 1 and Statement 2 are TRUE Statement 1 is FALSE and Statement 2 is TRUE Both Statement 1 and Statement 2 are FALSE Statement 1 is TRUE and Statement 2 is FALSE Both Statement 1 and Statement 2 are TRUE Statement 1 is FALSE and Statement 2 is TRUE Both Statement 1 and Statement 2 are FALSE Statement 1 is TRUE and Statement 2 is FALSE ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Buchholz relay is gas operated relay.It will operate when internal fault occurs in transformer.It is placed between tank to main conservator.It has two contacts trip and alarm.Use of Buchholz relay in distribution transformer is not economical.
MGVCL Exam Paper (30-07-2021 Shift 2) Transistors were used in ____ of Computers. First Generation Second Generation Third Generation Fourth Generation First Generation Second Generation Third Generation Fourth Generation ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A short-shunt dc compound generator supplies 50 A at 300 V. If the shunt field resistance, series field resistance and armature resistance are 30 Ω, 0.03Ω, and 0.05 Ω respectively, determine the e.m.f. generated. 306.5 V 308.5 V 304.5 V 302.5 V 306.5 V 308.5 V 304.5 V 302.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Vt = 300 VRse = 0.03 Ω, Rsh = 30 Ω, Ra = 0.05 ΩI_L = 50 AVsh = V + I_L*Rse = 300 + 50*0.03 = 301.5 VIsh = Vsh/Rsh = 301.5/30 = 10.05 AIa = Ish + I_L = 60.05 AEg = Vt + I_L*Rse + IaRaEg = 300 + 50*0.03 + 60.05*0.05 = 304.5025 V
MGVCL Exam Paper (30-07-2021 Shift 2) The line currents in amperes in phases a, b and c respectively are 500 +j150, 100 – j600 and – 300 + j600 referred to the same reference vector with phase sequence of abc. Find the symmetrical component of currents. Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (546.41+ j165.47) AIa₂ = (546.41-j165.47) A Ia₀ = (100+j50) AIa₁ = (200+ j107.75) AIa₂ = (200-j107.75) A Ia₀ = (100+j50) AIa₁ = (546.41+j165.47) AIa₂ = (-146.41-j65.47) A Ia₀ = (100+j50) AIa₁ = (146.41+ j65.47) AIa₂ = (546.41-j165.47) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Calculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 500 + j150 + (- 0.5 + j*0.866)*(100 – j6000) + (-0.5 - j*0.866)*(– 300 + j600)= (546.41 + j*156.47) AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= 500 + j150 + (- 0.5 - j*0.866)*(100 – j6000) + (-0.5 + j*0.866)*(– 300 + j600)= (-146.41 - j*56.47) AIao = 1/3*(Ia + Ib + Ic)= 1/3*(500 + j150 + 100 – j6000 – 300 + j600)= (100 + j*50) A
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