MGVCL Exam Paper (30-07-2021 Shift 2) Match the following shown in figure A = (i) & (3) A = (iii) & (2) B = (ii) & (1) C = (ii) & (1) A = (i) & (3) A = (iii) & (2) B = (ii) & (1) C = (ii) & (1) ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 4% 5% 3% 6% 4% 5% 3% 6% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
MGVCL Exam Paper (30-07-2021 Shift 2) જે ક્રિયા દર્શાવે પણ વાક્યનો અર્થ પૂર્ણ કરે એ એને ____કહેવાય. - આપેલ વિકલ્પોમાંથી યોગ્યનું ચયન કરી વાક્ય પૂર્ણ કરો ક્રિયા-વિભક્તિ વિભક્તિ કૃદંત નીપાત ક્રિયા-વિભક્તિ વિભક્તિ કૃદંત નીપાત ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following statement is TRUE? In the cables, sheaths are used to provide enough strength. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. The current carrying capacity of cables in DC is less than that in AC. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. In the cables, sheaths are used to provide enough strength. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. The current carrying capacity of cables in DC is less than that in AC. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors.
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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. 304.5 V 308.5 V 306.5 V 302.5 V 304.5 V 308.5 V 306.5 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
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