MGVCL Exam Paper (30-07-2021 Shift 2) If all the devices are connected to a central hub, then the topology is called Star Topology Tree Topology Bus Topology Tree Topology Star Topology Tree Topology Bus Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a test by Murray loop for ground fault on 600 m of cable having a resistance of 1.5 Ω/km, the faulty cable is looped with a sound cable of the same length and area of cross section. If the ratio of the other two arms of the testing network at balance is 3 : 1, find the distance of the fault from the testing end of cables. 100 m 400 m 300 m 200 m 100 m 400 m 300 m 200 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,d is distance in mQ is value of variable resistor in ohmP is variable resistance in ohmHere Q/P = 1/3 is given,d = Q/(P + Q)*Loop length= 1/4*(2*600)= 300 m
MGVCL Exam Paper (30-07-2021 Shift 2) The wind speed increases with height because of reduction of drag effect of the earth surface reduction in lift effect of the air reduction in the gravitational force enhancement of drag effect on the earth surface reduction of drag effect of the earth surface reduction in lift effect of the air reduction in the gravitational force enhancement of drag effect on the earth surface ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Wind speed increases as the height from the ground increases mainly due to the decrease in the friction produced by land terrain.
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 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = 5 kW and W2 = 25 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 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. 306.5 V 304.5 V 308.5 V 302.5 V 306.5 V 304.5 V 308.5 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) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ હુ ગીતને ગાઈશ મારાથી ગીત ગવાશે ANSWER DOWNLOAD EXAMIANS APP
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