MGVCL Exam Paper (30-07-2021 Shift 2) If all the devices are connected to a central hub, then the topology is called Tree Topology Star Topology Bus Topology Tree Topology Tree Topology Star Topology Bus Topology Tree Topology ANSWER DOWNLOAD EXAMIANS APP
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 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = -10 kW and W2 = 40 kW W1 = 10 kW and W2 = 20 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) 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 200 m 400 m 300 m 100 m 200 m 400 m 300 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) નીચે માંથી ક્યાં રુઢિપ્રયોગનો અર્થ 'છેતરવું' એવો થાય આડા ઉતરવું બે પાંદડે થવું હથેળીમાં ચાંદ દેખાડવો દસ મારવો આડા ઉતરવું બે પાંદડે થવું હથેળીમાં ચાંદ દેખાડવો દસ મારવો ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ 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. 304.5 V 308.5 V 302.5 V 306.5 V 304.5 V 308.5 V 302.5 V 306.5 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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