MGVCL Exam Paper (30-07-2021 Shift 3) A 3 pF capacitor is charged by a constant current of 2 pA for six seconds. The voltage across the capacitor at the end of charging will be 6 V 2 V 4 V 8 V 6 V 2 V 4 V 8 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation know that the charge stored in a capacitor is given as:Q = C*VQ = I*tV = (I*t)/C= (2μ*6)/(3μ)= 4 V
MGVCL Exam Paper (30-07-2021 Shift 3) The complex power injections at a bus are shown in figure Two lines connected drawing the powers as shown in figure. The complex power in the line-2 "S2" is 2-j5 2+j1 2+j5 2-j1 2-j5 2+j1 2+j5 2-j1 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Total power generate (Sg) = S1 + S2S2 = Sg - S1= 5 + 3j - 3 + 2j= 2 + 5j
MGVCL Exam Paper (30-07-2021 Shift 3) Which state is the largest producer of wind energy in India? Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu Andhra Pradesh Maharashtra Madhya Pradesh Tamil Nadu ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - State - Capaciy1 - Tamilnadu - 7.5 GW2 - Maharastra - 5 GW3 - Karnataka - 4.8 GW4 - Rajsthan - 4.3 GW
MGVCL Exam Paper (30-07-2021 Shift 3) Message ____ means that the data must arrive at the receiver exactly as sent. Confidentiality Authentication Integrity None ot these Confidentiality Authentication Integrity None ot these ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) If a cable of homogeneous insulation has maximum stress of 6 kV/mm, then the dielectric strength of insulation should be ____ 12 kV/mm 6 kv/mm 1.5 kV/mm 3 kV/mm 12 kV/mm 6 kv/mm 1.5 kV/mm 3 kV/mm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP If a cable have homogeneous insulation then the dielectric strength of insulation should be is equal to maximum dielectric strength.
MGVCL Exam Paper (30-07-2021 Shift 3) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end. 300 m 400 m 200 m 100 m 300 m 400 m 200 m 100 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
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