MGVCL Exam Paper (30-07-2021 Shift 3) The threshold voltage of an N-channel enhancement mode MOSFET is 0.5 V. When the device is biased at a gate voltage of 3 V, pinch-off would occur at a drain voltage of 2.5 V 4.5 V 1.5 V 3.5 V 2.5 V 4.5 V 1.5 V 3.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:Vds = Vgs - VtVds = 3 - 0.5Vds = 2.5 V
MGVCL Exam Paper (30-07-2021 Shift 3) A group of computers and other devices connected together is called a network and the concept of connected computers sharing resources is called ____ Routing Switching Networking Linking Routing Switching Networking Linking ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The sending end and receiving end voltages of the short transmission line are 150 kV and 120 kV respectively. Calculate its percentage voltage regulation. 30% 40% 20% 25% 30% 40% 20% 25% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,Vs - sending end voltage in kVVr - Receiving end voltage in kV.Voltage regulation = (Vs - Vr)/Vr= (150 - 120)/120= 0.25% R = 25 %
MGVCL Exam Paper (30-07-2021 Shift 3) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: globally, out of which over 200 millionQ: are now in India, andR: flipkart has 700 million usersS: growing exponentially RPSQ RPQS RSQP PQRS RPSQ RPQS RSQP PQRS ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The current flowing in to a balanced delta connected load through line 'a' is 10 A when the conductor of line ’b’ is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of ’abc’. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 0 A.Ia₁ = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 10 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. Ia₀ = 10 A.Ia₁ = (-5 + j2.44) A.Ia₂ = (5 - j2.44) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + j2.88) A. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(10 + 10∟-120°)= 5 + j*2.88 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(10 + 10∟120°)= 5 - j*2.88 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
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. 400 m 200 m 100 m 300 m 400 m 200 m 100 m 300 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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