MGVCL Exam Paper (30-07-2021 Shift 3) In which of the following situations, there is no need to provide directional overcurrent protection. single end fed, parallel feeder single end fed, single feeder double end fed, single feeder ring main single end fed, parallel feeder single end fed, single feeder double end fed, single feeder ring main ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Directional overcurrent relay protection is used for parallel feeder and ring main system.It is not used for the protection of radial system.
MGVCL Exam Paper (30-07-2021 Shift 3) Replace the braketed phrase grammatically and conceptually with the help of the given opation. If the given sentence is correct then select the option ’The given sentence is correct’.The corono virus continued to spread, with the development of infections advancing quickly continued to spreading, with the development of infections continued for spread, with the development of infections The given sentence is correct continue to spread, with the development of infections continued to spreading, with the development of infections continued for spread, with the development of infections The given sentence is correct continue to spread, with the development of infections ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In public sector, what is the full form of "LIC"? Life Insurers Life Insurance Company Life Insurance Corporation Lively Insurance Corporation Life Insurers Life Insurance Company Life Insurance Corporation Lively Insurance Corporation 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 = 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 = (-5 + j2.88) A.Ia₂ = (5 - j2.88) A. Ia₀ = 0 A.Ia₁ = (-1.66 + j2.88) A.Ia₂ = (1.66 + 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 = (-5 + j2.88) A.Ia₂ = (5 - j2.88) 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) 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 3.5 V 4.5 V 1.5 V 2.5 V 3.5 V 4.5 V 1.5 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) દ્રશ્ય' શબ્દનો વિરુદ્ધાર્થ શબ્દ આપેલો વિકલ્પોમાંથી કયો છે? દ્રષ્ટિવંત અદ્રશ્ય દ્રષ્ટિ દ્રશ્યમાન દ્રષ્ટિવંત અદ્રશ્ય દ્રષ્ટિ દ્રશ્યમાન ANSWER DOWNLOAD EXAMIANS APP
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