MGVCL Exam Paper (30-07-2021 Shift 1) ___ is the process of actuating equipment operation at remote locations. Supplementary control Supervisory control Data acquisition Data processing Supplementary control Supervisory control Data acquisition Data processing ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) જીવનમાં તડકોછાયો એતો એક ઘટમાળ છે.વાક્યમાં ક્યા સમાસનો ઉપયોગ કરવામાં આવ્યો છે? દ્વન્દ્વ તત્પુરુષ ઉપપદ કર્મધારય દ્વન્દ્વ તત્પુરુષ ઉપપદ કર્મધારય ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 15 Ω and the resistance of the resistor connected to the sound core was 45 Ω. Calculate the distance of the fault point from the test end. 75 m 450 m 150 m 225 m 75 m 450 m 150 m 225 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)= (15/60)*600= 150 m
MGVCL Exam Paper (30-07-2021 Shift 1) The Prime Minister Rozgar Yojana (PMRY) was launched in which year? 2005 1995 1993 2003 2005 1995 1993 2003 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 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 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
MGVCL Exam Paper (30-07-2021 Shift 1) An electric potential field is produced in air by point 1 μC and 4 μC located at (-2, 1, 5) and (1, 3, -1) respectively. The energy stored in the field is 10.28 mJ 12.50 mJ 2.57 mJ 5.14 mJ 10.28 mJ 12.50 mJ 2.57 mJ 5.14 mJ ANSWER DOWNLOAD EXAMIANS APP
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