MGVCL Exam Paper (30-07-2021 Shift 1) માણસ: ઘર :: પશુ : અંતરિક્ષ ગુફા પાતાળ આકાશ અંતરિક્ષ ગુફા પાતાળ આકાશ 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? ઉપપદ તત્પુરુષ દ્વન્દ્વ કર્મધારય ઉપપદ તત્પુરુષ દ્વન્દ્વ કર્મધારય 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 - 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 Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) 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) The expression for voltage regulation of a short transmission line for lagging power factor is given by: ___, where, 'I' is the line current, 'V_r' is the receiving end voltage, 'R' is the line resistance and 'X' is the line reactance. (IR cosφ_r - IX sinφ_r)/V_r (IX cosφ_r + IR sinφ_r)/V_r (IR cosφ_r + IX sinφ_r)/V_r (IX cosφ_r - IR sinφ_r)/V_r (IR cosφ_r - IX sinφ_r)/V_r (IX cosφ_r + IR sinφ_r)/V_r (IR cosφ_r + IX sinφ_r)/V_r (IX cosφ_r - IR sinφ_r)/V_r ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For lagging power factor = (IR*cosφ + Ixsinφ)/VrFor leading power factor = (IR*cosφ - Ixsinφ)/VrZero voltage regulation occurs on leading power factor only.
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 5.14 mJ 2.57 mJ 10.28 mJ 12.50 mJ 5.14 mJ 2.57 mJ 10.28 mJ 12.50 mJ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) A ___ is a device that forwards packets between networks by processing the routing information included in the packet. Firewall Router all the mentioned above Bridge Firewall Router all the mentioned above Bridge ANSWER DOWNLOAD EXAMIANS APP
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