MGVCL Exam Paper (30-07-2021 Shift 3) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વી + સમ વીસ + અમ વિસ્ + અમ વિ + સમ વી + સમ વીસ + અમ વિસ્ + અમ વિ + સમ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) What is the value of capacitance of a capacitor which has a voltage of 4 V and has 16 C of charge? 4 F 16 F 2 F 8 F 4 F 16 F 2 F 8 F ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:C = Q/V= 16/4= 4 F
MGVCL Exam Paper (30-07-2021 Shift 3) In the following question, one part of the sentence may have an error. Find out which part of the sentence has an error and select the option corresponding to it. If the sentence contains no error, Select "No error" option. (Avoid punctuation errors).(A) The boys campaigned / (B) not only in / (C) Mumbai also in Chennai / (D) NO ERROR A B C D A B C D ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) હું અહી આવી શકું' વાક્યને સ્થળવાચક ક્રીયાવીશેષણમાં ફેરવો. હું ચેન્નાઈ નહિ આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું નિયમિત રીતે અહી આવી શકું. હું ચેન્નાઈ નહિ આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું નિયમિત રીતે અહી આવી શકું. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A DC shunt generator has an induced voltage of 220 V on open circuit. When the machine is on load the terminal voltage is 200 V. Find the load current if the field resistance is 100 Ω and armature resistance is 0.2 Ω. 98 A 104 A 102 A 96 A 98 A 104 A 102 A 96 A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Voltage equation for Dc shunt generator,Eb = V + Ia*RaV = Ish*RshIsh = 200/100= 2 AEb = V + Ia*Ra20 = Ia*RaIa = 20/0.2= 100 AI = 100 - 2= 98 A
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 = (-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. 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. 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
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