MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Superscript Bold Italic Regular Superscript Bold Italic Regular ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Which one of the following is the largest windfarm located in India? Talcher Mundra Muppandal Vindhayachal Talcher Mundra Muppandal Vindhayachal ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rank - Windfarm name - Capacity1 - Muppandal windfarm - 1500 MW2 - Jaisalmer Wind Park - 1064 MW3 - Brahmanvel Wind Farm Dhule - 964 MW4 - Damanjodi Wind Farm - 99 MW5 - Tuppadahalli Wind Farm - 56.1 MW
MGVCL Exam Paper (30-07-2021 Shift 2) Find the word which is correctly spelt from the given options. Absurrd Wrinkel Maximum Sentimant Absurrd Wrinkel Maximum Sentimant ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A sudden, non power frequency change in the steady state condition of voltage or current that is bidirectional in polarity is called as Distortion Impulsive transient Harmonic Oscillatory transients Distortion Impulsive transient Harmonic Oscillatory transients ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Oscillatory transient is the a sudden, non power frequency change in the steady state condition of voltage or current that is bidirectional in polarity.
MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos√(LC) t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 35 kA 2500 A 25 kA 350 kA 35 kA 2500 A 25 kA 350 kA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
EXAMIANS