MGVCL Exam Paper (30-07-2021 Shift 3) Find the word which is correctly spelt from the given options. Dimnished Clasification Profitable Idantical Dimnished Clasification Profitable Idantical ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The complex power injections at a bus are shown in figure Two lines connected drawing the powers as shown in figure. The complex power in the line-2 "S2" is 2+j1 2-j5 2+j5 2-j1 2+j1 2-j5 2+j5 2-j1 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Total power generate (Sg) = S1 + S2S2 = Sg - S1= 5 + 3j - 3 + 2j= 2 + 5j
MGVCL Exam Paper (30-07-2021 Shift 3) A single phase full converter bridge, connected to 230 V, 50 Hz source is feeding a load R =10 Ω in series with a large inductance that makes the load current ripple free. For a firing angle of 450, calculate the rectification efficiency. 28.33% 76.66% 50.55% 63.66% 28.33% 76.66% 50.55% 63.66% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rectification efficiency = DC output power/AC input power
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 B A C D B A C D ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Continuous and rapid variations in the load current magnitude which causes voltage variations is known as Voltage distortion Flicker Harmonics Voltage sag Voltage distortion Flicker Harmonics Voltage sag ANSWER EXPLANATION DOWNLOAD EXAMIANS APP When voltage changes occur in rapid succession, with magnitudes large enough to cause lighting level variations.The human eye-brain response is most sensitive to periodic r.m.s. voltage changes that occur at around 8 - 10 cycles per second.
MGVCL Exam Paper (30-07-2021 Shift 3) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 2 hours. What is the permissible overload for a duration of two hours. Assume the permissible load kVA as a fraction of rated kVA is 1.43. 1430 kVA 1050 kVA 715 kVA 700 kVA 1430 kVA 1050 kVA 715 kVA 700 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload for transformer = Full load kVA*factor for permissible overload= 1000*1.43= 1430 kVA
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