MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? હસ્તમેળાપ વિવાહ પ્રેમાલાપ છુટાછેડા હસ્તમેળાપ વિવાહ પ્રેમાલાપ છુટાછેડા ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The operating time of an inverse definite minimum time (IDMT) overcurrent relay is ____ proportional to the plug-setting multiplier (PSM) and____proportional to the time-multiplier setting (TMS). inversely, directly inversely, inversely directly, inversely directly, directly inversely, directly inversely, inversely directly, inversely directly, directly ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a short circuit test on circuit breaker, time to reach the peak re-striking voltage is 50 µs with the peak restriking voltage of 100 kV. Determine average RRRV (Rate of rise re-striking voltage)? 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec 1 x 10⁶ kV/sec 2 x 10⁶ kV/sec 3 x 10⁶ kV/sec 4 x 10⁶ kV/sec ANSWER EXPLANATION DOWNLOAD EXAMIANS APP RRRV = Peak value of restriking voltage/time to reach the peak re-striking voltageRRRV = 100*1000/(50 μs)RRRV = 2*10⁶ kV/sec
MGVCL Exam Paper (30-07-2021 Shift 2) Choose the word which best expresses the similar meaning of the given word "LUCID" Clear Sensible Puzzled Cloudy Clear Sensible Puzzled Cloudy ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A THHN conductor will have a 60˚C termination on one end and a 75˚C termination on the other, which ampacity column will be used? None of these 60˚C column 75˚C column 90˚C column None of these 60˚C column 75˚C column 90˚C column ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Conductor with higher temperature ratings, provide the ampacity is determined based on the 60°C ampacity of the conductor.
MGVCL Exam Paper (30-07-2021 Shift 2) A 1000 kVA ONAN cooled transformer has a load of 500 kVA throughout the day except for a period of 4 hours. What is the permissible overload for a duration of four hours. Assume the permissible load kVA as a fraction of rated kVA is 1.24. 1240 kVA 620 kVA 2480 kVA 930 kVA 1240 kVA 620 kVA 2480 kVA 930 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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