MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Regular Bold Superscript Italic Regular Bold Superscript Italic ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) In a single phase energy meter, the shunt electromagnet is energized by a ____of ____ pressure coil, thin wire of more turns current coil, thick wire of few turns current coil, thin wire of more turns pressure coil, thick wire of few turns pressure coil, thin wire of more turns current coil, thick wire of few turns current coil, thin wire of more turns pressure coil, thick wire of few turns ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In single phase energy meter,Current coil is made up from thick wire having few no. of turns.Pressure coil is made up from thin wire having more no. of turns.
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. 350 kA 35 kA 2500 A 25 kA 350 kA 35 kA 2500 A 25 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
MGVCL Exam Paper (30-07-2021 Shift 2) નીચે માંથી ક્યાં રુઢિપ્રયોગનો અર્થ 'છેતરવું' એવો થાય આડા ઉતરવું હથેળીમાં ચાંદ દેખાડવો દસ મારવો બે પાંદડે થવું આડા ઉતરવું હથેળીમાં ચાંદ દેખાડવો દસ મારવો બે પાંદડે થવું ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Transistors were used in ____ of Computers. First Generation Second Generation Fourth Generation Third Generation First Generation Second Generation Fourth Generation Third Generation ANSWER DOWNLOAD EXAMIANS APP
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. 2480 kVA 1240 kVA 620 kVA 930 kVA 2480 kVA 1240 kVA 620 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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