MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is an application layer service? Mail service Network virtual terminal File transfer, access and management Mail service Mail service Network virtual terminal File transfer, access and management Mail service ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) The expression for voltage regulation of a short transmission line for leading power factor is given by: ____, where, 'I' is the line current, 'V_r' is the receiving end voltage, 'R' is the line resitance and 'X' is the line reactance. (IXcosφ_r - IRsinφ_r)/V_r (IRcosφ_r - IXsinφ_r)/V_r (IRcosφ_r + IXsinφ_r)/V_r (IXcosφ_r + IRsinφ_r)/V_r (IXcosφ_r - IRsinφ_r)/V_r (IRcosφ_r - IXsinφ_r)/V_r (IRcosφ_r + IXsinφ_r)/V_r (IXcosφ_r + IRsinφ_r)/V_r ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For lagging power factor = (IR*cosφ + IXsinφ)/V_rFor leading power factor = (IR*cosφ - IXsinφ)/V_rZero voltage regulation occurs on leading power factor only.
MGVCL Exam Paper (30-07-2021 Shift 2) પાણી : પાણીગ્રહણ :: હસ્ત : ??? હસ્તમેળાપ પ્રેમાલાપ છુટાછેડા વિવાહ હસ્તમેળાપ પ્રેમાલાપ છુટાછેડા વિવાહ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Vikram Seth is associated with which of the following fields? Sports Politician Music Literature Sports Politician Music Literature ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 3-phase motor load has a p.f. of 0.397 lagging. The two wattmeter method used to measure power show the input as 30 kW. Find the reading on each wattmeter. W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW W1 = -10 kW and W2 = 40 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -5 kW and W2 = 35 kW ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Here,Cosφ = 0.397P = 30 kW = √3*V_L*I_L*cosφV_L*I_L = 30kW/(√3*0.397) = 43628.48 VAReadings of wattmeter W1 = V_L*I_L*cos(30+φ)Readings of wattmeter W2 = V_L*I_L*cos(30-φ)
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. 930 kVA 1240 kVA 620 kVA 2480 kVA 930 kVA 1240 kVA 620 kVA 2480 kVA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Permissible overload = Full load kVA*factor for permissible load= 1000*1.24= 1240 kVA
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