MGVCL Exam Paper (30-07-2021 Shift 2) હુ ગીત ગાઈશ' ને કર્મણીમાં ફેરવો. ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ગીતને હુ ગાઈશ ગીતને હુ જ ગાઇશ મારાથી ગીત ગવાશે હુ ગીતને ગાઈશ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) Recently, which country has withdrawn the 'Generalized System of Preferences' with India? Pakistan United States of America Russia Peru Pakistan United States of America Russia Peru ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.3 p.u. and j0.2 p.u. respectively. The generator no load voltage is 1.1 p.u. and that of infinite bus is 1.0 p.u. The inertia constant of the generator is 4 MW-sec/MVA. Determine the frequency of natural oscillations if the generator is loaded to 60% of its maximum power transfer capacity and small perturbation in power is given. 1.42 Hz 1.32 Hz 1.52 Hz 1.62 Hz 1.42 Hz 1.32 Hz 1.52 Hz 1.62 Hz ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Frequency of natural oscillation is given by,fn = {((dPe/dδ)at(δo))/M)}dPe/dδ = ((V1*V2)/X*(cosδ))= (11/05)*cosδ= (11/0.5)*0.5M = (H*s)/(πf)= 4/(50π)fn = (1.76*(3/50π))= 9.4 rad/sec = 1.32 Hz
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 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 kW W1 = -5 kW and W2 = 35 kW W1 = 5 kW and W2 = 25 kW W1 = 10 kW and W2 = 20 kW W1 = -10 kW and W2 = 40 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) Which one of the following statement is TRUE? In the cables, sheaths are used to provide enough strength. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. The current carrying capacity of cables in DC is less than that in AC. In the cables, sheaths are used to provide enough strength. In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors. Underground cables are laid at sufficient depth to avoid being unearthed easily due to removal of soil. The current carrying capacity of cables in DC is less than that in AC. ANSWER EXPLANATION DOWNLOAD EXAMIANS APP In the cables, the location of fault is usually found out by comparing the capacitances of insulated conductors.
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is not a Font style in Microsoft Word? Bold Regular Italic Superscript Bold Regular Italic Superscript ANSWER DOWNLOAD EXAMIANS APP
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