MGVCL Exam Paper (30-07-2021 Shift 3) Choose the word which expresses nearly the opposite meaning of the given word "DILIGENT " Busy Lasy Active Careful Busy Lasy Active Careful ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A group of computers and other devices connected together is called a network and the concept of connected computers sharing resources is called ____ Routing Networking Linking Switching Routing Networking Linking Switching ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Fill in the blanks with suitable Preposition from the given alternatives.BCCI receives____1000 applications for Team India head coach position since over for against since over for against ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) હું અહી આવી શકું' વાક્યને સ્થળવાચક ક્રીયાવીશેષણમાં ફેરવો. હું નિયમિત રીતે અહી આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું ચેન્નાઈ નહિ આવી શકું. હું નિયમિત રીતે અહી આવી શકું. હું કદી પણ નહિ આવી શકું. હું આજે નહિ આવી શકું. હું ચેન્નાઈ નહિ આવી શકું. ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) In a series RLC circuit, the supply voltage is 230 V at 50 Hz. The resonant current is 2 A at the resonant frequency of 50 Hz. Under the resonant condition, the voltage across the capacitor is measured to be equal to 460 V. What are the values of L and C? 0.73 mH and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF 0.73 H and 13.85 pF 0.73 mH and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF 0.73 H and 13.85 pF ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Vc = I*Xc460 = 2*(1/(2*ᴨ*50*C))C = 2/(460*2*50*ᴨ)= 1.385*10⁻⁵ F= 13.85 FV_L = I*X_LL = V/(I*2*ᴨ*f)= 460/(2*ᴨ*50*2)= 0.7345 H
MGVCL Exam Paper (30-07-2021 Shift 3) A 50 Hz synchronous generator is connected to an infinite bus through a line. The p.u. reactances of generator and the line are j0.4 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. 0.8 Hz 1.6 Hz 1.2 Hz 1.8 Hz 0.8 Hz 1.6 Hz 1.2 Hz 1.8 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))= (1.1/0.6)*cosδ= (1.1/0.6)*0.5= 0.91M = (H*s)/(πf)= 4/(50π)fn = (0..91/(4/50π))= 8.9 rad/sec= 1.2 Hz
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