MGVCL Exam Paper (30-07-2021 Shift 2) Tree Topology Web layout Paper size Page Orientation Font style Web layout Paper size Page Orientation Font style ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A three-phase, 4-pole, 50 Hz induction motor runs at 1425 rpm. Find out the percentage slip of the induction motor. 6% 4% 5% 3% 6% 4% 5% 3% ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Slip, s = (Ns - Nr)/NsNs = 120f/PNs = 1500 rpms = (1500 - 1425)/1500s = 0.05%s = 5%
MGVCL Exam Paper (30-07-2021 Shift 2) What is the full form of UN in 'UN Human Rights Prize'? Unity Nation United Nations Uniform Nacional Uniformed Nations Unity Nation United Nations Uniform Nacional Uniformed Nations ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A coil having an inductance of 100 mH is magnetically coupled to another coil having an inductance of 900 mH. The coefficient of coupling between the coils is 0.45. Calculate the equivalent inductance if the two coils are connected in series opposing. 730 mH 1270 mH 1135 mH 932 mH 730 mH 1270 mH 1135 mH 932 mH ANSWER EXPLANATION DOWNLOAD EXAMIANS APP M = k√(L₁L₂)M= 0.45√(100*900)M = 135 mHL_eq for opposite connection of coils = L₁ + L₂ - 2ML_eq = 900 + 100 - 2*135L_eq = 730 mH
MGVCL Exam Paper (30-07-2021 Shift 2) 1. આપને મુખ્યમત્રીએ ખાસ કામ માટે નિયુક્ત કર્યા છે.2. તેનાથી આંખો ઝીણી કરાઈ3. પ્રધાનમંત્રીએ સ્કુલનું ઉદ્રઘાટન કર્યું.આરોપીએ બધા ગુના કબુલ કરી લીધા.ઉક્ત વાક્યોમાં કેટલા કર્મણી વાક્યો પ્રયોગના છે? કુલ 1 બધા જ વાક્યો કર્મણી વાક્ય પ્રયોગના છે બધા જ કર્તરી વાક્ય પ્રયોગના છે કુલ 2 કુલ 1 બધા જ વાક્યો કર્મણી વાક્ય પ્રયોગના છે બધા જ કર્તરી વાક્ય પ્રયોગના છે કુલ 2 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 1.62 Hz 1.32 Hz 1.52 Hz 1.42 Hz 1.62 Hz 1.32 Hz 1.52 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