MGVCL Exam Paper (30-07-2021 Shift 3) The maximum value of power factor is ____ and it exists in a pure____ circuit. 0.866, resistive 0.866, inductive 1, inductive 1, resistive 0.866, resistive 0.866, inductive 1, inductive 1, resistive ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Value of power factor is varies from 0 to 1.R→PF = 1L→PF = 0 lagC→PF = 0 leadR-L→0 lag < PF < 1R-C→0 lead < PF < 1R-C-L→depends on value of R, L and C. PF = R/Z
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 ____ Linking Routing Switching Networking Linking Routing Switching Networking ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) સારું:નરસું::શુક્લ: ? અવિશાની અમર પરમ કૃષ્ણ અવિશાની અમર પરમ કૃષ્ણ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Murray loop test is performed on a faulty cable 300 m long. At balance, the resistance connected to the faulty core was set at 20 Ω and the resistance of the resistor connected to the sound core was 40 Ω. Calculate the distance of the fault point from the test end. 100 m 400 m 200 m 300 m 100 m 400 m 200 m 300 m ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Where,R2 is the resistance connected to the faulty core in ohmR2 is the resistance of the resistor connected to the sound core in ohmR1 Distance of fault location (Lx) = R2/(R1 + R2)*(2*L)= (20/60)*600= 200 m.
MGVCL Exam Paper (30-07-2021 Shift 3) 1. મોટેભાગે 'જાણે' શબ્દ હોય ત્યારે,ઉત્પ્રેક્ષા અલંકાર બને છે.૨. જયારે ઉપમેય અને ઉપમાન એક જ હોય ત્યારે ઉપમા અલંકાર બને છે.સરખાવવામાં આવેલ બે શબ્દોની વચ્ચે 'જયારે', 'જેવો', 'જેવી' જેવા શબ્દો આવે ત્યારે રૂપક અલંકાર બને.જયારે ટીકા કે નિંદા કે વ્યંગના રૂપે પ્રશંસા કરાય ત્યારે અતિશયોક્તિ અલંકાર બને.ઉપરોક્ત વિધાનોમાંથી કયું/ક્યાં વિધાનો સાચા છે. માત્ર 1 માત્ર 4 2, 4 1, 3 માત્ર 1 માત્ર 4 2, 4 1, 3 ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) A three phase four pole 50 Hz induction motor has a rotor resistance of 0.02 Ω/phase and stand-still reactance of 0.5 Ω/phase. Calculate the speed at which the maximum torque is developed. 1440 rpm 1525 rpm 1500 rpm 1475 rpm 1440 rpm 1525 rpm 1500 rpm 1475 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Condition for maximum torque,Sm = R2/X2= 0.04Ns = (120*f)/P= (120*50)/4= 1500 rpmS = (Ns - Nr)/NsNr = Ns(1 - Sm)= 1500*(1 - 0.04)= 1440 rpm
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