MGVCL Exam Paper (30-07-2021 Shift 3) The current in the coil of a large electromagnet falls from 6 A to 2 A in 10 ms. The induced emf across the coil is 100 V. Find the self-inductance of the coil. 0.25 H 0.75 H 1.25 H 0.5 H 0.25 H 0.75 H 1.25 H 0.5 H ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation of self inductance:V = Ls*(di/dt)Ls = V/(di/dt)= 100/((6 - 2)/(0.01))= 0.250 H= 250 mH
MGVCL Exam Paper (30-07-2021 Shift 3) વિષમ' ની સંધી છૂટી પાડો. - કયો વિકલ્પ સાચો છે? વિસ્ + અમ વીસ + અમ વી + સમ વિ + સમ વિસ્ + અમ વીસ + અમ વી + સમ વિ + સમ ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) Choose the best option from the given alternatives which can be substituted for the given word/sentence.Code of diplomatic etiquette and precedence Wardrobe Protocol Sheath Secular Wardrobe Protocol Sheath Secular ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 3) The threshold voltage of an N-channel enhancement mode MOSFET is 0.5 V. When the device is biased at a gate voltage of 3 V, pinch-off would occur at a drain voltage of 3.5 V 4.5 V 1.5 V 2.5 V 3.5 V 4.5 V 1.5 V 2.5 V ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Equation is given by:Vds = Vgs - VtVds = 3 - 0.5Vds = 2.5 V
MGVCL Exam Paper (30-07-2021 Shift 3) Rearrange the following to form a meaningful sentence and find the most logical order from the given options.P: globally, out of which over 200 millionQ: are now in India, andR: flipkart has 700 million usersS: growing exponentially RSQP RPSQ PQRS RPQS RSQP RPSQ PQRS RPQS 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 0.73 H and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 pF 0.73 mH and 13.85 pF 0.73 H and 13.85 pF 0.53 H and 15.83 pF 0.53 mH and 15.83 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
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