MGVCL Exam Paper (30-07-2021 Shift 2) In the circuit of figure, the switch is closed to position-1 for a long time. At t = 0 the switch position is changed from postion-1 to postion-2. Find an expression for the current i(t). i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A i(t) = (E/R) sin√(LC) t A i(t) = (E/R) cos√(LC) t A i(t) = (E/R) sin[1/√(LC)]t A i(t) = (E/R) cos[1/√(LC)]t A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Standard eqution of current for this condition,i(t) = i(0+)*cos(wt)i(0+) = E/RFor LC circuit,w = 1/√(LC)i(t) = (E/R)*cos((1/√(LC))*t) A
MGVCL Exam Paper (30-07-2021 Shift 2) A delta-connected balanced three-phase load is supplied from a three-phase, 400 V supply. The line current is 20 A and the power taken by the load is 10 kW. Find the power factor. 0.72 0.62 0.82 0.52 0.72 0.62 0.82 0.52 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP P = 10 kWVL = 400 VIL = 20 A3- phse power, P = √3*VL*IL*cosφcosφ = 10*1000/(√3*400*20)= 0.7216
MGVCL Exam Paper (30-07-2021 Shift 2) A circular loop has its radius increasing at a rate of 2 m/s. The loop is placed perpendicular to a constant magnetic field of 0.1 Wb/m². When the radius of the loop is 2 m, the emf induced in the loop will be 0.8π V 0.2π V 0.4π V zero 0.8π V 0.2π V 0.4π V zero ANSWER EXPLANATION DOWNLOAD EXAMIANS APP emf = dΨ/dt = change of flux linkageemf = dф/dtemf = d(B*Area)/dtemf = B*d(area)/dtemf = B*d(πr²)/dtemf = Bπ*2r*dr/dtemf = Bπ*2r*change of radius per unit timeemf = 0.1*π*2*2emf = 4π
MGVCL Exam Paper (30-07-2021 Shift 2) Which of the following is a national organisation monitoring the doping control programme in sports in India? National Anti-Doping Agency Primary Anti-Doping Agency Indian Anti-Doping Agency International Anti-Doping Protocol National Anti-Doping Agency Primary Anti-Doping Agency Indian Anti-Doping Agency International Anti-Doping Protocol ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 2) A circuit breaker is rated 2500 A, 2000 MVA, 33 kV, 3 sec, 3-phase oil circuit breaker. Determine the breaking current. 25 kA 35 kA 2500 A 350 kA 25 kA 35 kA 2500 A 350 kA ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Breaking capacity of circuit breaker = √3*VL*IbVL - line voltage in voltIL - Line current/breaking in ampereVL = 33 kVS = 2000 MVAIb = 30*1000 (in kVA)/(√3*33)Ib = 34.99 kA
MGVCL Exam Paper (30-07-2021 Shift 2) સ્વાધીન' નો વિરોધીભાસી શબ્દ કયો છે? અસ્વાધીન કુવાધીન પરાધીન નીસ્વધીન અસ્વાધીન કુવાધીન પરાધીન નીસ્વધીન ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS