MGVCL Exam Paper (30-07-2021 Shift 1) The frequency of the emf in the stator of a 4-pole induction motor is 50 Hz, and that in the rotor is 2 Hz. What is the slip and at what speed is the motor running? 4%, 1455 rpm 3%, 1440 rpm 3%, 1455 rpm 4%, 1440 rpm 4%, 1455 rpm 3%, 1440 rpm 3%, 1455 rpm 4%, 1440 rpm ANSWER EXPLANATION DOWNLOAD EXAMIANS APP Rotor frequency, f_r = s*f_ss = 2/50s = 0.04Ns = 120f/PNs = 1500 rpms = (Ns - Nr)/NsNr = Ns(1 - s)Nr = 1500*(1 - 0.04)Nr = 1440 rpm
MGVCL Exam Paper (30-07-2021 Shift 1) An RLC series circuit consists of R = 16 Ω, L = 5 mH and C = 2 μF. Calculate the quality factor. 4.125 2.125 3.125 5.125 4.125 2.125 3.125 5.125 ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For series RLC circuit,Q = (1/R)*√(L/C)Q = 1/16*√(5 mH/2 μF)Q = 50/16Q = 3.125
MGVCL Exam Paper (30-07-2021 Shift 1) Which of the following malicious programs does not replicate automatically? Trap doors Trojan horse Virus Worm Trap doors Trojan horse Virus Worm ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) If a 415 V, 50 Hz, star-connected, three-phase squirrel-cage induction motor is operated from a 415 V, 70 Hz supply, the torque that the motor can now provide while drawing rated current from the supply increases or reduces depending upon the rotor resistance increases remains the same reduces increases or reduces depending upon the rotor resistance increases remains the same reduces ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) A ___ is a device that forwards packets between networks by processing the routing information included in the packet. Firewall Bridge Router all the mentioned above Firewall Bridge Router all the mentioned above ANSWER DOWNLOAD EXAMIANS APP
MGVCL Exam Paper (30-07-2021 Shift 1) The current flowing to a balanced delta connected load through line 'a' is 15 A when the conductor of line 'b' is open. With the current in line 'a' as reference, compute the symmetrical components of the line currents. Assume the phase sequence of 'abc'. Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (7.5 - j 4.33) A Ia₀ = 0 AIa₁ = (7.5 - j13) AIa₂ = (7.5 + j13) A Ia₀ = 0 AIa₁ = (7.5 + j4.33) AIa₂ = (-7.5 - j4.33) A Ia₀ = 0 AIa₁ = (-7.5 - j13) AIa₂ = (-7.5 - j4.33) A ANSWER EXPLANATION DOWNLOAD EXAMIANS APP For the given question,Ib = 0Ia = -IcCalculation:Ia1 = 1/3*(Ia + a*Ib + a²*Ic)= 1/3*(15 + 15∟-120°)= 7.5 + j*4.33 AIa2 = 1/3*(Ia + a²*Ib + a*Ic)= (1/3)*(15 + 15∟120°)= 7.5 - j*4.33 AIao = (1/3)*(Ia + Ib + Ic)= (Ia - Ia)= 0
EXAMIANS