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MGVCL Exam Paper (30-07-2021 Shift 1)
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Three equal star-connected inductors take 8 kW at 0.8 power factor when connected to a 460 V three - phase 3-wire supply. If one of the inductor in phase – Y is short circuited, then the line current in phase-Y is ___ Assume RYB sequence.

Here, Y phase is shorted,

Equation of 3-phase power = √3*VL*IL*cosφ

IL = P/(√3*VL*cosφ)

= 8000/(√3*400*0.8)

= 12.55 A

Phase impedance = Vp/Ip

= 460/(√3*12.55)

= 21.16 ohm

cosφ = 0.8, So φ = 37 degree.

Here Impedance of Y phase is shorted,

So, Neutral point and Y phase at same potential,

Three line voltages are given by,

Vry = 460∟0°

Vyb = 460∟120°

Vbr = 460∟-120°

Here N and Y phase at same potential,

Ir = (460∟-120°)/(21.16∟37°)

= -21.7∟83° A

Ib = (-460∟120°)/(21.16∟37°)

= 21.7∟-157° A

Ib = -(Ir + Iy)

= -21.7∟83° + 21.7∟-157°

= 37.6∟53° A

MGVCL Exam Paper (30-07-2021 Shift 1)
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A single phase motor connected to 400 V, 50 Hz supply takes 30 A at a power factor of 0.7 lagging. Calculate the capacitance required in parallel with the motor to raise the power factor to 0.9 lagging.

For single phase induction motor,

P = VIcosφ

P = 400*30*0.7

P = 8400 W

Q1 = P*tan(φ) = 8400*1.020

Q = 8568.82 VAR

Now power factor change to 0.9

So, φ' = 25.84 degree

Q = 8400*tan(25.84)

Q = 4067.95

Qnet = Q - Q'

Qnet = 8568.82 - 4067.95

Qnet = 4500.8

C = V²/(2ᴨf)

C = (400*400)/(2ᴨ*50)

C = 8.95*10⁻⁵

C = 89.5 μF