EE MCQ

SSC JE Electrical 2019 with solution SET-1
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The resistive component of the rotor with resistance R2 offered to the backward rotating flux wave in a single-phase motor runs with slip s is given as

The flux rotating in the clockwise direction (i.e. the direction of spinning) is called the forward rotating flux (φf) and that in the other direction is called the backward rotating flux (φb). The forward rotating flux has synchronous speed Ns (= 120 f/P) and the synchronous speed of rotating backward flux (anticlockwise) is –Ns.

Slip in forward Direction:-

Sf = (Ns – N)/Ns = s

Slip in backward Direction:-

Sb = (–Ns – N)/–Ns = s

Sb = (2Ns + Ns + N)/Ns

Sb = (2 – s)

At standstill, N = 0 so that Sf = Sb = 1. For forward rotating flux, the slip is s (less than 1) and for backward rotating flux, the slip is 2 – s (greater than 1). We know that in a 3-phase induction motor, the torque developed is directly proportional to effective rotor resistance.

The effective rotor resistance for single-phase motor is R2/s in the forward direction and R2/2 – s in the backward direction.

Related Question

SSC JE Electrical 2019 with solution SET-1
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In an AC network, the load connected is (10 + j10). The phase relation between the voltage applied and the current through the load is:

L0ad impedance Z = R + jX

z = 10 + 10j

Phase angle θ= tan−1(IL/IR)

θ= tan−1(10/10)

tanθ = 1

tanθ =45°

SSC JE Electrical 2019 with solution SET-1
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Three lamps are connected in series across a 120V supply and take a current of 1.5A. If the resistance of two of the lamps is 30Ω, what is the resistance of the third lamp?

Method 1

Supply Voltage = 120 V

Current I = 1.5 A

Total Resistance R = V/I = 120/1.5 = 80 Ω

Now resistance of 3 lamp be R1, R2, & R3

Req = R1 + R2 + R3

80 = 30 + 30 + R3

R3 = 20 Ω

Method 2

Applying KVL on the above circuit

120 = 1.5(R1 + R2) + 1.5R3

120 = 1.5(30 + 30) + 1.5R3

120 = 90 + 1.5R3

R3 = 30/1.5

R3 = 20 Ω

SSC JE Electrical 2019 with solution SET-1
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If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

At full load copper Loss = I2R

At half load copper Loss = (I/2)2 × R = I2/4 × R

400 = I2/4 × R

I2R = 4 × 400

I2R = Full load copper Loss = 1600 W

SSC JE Electrical 2019 with solution SET-1
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A 200 V d.c machine as Ra = 0.5 Ω and its full-load Ia= 20A. Determine the induced e.m.f when the machine acts as a motor

Given Data

Voltage Va = 200 v

Armature Resistance Ra = 0.5Ω

Armature Current Ia = 20 A

Induced EMF = Ea = ?

The Induced EMF of a DC machine working as a Motor is

Ea = Va − IaRa

Ea = 200 − 20 × 0.5

Ea = 190 V