EE MCQ

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°

Related Question

SSC JE Electrical 2019 with solution SET-1
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Identify the important feature of a DC series motor

In DC series motor Torque (Ta) increase as the Square of armature current (Ia) Ta ∝ Ia2. So DC motor provides high starting torque.

SSC JE Electrical 2019 with solution SET-1
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A three-phase induction machine, when it is operating like a motor, the range of slip is given by

Torque speed characteristic of induction motor

Motoring mode: 0 ≤ s ≤ 1

For this range of slip, the load resistance in the circuit is positive, i.e. torque developed is in the direction in which the rotor rotates.

In this region the value of slip lies between 0 to 11 i.e., slip is positive.

The motor rotates in the same direction as that of a rotating magnetic field.

At s = 0 (synchronous speed), the torque produced by the motor is zero because the induced voltage in the rotor is zero when N= Ns.

The torque increases as the slip increases while the air gap flux remains constant.

The torque-slip characteristic from no-load to somewhat beyond full-load is almost linear.

Generating mode: s <0

In this operating mode, the slip s is negative i.e., s < 0. The slip will be negative if and only if the rotor speed N is greater than the synchronous speed Ns (N> Ns). However, the rotor and R.M.F both rotate in the same direction.

In this region, the motor acts as a generator and return the power back to a.c. source.

Braking or plugging Mode: s > 1

In this region, the value of slip is greater than 1 and the rotor rotates in the opposite direction of the rotating magnetic field.

This is achieved by interchanging any two phases of the stator supply.

SSC JE Electrical 2019 with solution SET-1
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Hysteresis loss is NOT a function of

Hysteresis Loss = Kh × BM1.67 × f × v watts

where

Kh = Hysteresis constant depends upon the material

Bm = Maximum flux density

f = frequency

v = Volume of the core

Hence the hysteresis loss does not depend upon the ambient temperature.

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