GATE EE Solved Problems

GATE 2017 Electrical Engineering (EE) Electrical Machines (2017)

Solved problems

Author: Prof. Mithun Mondal Subject: Electrical Machines Year: 2017 Total Questions: 8
Section 01

1-Mark Questions

QQuestion 1 1 Mark

A single-phase transformer has a turns ratio of 1:2. If the primary voltage is 110V, what is the secondary voltage?

AOptions

  1. 55V
  2. 110V
  3. 220V
  4. 440V

SSolution

For a transformer with turns ratio 1:2, the secondary voltage is twice the primary voltage. \(V_s = \frac{N_s}{N_p} \times V_p = 2 \times 110V = 220V\) Correct answer: C.

QQuestion 2 1 Mark

In a synchronous motor, the power factor can be controlled by varying:

AOptions

  1. Field current
  2. Armature current
  3. Supply voltage
  4. Load torque

SSolution

In a synchronous motor, the power factor is controlled by varying the field excitation (field current). Under-excitation leads to lagging power factor, over-excitation leads to leading power factor. Correct answer: A.

Section 02

2-Mark Questions

QQuestion 3 2 Mark

In a 3-phase induction motor, the synchronous speed is 1500 RPM and the rotor speed is 1440 RPM. Calculate the slip.

AOptions

  1. 0.04
  2. 0.06
  3. 0.08
  4. 0.10

SSolution

Slip \(s = \frac{N_s - N_r}{N_s} = \frac{1500 - 1440}{1500} = \frac{60}{1500} = 0.04\) Correct answer: A.

QQuestion 4 2 Mark

A DC shunt motor has an armature resistance of 0.5 Ω and field resistance of 100 Ω. The supply voltage is 220V. Calculate the no-load armature current if the no-load losses are 500W.

AOptions

  1. 2.27A
  2. 2.50A
  3. 3.00A
  4. 3.27A

SSolution

The field circuit draws \(I_f = \frac{220}{100} = 2.2\) A, which is separate from the armature and is dissipated in the field resistance alone.

At no load the armature input covers the 500 W of rotational and core loss:

\[VI_a - I_a^2R_a = 500 \quad\Rightarrow\quad 220I_a - 0.5I_a^2 = 500\]
\[I_a = 2.28 \text{ A}\]

The armature copper loss is negligible here, so \(I_a \approx \frac{500}{220} = 2.27\) A. Correct answer: A.

QQuestion 5 2 Mark

The efficiency of a transformer at full load with 0.8 power factor lagging is 96%. What will be the efficiency at half load with the same power factor?

AOptions

  1. 94.5%
  2. 95.2%
  3. 95.8%
  4. 96.2%

SSolution

Let the rating be \(S\). Full-load output is \(0.8S\), so the total full-load loss is

\[W_i + W_c = 0.8S\left(\frac{1}{0.96} - 1\right) = 0.0333S\]

Take the usual design condition that the transformer is most efficient at full load, i.e. \(W_i = W_c = 0.01667S\).

At half load the copper loss falls to a quarter while the iron loss is unchanged:

\[\eta_{1/2} = \frac{0.4S}{0.4S + W_i + 0.25W_c} = \frac{0.4S}{0.4S + 0.02083S} = 0.950\]

That is 95.0%, and the nearest option is B.

QQuestion 6 2 Mark

A 3-phase, 4-pole, 50Hz induction motor has a rotor resistance of 0.1Ω per phase. At what rotor speed will the maximum torque occur?

AOptions

  1. 1425 RPM
  2. 1450 RPM
  3. 1475 RPM
  4. 1485 RPM

SSolution

Synchronous speed: \(N_s = \frac{120 \times 50}{4} = 1500\) RPM Maximum torque occurs when \(s = \frac{R_}{X_2}\) For typical induction motors, this occurs at approximately 5% slip. \(N_r = N_s(1-s) = 1500(1-0.05) = 1425\) RPM Correct answer: A.

QQuestion 7 2 Mark

The starting torque of a 3-phase squirrel cage induction motor can be improved by:

AOptions

  1. Increasing rotor resistance
  2. Decreasing rotor resistance
  3. Increasing supply voltage
  4. Using deep bar rotors

SSolution

Starting torque rises with rotor resistance, but a squirrel-cage rotor is short-circuited internally and has no slip rings, so no external resistance can be inserted in it. The resistance has to be raised by rotor design instead. In a deep-bar rotor the rotor frequency at standstill equals the supply frequency, skin effect crowds the current into the top of the bar and the effective rotor resistance is high, giving a large starting torque; as the machine runs up the rotor frequency falls, the current spreads over the full bar depth and the resistance returns to a low value for efficient running. Correct answer: D.

QQuestion 8 2 Mark

A single-phase induction motor requires:

AOptions

  1. Starting winding only
  2. Running winding only
  3. Both starting and running windings
  4. External starting device

SSolution

Single-phase induction motors are not self-starting due to lack of rotating magnetic field. They require both starting winding (auxiliary) and running winding (main) to create phase difference. Correct answer: C.