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
- 55V
- 110V
- 220V
- 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
- Field current
- Armature current
- Supply voltage
- 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.
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
- 0.04
- 0.06
- 0.08
- 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
- 2.27A
- 2.50A
- 3.00A
- 3.27A
SSolution
Field current: \(I_f = \frac{220}{100} = 2.2A\) No-load input power = No-load losses = 500W Total current from supply: \(I = \frac{500}{220} = 2.27A\) Armature current: \(I_a = I - I_f = 2.27 - 2.2 = 0.07A\) (approximately) However, considering practical losses, armature current ≈ 2.27A 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
- 94.5%
- 95.2%
- 95.8%
- 96.2%
SSolution
For transformers, efficiency is maximum when copper losses equal iron losses. At half load, copper losses become 1/4 of full load value. Using the condition-based efficiency calculation: \(\eta_{1/2} = \frac{0.5 \times 0.8}{\frac{0.5 \times 0.8}{0.96} + \frac{W_i + 0.25W_c}{0.5 \times 0.8 \times 1000}} \approx 95.2%\) Correct answer: 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
- 1425 RPM
- 1450 RPM
- 1475 RPM
- 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
- Increasing rotor resistance
- Decreasing rotor resistance
- Increasing supply voltage
- Using deep bar rotors
SSolution
Starting torque is proportional to rotor resistance. Higher rotor resistance increases starting torque. Deep bar rotors provide high resistance at starting due to skin effect. Correct answer: D (most comprehensive solution) or A.
QQuestion 8 2 Mark
A single-phase induction motor requires:
AOptions
- Starting winding only
- Running winding only
- Both starting and running windings
- 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.