SECTION A – Short Answers (2 × 10 = 20 Marks)
(a) A 6-pole alternator rotates at 1000 rpm. What is the frequency of generated voltage?
f=P×N120=6×1000120=50 Hzf = \frac{P \times N}{120} = \frac{6 \times 1000}{120} = 50\ \text{Hz}f=120P×N=1206×1000=50 Hz
Frequency = 50 Hz
(b) Methods for predetermining voltage regulation of a 3-phase alternator
EMF (Synchronous impedance) method MMF (Ampere-turn) method
ZPF (Zero Power Factor) or Potier method ASA (American Standards Association) method
(c) What are the V-curves of a synchronous motor?
The V-curves are plots of armature current vs field current at constant mechanical load.
They show that armature current is minimum at unity power factor and increases for both leading & lagging conditions (curve resembles “V” shape).
(d) Conditions for parallel operation of two single-phase alternators
Equal voltage magnitude Equal frequency
Equal phase sequence Equal phase angle (in phase)
(e) Why are rotor slots of an induction motor skewed?
To reduce magnetic hum and noise. To minimize cogging (magnetic locking).
To smooth torque and improve performance.
(f) Condition for producing maximum torque in a 3-phase induction motor
R2=X2R_2 = X_2R2=X2
(where R2R_2R2 = rotor resistance per phase, X2X_2X2 = rotor reactance per phase).
At this condition, torque is maximum and independent of rotor resistance.
(g) Define cogging in an induction motor
Cogging (Magnetic Locking):
When rotor and stator teeth align and magnetic locking occurs, preventing the motor from starting.
Occurs when stator & rotor slots are equal in number.
(h) Why is power factor of induction motor low at starting?
At starting, rotor speed is zero ⇒ slip s=1s = 1s=1.
Rotor current frequency is high ⇒ rotor reactance high ⇒ lagging current ⇒ low power factor.
(i) Different types of single-phase induction motors
Split-phase induction motor Capacitor start motor
Capacitor start–capacitor run motor Permanent split capacitor (PSC) motor
Shaded pole motor
(j) Modifications in a DC series motor to operate on AC supply
Use laminated core to reduce eddy current losses.
Add compensating winding to neutralize armature reaction.
Use high reactance field winding for reduced current.
Add interpoles for sparkless commutation.
SECTION B – Descriptive Questions (Any 3 × 10 = 30 Marks)
(a) Rotating Field System in Alternators
Preferred over stationary field because: Slip rings carry low DC current only.
Lighter, smaller rotor. Easier insulation and cooling.
Types of Rotors:
Salient pole rotor (Hydro generators) – large diameter, short axial length.
Non-salient (Cylindrical) rotor (Turbo alternators) – small diameter, long axial length.
(b) Working of Synchronous Motor
When connected to 3-phase supply, the stator creates a rotating magnetic field (RMF).
The rotor field (DC excitation) locks with RMF → synchronous speed.
Starting: Motor is not self-starting → use damper winding (acts as squirrel cage motor initially).
(c) Numerical – 3-Phase Induction Motor
Given:
R2=0.5Ω,X2=5Ω,6−pole,50HzR_2 = 0.5Ω, X_2 = 5Ω, 6-pole, 50HzR2=0.5Ω,X2=5Ω,6−pole,50Hz
(i) Synchronous speed Ns=120f/P=1000 rpmN_s = 120f/P = 1000\ rpmNs=120f/P=1000 rpm
(ii) For maximum torque: s=R2/X2=0.1⇒N=900 rpms = R_2/X_2 = 0.1 \Rightarrow N = 900\ rpms=R2/X2=0.1⇒N=900 rpm.
(iii) Ratio Tmax/Tst=(sst/smax)=(1/0.1)=10T_{max}/T_{st} = (s_{st}/s_{max}) = (1/0.1) = 10Tmax/Tst=(sst/smax)=(1/0.1)=10.
For half of max torque → use Rext=R2(1/√2−1)R_{ext} = R_2(1/√2 - 1)Rext=R2(1/√2−1).
(d) Star–Delta and Autotransformer Starters
Star–Delta Starter: Reduces voltage per phase by √3, limits starting current to 1/3 of DOL.
Autotransformer Starter: Reduces applied voltage by tapping ratio; better torque per ampere of line current.
(e) Shaded Pole & Permanent Split Capacitor Motor
Shaded Pole Motor: Small shaded copper band creates rotating field; simple but low torque.
PSC Motor: Capacitor permanently connected in series with auxiliary winding; quiet, reliable, used in fans.
SECTION C – Long Questions (Any 1 from each)
3(a) Alternator EMF Calculation
Given data → use:
E=4.44×f×N×Φ×Kd×KpE = 4.44 \times f \times N \times \Phi \times K_d \times K_pE=4.44×f×N×Φ×Kd×Kp
where Kd=sin(mα/2)msin(α/2)K_d = \frac{\sin(m\alpha/2)}{m\sin(\alpha/2)}Kd=msin(α/2)sin(mα/2), Kp=cos(β/2)K_p = \cos(\beta/2)Kp=cos(β/2).
Substitute f=50, m=2, β=180−150=30°f=50,\ m=2,\ \beta=180-150=30°f=50, m=2, β=180−150=30° to find flux per pole.
4(a) Load Sharing Between Parallel Generators
Load shared ∝ drop characteristics:
P1P2=Δf2Δf1\frac{P_1}{P_2} = \frac{Δf_2}{Δf_1}P2P1=Δf1Δf2
Use given frequency regulation to calculate 3000 kW load distribution and maximum load limit without overloading.
5(a) Torque–Slip Characteristic
At start (s = 1): Torque ∝ R₂.
At max torque: s=R2/X2s = R_2/X_2s=R2/X2.
Operating region: 0 < s < s_max → stable.
Graph: Starting → Rising → Peak → Falling torque curve.
6(a) Speed Control Methods of Induction Motor
Consequent Pole Method: Change stator pole number → changes synchronous speed.
V/f Method: Maintain constant flux by keeping V/fV/fV/f constant → used with variable frequency drives (VFDs).
7(a) Tests on Single Phase Induction Motor
No-load test: Determines no-load current, losses, power factor.
Blocked-rotor test: Determines equivalent circuit parameters (R₁, R₂, X₁, X₂).
7(b) Double Cage Induction Motor
Has two rotor cages:
Outer cage: High resistance, low reactance → better starting torque.
Inner cage: Low resistance, high reactance → good efficiency at running speed.
Equivalent circuit = two parallel rotor circuits.
