answers_exam_style/2020_answer.md

25.6 KB · commit d2f3b6e

← 2019 Answer | 🏠 Index | 2021 Answer →

ECE 2207: 2020 Semester Final: Exam Style Answers

RUET · ECE Dept · 2nd Year Odd Semester 2020 Full Marks: 72 · Time: 3 Hours · Attempt any 6 (3 from each section)

SECTION - A (Transformers: Q1 to Q4)

Question 1

(a) What is the purpose of laminating the core in a transformer? [02]

The iron core sits in an alternating magnetic field. This induces EMFs in the core itself, driving circulating currents called eddy currents. These currents cause $I^{2} R$ heating and waste energy.

Laminating the core (cutting it into thin sheets insulated from each other) breaks the low-resistance path for eddy currents. Each lamination has high resistance across its thickness. So eddy currents are confined to each thin sheet. Since power loss $\propto t^{2}$ (lamination thickness), thin laminations drastically reduce eddy current losses.

Typical lamination thickness: 0.3–0.5 mm for power frequency (50/60 Hz) transformers.

(b) Derive the equivalent circuit of a single-phase two-winding transformer. [04]

Exact Equivalent Circuit of Transformer

Step 1: Ideal transformer with no losses, no leakage: $\frac{V _{1}}{V _{2}} = \frac{N _{1}}{N _{2}} = a, I_{1} = \frac{I _{2}}{a}$

Step 2: Add core loss and magnetizing current (shunt branch): Primary draws no-load current $I_{0} = I_{c} + j I_{m}$ even at no load.

$I_{c}$ in phase with $V_{1}$ : represented by $R_{c} = V_{1} / I_{c}$ in shunt.
$I_{m}$ lags $V_{1}$ by 90°: represented by $X_{m} = V_{1} / I_{m}$ in shunt.

Step 3: Add primary winding resistance and leakage reactance: Series elements $R_{1}$ and $j X_{1}$ on the primary side.

Step 4: Refer secondary to primary: Replace $R_{2}$ , $j X_{2}$ with $a^{2} R_{2} = R_{2}^{'}$ , $j a^{2} X_{2} = j X_{2}^{'}$ on the primary side.

Step 5: Final approximate equivalent circuit (shunt branch at input):

$V_{1} \to [R_{1} + j X_{1} + R_{2}^{'} + j X_{2}^{'}] \to E_{1}$

Shunt branch ( $R_{c} ∥ j X_{m}$ ) connected across $V_{1}$ .

For simplicity, combine series elements: $R_{01} = R_{1} + R_{2}^{'}, X_{01} = X_{1} + X_{2}^{'}$

Approximate Equivalent Circuit with Shunt Branch at Input

(c) Explain the working of a transformer under no-load with neat sketch. [03]

When primary voltage $V_{1}$ is applied with secondary open ( $I_{2} = 0$ ):

Primary current $I_{0}$ flows (small, 2–10% of rated).
$I_{0}$ has two components:
- Magnetizing component $I_{m}$ (90° lagging from $V_{1}$ ): sets up the alternating core flux $Φ_{m}$ .
- Core-loss component $I_{c}$ (in phase with $V_{1}$ ): supplies hysteresis and eddy current losses.
Core flux $Φ_{m}$ induces primary EMF $E_{1} = 4.44 f N_{1} Φ_{m}$ (opposing $V_{1}$ ).
Same flux induces secondary EMF $E_{2} = 4.44 f N_{2} Φ_{m}$ .
Since secondary is open, $V_{2} = E_{2}$ . No output current flows.

$I_{0} = I_{c}^{2} + I_{m}^{2}, cos ϕ_{0} = \frac{I _{c}}{I _{0}} = \frac{P _{0}}{V _{1} I _{0}}$

No-load test circuit

(d) Transformer: turns ratio 4:1 step-down. No-load: 10A at pf 0.2 lag. Secondary load: 200A at 0.85 pf lag. Find primary current and pf. [03]

Given: $I_{0} = 10$ A, $cos ϕ_{0} = 0.2$ , $a = 4$ (step-down), $I_{2} = 200$ A, $cos ϕ_{2} = 0.85$ lag.

No-load current components: $I_{c} = I_{0} cos ϕ_{0} = 10 \times 0.2 = 2.0 A$ $I_{m} = I_{0} sin ϕ_{0} = 10 \times sin (78.46°) = 10 \times 0.9798 = 9.798 A$

Secondary current referred to primary: $I_{2}^{'} = I_{2} / a = 200/4 = 50$ A at $ϕ_{2} = 31.79°$ lag.

Phasor addition (taking $V_{1}$ as reference, lagging components negative):

Component	In-phase (A)	Quadrature (A)
$I_{0}$	$+ 2.0$	$- 9.798$
$I_{2}^{'}$	$+ 50 \times 0.85 = + 42.5$	$- 50 \times 0.527 = - 26.35$
$I_{1}$ total	$+ 44.5$	$- 36.15$

$I_{1} = 44. 5^{2} + 36.1 5^{2} = 1980.25 + 1306.82 = 3287.07 = 57.33 A$

$cos ϕ_{1} = \frac{44.5}{57.33} = 0.776 lagging$

Question 2

(a) Compare two-winding transformer and auto-transformer. [03]

Feature	Two-Winding Transformer	Auto-Transformer
Windings	Two separate windings	Single winding with a tap
Isolation	Primary and secondary are electrically isolated	No galvanic isolation
Copper used	More	Less (saving = $1 - k$ fraction)
Efficiency	Slightly lower	Higher (part of power conducted directly)
Size/Weight	Larger	Smaller and lighter
Cost	Higher	Lower
Voltage ratio	Any ratio practical	Better for close ratios ( $k \approx 1$ )
Short-circuit current	Limited by leakage	Higher (less impedance)
Application	Power transmission, isolation needed	Starters, lab variacs, close-ratio power

(b) Prove: copper saved in auto-transformer = $(1 - k)$ times that in ordinary transformer. [04]

For a two-winding transformer of rating $V A$ , secondary voltage $V_{2}$ , secondary current $I_{2}$ :

Total copper used $\propto$ total conductor volume $\propto N_{1} I_{1} + N_{2} I_{2}$
Since $N_{1} I_{1} = N_{2} I_{2} = S / V$ (approximate for ideal): copper $\propto 2 \cdot N \cdot I \propto$ total ampere-turns.

For an auto-transformer with $k = V_{2} / V_{1}$ (step-down, $k < 1$ ):

The common winding (shared section) carries current $(I_{2} - I_{1})$ . The series winding carries current $I_{1}$ .

Copper in auto-transformer: $W_{a u t o} \propto N_{1} I_{1} + N_{2} (I_{2} - I_{1})$

$= N_{1} I_{1} + N_{2} I_{2} - N_{2} I_{1} = N_{1} I_{1} (1 + \frac{N _{2}}{N _{1}}) - N_{2} I_{1}$

Since $N_{1} I_{1} = N_{2} I_{2}$ (approximately): $W_{a u t o} \propto N_{2} (I_{2} - I_{1}) + N_{1} I_{1} = ampere-turns of common section + series section$

Ratio of copper used: $\frac{W _{a u t o}}{W _{or d ina r y}} = 1 - k$

Therefore: $Copper saved = W_{or d ina r y} - W_{a u t o} = W_{or d ina r y} - (1 - k) W_{or d ina r y} = k \cdot W_{or d ina r y}$

Or equivalently: copper in auto-transformer is $(1 - k)$ fraction of ordinary transformer copper. (Proved)

(c) Draw circuit diagrams for OC and SC tests on a single-phase transformer. [02]

1. Open-Circuit (OC) / No-Load Test:

Connections: LV winding connected to rated supply voltage ( $V_{1}$ ) via Voltmeter ( $V$ ), Ammeter ( $A$ ), and LPF Wattmeter ( $W$ ). HV winding is left open-circuited.
Yields: Core loss ( $W_{0} \approx P_{cor e}$ ) and shunt branch parameters ( $R_{0}, X_{0}$ ).

Open-Circuit (OC) Test Circuit Diagram

2. Short-Circuit (SC) / Impedance Test:

Connections: LV winding is solidly short-circuited with a thick conductor strip. Reduced AC voltage ( $5 - 10%$ rated) is applied to HV winding via a Variac until rated current ( $I_{sc}$ ) circulates.
Yields: Full-load copper loss ( $W_{sc} \approx P_{c u}$ ) and equivalent series impedance ( $R_{01}, X_{01}, Z_{01}$ ).

Short-Circuit (SC) Test Circuit Diagram

(d) No-load test: Primary = 220V, Secondary = 110V, $I_{0} = 0.5$ A, Power = 30W. Find: (i) magnetizing current, (ii) loss component, (iii) iron loss. [03]

(Same as Q1(d) above: identical data and method.)

$P_{F e} = 30 W, I_{c} = 0.136 A, I_{m} = 0.481 A$

Question 3

(a) Why are transformers rated in kVA? [02]

Transformer losses are:

Iron (core) loss: Depends on supply voltage only. Independent of load current.
Copper loss: Depends on current squared ( $I^{2} R$ ). Depends on load current, not on load power factor.

The total loss, and hence temperature rise and efficiency, depend on voltage and current: not on the power factor of the load. Since different loads connected to the same transformer have different power factors, the transformer can handle a given $V \times I$ product regardless of whether the load is resistive, inductive, or capacitive. So its rating is expressed in volt-amperes (VA or kVA), not in watts (kW).

(b) What is an auto-transformer? Show copper saving compared to ordinary transformer. [05]

Auto-transformer: A transformer in which a single winding acts as both primary and secondary. A tap on the winding divides it into a series section and a common section. Power is transferred partly by conduction (direct electrical connection) and partly by induction (magnetic coupling).

$k = \frac{V _{2}}{V _{1}} = \frac{N _{2}}{N _{1}} (transformation ratio)$

Copper saving proof:

In an ordinary transformer, total copper $W_{T} \propto (N_{1} I_{1} + N_{2} I_{2})$ .

For the same kVA rating: $V_{1} I_{1} = V_{2} I_{2} = S$

In an auto-transformer:

Series winding carries $I_{1}$ , has $(N_{1} - N_{2})$ turns → copper $\propto (N_{1} - N_{2}) I_{1}$
Common winding carries $(I_{2} - I_{1})$ , has $N_{2}$ turns → copper $\propto N_{2} (I_{2} - I_{1})$

Total auto copper: $W_{A} \propto (N_{1} - N_{2}) I_{1} + N_{2} (I_{2} - I_{1}) = N_{1} I_{1} - N_{2} I_{1} + N_{2} I_{2} - N_{2} I_{1}$ $= N_{1} I_{1} + N_{2} I_{2} - 2 N_{2} I_{1} = W_{T} - 2 N_{2} I_{1}$

Since $N_{2} / N_{1} = k$ and $I_{2} / I_{1} = 1/ k$ :

$\frac{W _{A}}{W _{T}} = 1 - k$

$Copper saved = k \cdot W_{T}$

Greater saving for $k \to 1$ (small voltage difference). Auto-transformers are most economical when the voltage ratio is close to 1 (e.g., 400/415V boosters, 3300/11000V is less efficient as auto-transformer).

(c) Step-by-step equivalent circuit of a transformer referred to primary1. Ideal Transformer Core Model:** Start with an ideal core (zero winding resistance, zero leakage flux, infinite permeability, zero core loss). $E_{1} = a E_{2}, I_{1} = I_{2} / a where a = N_{1} / N_{2}$ Step 1: Ideal Transformer Model

2. Winding Resistances and Leakage Reactances: Add practical winding series resistance ( $R_{1}, R_{2}$ ) and leakage reactance ( $X_{1}, X_{2}$ ) on both sides. $V_{1} = E_{1} + I_{1} (R_{1} + j X_{1}), E_{2} = V_{2} + I_{2} (R_{2} + j X_{2})$ Step 2: Practical Windings with Resistance and Leakage Reactance

3. Core Excitation Shunt Branch: Add a parallel branch across $E_{1}$ to model core iron loss ( $R_{c}$ ) and magnetizing reactance ( $X_{m}$ ). The total no-load current is $I_{0} = I_{c} + I_{m}$ . Step 3: Complete Circuit Model with Physical Core Shunt Branch

4. Transferring Secondary to Primary (Exact Equivalent Circuit): To eliminate the ideal transformer, transfer secondary parameters to the primary using $a^{2}$ : $R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, V_{2}^{'} = a V_{2}, I_{2}^{'} = I_{2} / a$ Step 4: Exact Equivalent Circuit

5. Approximate Equivalent Circuit: Since $I_{0}$ is small and the primary voltage drop $I_{0} (R_{1} + j X_{1})$ is negligible, move the shunt branch to the primary terminals. The series impedances combine to $R_{01} = R_{1} + R_{2}^{'}$ and $X_{01} = X_{1} + X_{2}^{'}$ . Step 5: Approximate Equivalent Circuit

Question 4

(a) Fields of application of auto-transformer. [03]

Starting of induction motors: Reduced-voltage starting (auto-transformer starter).
Laboratory variacs: Variable voltage AC supplies for testing.
Power transmission inter-ties: Close-voltage-ratio interconnections between two power systems (e.g., 400kV/345kV).
Railway traction: 25kV/12.5kV boosters along the track.
Voltage stabilizers: Automatic voltage regulators for small consumers.
Fluorescent lamp ballasts and dimmers: Voltage adjustment.

(b) 3-phase supply continuity with one burnt phase. Open-delta 57.7% proof. [05]

Open-Delta (V-V) Connection Circuit and Phasor Diagram

Yes: continuity is possible using the open-delta (V-V) connection.

If one transformer in a Δ-Δ bank burns out, the remaining two can be reconnected in open-delta. Three-phase power still reaches the load, though at reduced capacity.

Proof that open-Δ = 57.7% of closed-Δ:

Let each transformer be rated $S$ kVA.

Closed-Δ (3 transformers): Total capacity $= 3 S$ kVA.

Open-Δ (2 transformers): Each transformer carries rated current $I$ at rated voltage $V$ . Each transformer delivers $S = V I$ kVA. But for a balanced 3-phase load, the power factor of each transformer in open-Δ differs:

One transformer supplies at power factor $cos (30° + ϕ)$
The other at $cos (30° - ϕ)$

For unity load pf ( $ϕ = 0$ ): Each transformer works at pf $= cos 30° = 3 /2$ .

Output of each transformer $= V \cdot I \cdot \frac{3}{2}$

Total open-Δ output $= 2 \times V \cdot I \cdot \frac{3}{2} = 3 \cdot V I = 3 \cdot S$

$\frac{Open-Δ}{Closed-Δ} = \frac{3 S}{3 S} = \frac{1}{3} = 0.577 = 57.7%$

(Proved)

(c) Two 25 kVA transformers in open-Δ, supply 220V balanced 3-phase load. [04]

(i) Total load without overloading:

Open-Δ rating $= 3 \times 25 = 43.3$ kVA.

Check: Each transformer handles 25 kVA. Two transformers in open-Δ: $2 \times 25 \times cos 30° = 2 \times 25 \times 0.866 = 43.3$ kVA.

$Load without overloading = 43.3 kVA$

(ii) When third 25 kVA transformer closes the Δ:

Closed-Δ rating $= 3 \times 25 = 75$ kVA.

$Total load with closed-Δ = 75 kVA$

SECTION - B (Induction Motors: Q5 to Q8)

Question 5

(a) Classify AC motors. [02]

AC Motor classification tree showing Synchronous Motor and Asynchronous Motor

(b) Why is an asynchronous motor treated as a rotating transformer? [04]

Induction motor as a generalized rotating transformer showing stator primary, air gap, and short-circuited rotor secondary

Feature	Transformer	Induction Motor (Asynchronous Motor)
Primary	Primary winding	Stator winding
Secondary	Secondary winding	Rotor bars/winding
Medium	Static iron core	Rotating air gap
Power transfer	By mutual induction	By rotating magnetic field
Secondary circuit	Short-circuited (loaded)	Short-circuited (end rings)
Secondary current	Load current	Rotor current → produces torque

The stator is the "primary": it receives power from the supply. The rotor is the "secondary": it receives power by induction and converts it to mechanical work. At standstill, a 3-phase IM is essentially a 3-phase transformer with a short-circuited secondary. When the rotor runs at slip $s$ , the rotor EMF = $s E_{2}$ and rotor frequency = $s f$ : exactly like a transformer operating at reduced frequency.

(c) How can the equivalent circuit model of an induction motor be obtained? [06]

1. Stator Model and Standstill Rotor ( $s = 1$ ): At standstill, the motor acts like a transformer. Stator has $R_{1}, X_{1}$ and shunt branch $R_{c}, X_{m}$ . Rotor has $R_{2}$ and $X_{2}$ at frequency $f$ . Step 1: Stator and rotor transformer model

2. Rotor at Running Slip $s$ : Rotor frequency is $s f$ , induced EMF is $s E_{2}$ , and reactance is $s X_{2}$ . $I_{2} = \frac{s E _{2}}{R _{2} + j s X _{2}}$ Step 2: Rotor circuit at slip s

3. Frequency Transformation: Divide the current equation by $s$ to refer the rotor to stator frequency $f$ : $I_{2} = \frac{E _{2}}{R _{2} / s + j X _{2}}$ The rotor resistance is modeled as a variable resistance $R_{2} / s$ . Step 3: Frequency transformation

4. Power Separation: Split $R_{2} / s$ into actual copper loss and mechanical load components: $\frac{R _{2}}{s} = R_{2} + R_{2} (\frac{1 - s}{s})$ $R_{2}$ causes heat ( $P_{c u}$ ), and $R_{L} = R_{2} (1 - s) / s$ represents gross mechanical power ( $P_{m}$ ). Step 4: Power separation

5. Exact Equivalent Circuit (Referred to Stator): Refer rotor parameters to stator using turns ratio $a = N_{1} / N_{2}$ : $R_{2}^{'} = a^{2} R_{2}, X_{2}^{'} = a^{2} X_{2}, R_{L}^{'} = R_{2}^{'} (\frac{1 - s}{s})$ Step 5: Exact Equivalent Circuit

6. Approximate Equivalent Circuit: Since stator voltage drop is small, the shunt branch can be shifted to the input terminals. Step 6: Approximate per-phase equivalent circuit

Question 6

(a) How can the direction of rotation of a 3-phase IM be reversed? [02]

Interchange any two of the three supply phase connections to the stator terminals. For example, swap phase A and phase B connections.

This reverses the phase sequence (A-B-C → B-A-C), which reverses the direction of the rotating magnetic field. Since the rotor follows the RMF, the rotor also reverses direction.

(b) Starting from equivalent circuit, derive power equations of an induction motor. [04]

Approximate Equivalent Circuit of Induction Motor

From the equivalent circuit, per-phase power flow:

Stator input power: $P_{1} = 3 V_{1} I_{1} cos ϕ_{1}$

Stator copper loss: $P_{s, C u} = 3 I_{1}^{2} R_{1}$

Stator iron loss: $P_{F e} = 3 V_{1}^{2} / R_{c} \approx 3 E_{1}^{2} / R_{c}$

Air-gap power (power transferred to rotor): $P_{g} = P_{1} - P_{s, C u} - P_{F e} = 3 I_{2}^{2} \cdot \frac{R _{2}}{s}$

Rotor copper loss: $P_{r, C u} = 3 I_{2}^{2} R_{2} = s \cdot P_{g}$

Gross mechanical power: $P_{m} = P_{g} - P_{r, C u} = P_{g} (1 - s) = 3 I_{2}^{2} R_{2} \frac{1 - s}{s}$

Net output power: $P_{o u t} = P_{m} - P_{f r i c t i o n + w in d a g e}$

Summary ratios: $P_{g} : P_{r, C u} : P_{m} = 1 : s : (1 - s)$

(c) Why is a synchronous motor not self-starting? [02]

A synchronous motor requires the rotor to rotate in synchronism with the stator RMF. At starting:

The stator RMF immediately rotates at synchronous speed $N_{s} = 120 f / P$ rpm.
The rotor (with field winding energized) is at rest.
The stator field grabs the rotor poles and tries to pull them around.
But the rotor has inertia. Before it can respond, the stator field has already rotated 180° and is now pulling in the opposite direction.
Net average torque over one cycle = 0.

So a synchronous motor cannot self-start. A damper winding (squirrel-cage bars on the rotor pole faces) is used to start it as an induction motor. Then the field winding is energized to pull it into synchronism.

(d) 400V, 50 Hz, 6-pole, 3-φ IM. Rotor power input = 75 kW. Rotor EMF makes 100 alternations per minute. Find: (i) slip, (ii) rotor speed, (iii) rotor Cu losses per phase, (iv) mechanical power. [04]

Given: $P_{g} = 75$ kW, $P = 6$ , $f = 50$ Hz

Rotor EMF frequency: 100 alternations per minute = 100/60 Hz $= 5/3$ Hz

(i) Slip: $f_{r} = s f ⟹ s = \frac{f _{r}}{f} = \frac{100/60}{50} = \frac{100}{3000} = 0.0333 = 3.33%$

(ii) Synchronous speed: $N_{s} = \frac{120 \times 50}{6} = 1000 rpm$

Rotor speed: $N = N_{s} (1 - s) = 1000 (1 - 0.0333) = 966.7 rpm$

(iii) Rotor copper losses (total): $P_{r, C u} = s \times P_{g} = 0.0333 \times 75000 = 2500 W$

Per phase: $P_{r, C u, phase} = \frac{2500}{3} = 833.3 W$

(iv) Mechanical power: $P_{m} = (1 - s) P_{g} = (1 - 0.0333) \times 75000 = 0.9667 \times 75000 = 72500 W = 72.5 kW$

Question 7

(a) Describe any one method for making single-phase IM self-starting. [03]

Capacitor-start method:

A single-phase IM has a main winding (M) and an auxiliary (starting) winding (A) placed 90° apart in space. A capacitor is connected in series with the auxiliary winding.

The capacitor advances the phase of the auxiliary winding current. If the capacitor value is chosen correctly, the auxiliary current leads the main current by nearly 90° in time. This 90° time-phase shift between $I_{m}$ and $I_{a}$ , combined with the 90° space separation of the windings, produces a rotating magnetic field. This RMF develops a starting torque.

Once the motor reaches about 75% of synchronous speed, a centrifugal switch opens and disconnects the auxiliary winding. The motor continues to run on the main winding alone.

Starting torque: $\approx 200$ –400% of full-load torque.

(b) Show that maximum torque varies proportionally with $V^{2}$ . [04]

The rotor EMF at standstill $E_{2} \propto V$ (supply voltage), since $E_{2} = K \cdot V$ (transformation ratio).

Maximum torque: $T_{m a x} = \frac{k E _{2}^{2}}{2 X _{2}}$

Since $E_{2} \propto V$ : $T_{m a x} \propto E_{2}^{2} \propto V^{2}$

$T_{m a x} \propto V^{2}$

This means a 10% voltage drop reduces maximum torque by about 19% (since $(0.9)^{2} = 0.81$ , a drop to 81% of original). Voltage sags are therefore very damaging to motor performance. (Shown)

(c) Prove that star-delta starter is equivalent to an auto-transformer of ratio $1/ 3$ (58%). [05]

Direct-on-line (DOL) starting: delta connection:

Each stator phase sees full line voltage $V_{L}$ . Per-phase starting impedance $Z_{s}$ . Starting current per phase: $I_{ϕ, D O L} = \frac{V _{L}}{Z _{s}}$

Line current (delta connection): $I_{L, D O L} = 3 I_{ϕ, D O L} = \frac{3 V _{L}}{Z _{s}}$

Star-delta starting: star connection:

Each phase sees $V_{L} / 3$ . Starting current per phase: $I_{ϕ, Y} = \frac{V _{L} / 3}{Z _{s}} = \frac{V _{L}}{3 Z _{s}}$

Line current (star connection) = phase current: $I_{L, Y} = I_{ϕ, Y} = \frac{V _{L}}{3 Z _{s}}$

Ratio of starting line currents: $\frac{I _{L, Y}}{I _{L, D O L}} = \frac{V _{L} / ( 3 Z _{s} )}{3 V _{L} / Z _{s}} = \frac{1}{3}$

Starting current is reduced to $1/3$ of DOL current.

Auto-transformer equivalence:

For an auto-transformer with ratio $x = V_{2} / V_{1}$ , the supply current is $x^{2}$ times the DOL current.

Here: $x^{2} = 1/3 ⟹ x = 1/ 3 = 0.577 \approx 58%$

$Star-delta starter \equiv Auto-transformer starter with ratio \frac{1}{3} = 57.7%$

(Proved)

Question 8

(a) Define: (i) Plugging, (ii) Slip. [02]

(i) Plugging: An electric braking method. The phase sequence of the stator supply is reversed while the motor is running. The motor develops a torque opposing its current direction of rotation. The motor decelerates rapidly. The supply must be disconnected when speed reaches zero, otherwise the motor reverses.

(ii) Slip: The fractional difference between synchronous speed and rotor speed: $s = \frac{N _{s} - N}{N _{s}}$

At standstill: $s = 1$ . At synchronous speed: $s = 0$ (never reached in practice). Normal full-load: $s = 0.02$ – $0.05$ (2–5%).

(b) Derive torque-slip characteristics of 3-phase IM and explain. [03]

Torque-Speed characteristics

Torque equation: $T = \frac{k s E _{2}^{2} R _{2}}{R _{2}^{2} + s ^{2} X _{2}^{2}}$

Key points on the curve:

At $s = 0$ (synchronous speed): $T = 0$ .
As $s$ increases from 0: torque increases (because $s E_{2}^{2} R_{2}$ grows faster than denominator initially).
At $s = s_{m T} = R_{2} / X_{2}$ : $T = T_{m a x}$ (maximum torque, pull-out torque).
For $s > s_{m T}$ : torque decreases (denominator grows faster).
At $s = 1$ (standstill): $T = T_{s t}$ (starting torque, usually 1.5–2 × full-load torque for typical motors).

The motor operates stably only in the region $0 < s < s_{m T}$ (positive slope of T-s curve). In this region, if load increases, speed drops (slip increases), torque increases to meet the load: a stable equilibrium. In the region $s > s_{m T}$ , the motor is unstable and will stall.

(c) Explain the double-field revolving theory for single-phase IM. [04]

Resolution of alternating flux into two oppositely rotating fields

A single-phase stator winding carries alternating current $i = I_{m} sin ω t$ . It creates a pulsating magnetic flux along one fixed axis: $Φ = Φ_{m} sin ω t$

Decomposition: This pulsating flux can be resolved into two equal halves rotating in opposite directions:

$Φ = Forward field \frac{Φ _{m}}{2} sin (ω t - θ) + Backward field \frac{Φ _{m}}{2} sin (ω t + θ)$

In vector form:

Forward RMF ( $Φ_{f}$ ): Magnitude $Φ_{m} /2$ , rotates at $+ N_{s}$ (counterclockwise).
Backward RMF ( $Φ_{b}$ ): Magnitude $Φ_{m} /2$ , rotates at $- N_{s}$ (clockwise).

Torques produced:

At rotor speed $N$ , slip for forward field $s_{f} = (N_{s} - N) / N_{s} = s$ . Slip for backward field $s_{b} = (N_{s} + N) / N_{s} = (2 - s)$ .

$T_{f} = f (s)$ : forward torque (positive direction)
$T_{b} = f (2 - s)$ : backward torque (negative direction)

At standstill ( $s = 1$ ): $s_{f} = 1$ , $s_{b} = 1$ , so $T_{f} = T_{b}$ . Net torque $= 0$ .

When running (given a push): At $s < 1$ : $s_{f} < s_{b}$ , so $T_{f} > T_{b}$ . Net torque $> 0$ in forward direction. Motor continues to run.

Torque-speed characteristic under double-field revolving theory showing zero starting torque

Conclusion: A single-phase IM has zero starting torque. It needs an auxiliary starting arrangement.

(d) How is the V-curve for a synchronous motor obtained? [03]

A V-curve shows the relationship between armature current $I_{a}$ and field current $I_{f}$ at constant load (constant mechanical output).

Procedure:

Run the synchronous motor at constant mechanical load.
Vary the field current $I_{f}$ from underexcited (low $I_{f}$ ) to overexcited (high $I_{f}$ ).
Record armature current $I_{a}$ at each field current setting.

Shape of curve:

At unity power factor (certain value of $I_{f}$ ): $I_{a}$ is minimum. This is the bottom of the V.
Below unity pf (underexcited): $I_{a}$ increases and lags voltage (lagging pf).
Above unity pf (overexcited): $I_{a}$ increases and leads voltage (leading pf, acts as capacitor bank).

Source: PrevYearQuestions/2020.md