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ECE 2207: Electrical Machines I

CT Questions: All Answers: Exam Style

Department: ECE, RUET | Session: 2023-24

How to use this file: Concise, exam-ready answers. Every key step is present without extra commentary. Focus on structure and critical equations.

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CT-01 (19/07/2026): Induction Motors

Q1. Prove that 3-φ stator flux rotates at synchronous speed. [10 Marks]

Setup: Three windings 120° apart in space. Balanced 3-phase supply gives:

$${\Phi }_R={\Phi }_m\sin (\omega t),{\Phi }_Y={\Phi }_m\sin (\omega t-120°),{\Phi }_B={\Phi }_m\sin (\omega t+120°)$$

At four key instants:

ωt	Φ_R	Φ_Y	Φ_B	Resultant
0°	0	−(√3/2)Φ_m	+(√3/2)Φ_m	1.5Φ_m ↑
60°	+(√3/2)Φ_m	−(√3/2)Φ_m	0	1.5Φ_m (60° rotated)
120°	+(√3/2)Φ_m	0	−(√3/2)Φ_m	1.5Φ_m (120° rotated)
180°	0	+(√3/2)Φ_m	−(√3/2)Φ_m	1.5Φ_m (180° rotated)

General proof (component method):

Resolve along X and Y axes:

$${\Phi }_x=\frac{3}{2}{\Phi }_m\sin (\omega t),{\Phi }_y=-\frac{3}{2}{\Phi }_m\cos (\omega t)$$

Magnitude: $${\Phi }_r=\sqrt{{\Phi }_x^2+{\Phi }_y^2}=\frac{3}{2}{\Phi }_m=1.5{\Phi }_m\text{(constant)}$$

Space angle: $\theta =\omega t-90°$, so $\frac{d\theta }{dt}=\omega =2\pi f$

$$\boxed{N_s=\frac{120f}{P}\text{ rpm}}$$

Conclusion: Resultant flux = $1.5{\Phi }_m$ (constant magnitude), rotates at synchronous speed $N_s$. (Proved)

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Q2. Explain the basic operating principle of an induction motor. [10 Marks]

1. RMF is produced. 3-phase supply creates a rotating magnetic field of constant magnitude $1.5{\Phi }_m$ at speed $N_s=\frac{120f}{P}$.

2. Field cuts rotor bars. At start $N=0$, relative speed $=N_s$. The rotating field sweeps across stationary rotor conductors.

3. EMF is induced. By Faraday's Law, $e=Bl{v}_{\text{rel}}$ is induced in each rotor bar.

4. Rotor current flows. End rings (squirrel cage) or external resistors (wound rotor) close the circuit. Induced EMF drives rotor current.

5. Torque is developed. By Lorentz force law, $F=BIl$ acts on current-carrying bars inside the stator field.

6. Rotor spins (Lenz's Law). Rotor turns in the same direction as the RMF to reduce relative motion.

7. Cannot reach $N_s$. If $N=N_s$: relative speed = 0 → EMF = 0 → current = 0 → torque = 0. Friction slows the rotor. Always $N<N_s$.

$$s=\frac{N_s-N}{N_s},f_r=sf$$

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CT-02 (03/08/2026): Induction Motors

Q1. Justify: Maximum torque is independent of R₂. [10 Marks]

Torque equation: $$T=\frac{k\cdot s{E}_2^2{R}_2}{R_2^2+s^2{X}_2^2},k=\frac{3}{2\pi N_s}$$

Slip at max torque: Differentiate w.r.t. $s$, set to zero:

$$\frac{dT}{ds}=0⟹R_2^2-s^2{X}_2^2=0⟹\boxed{s_{mT}=\frac{R_2}{X_2}}$$

Peak slip depends on $R_2$. Increasing $R_2$ shifts the torque peak to higher slip.

Value of $T_{\max }$: Substitute $s=R_2/X_2$:

$$T_{\max }=\frac{k\cdot \frac{R_2}{X_2}\cdot E_2^2\cdot R_2}{R_2^2+\frac{R_2^2}{X_2^2}\cdot X_2^2}=\frac{k\cdot \frac{R_2^2}{X_2}\cdot E_2^2}{2R_2^2}=\boxed{\frac{k{E}_2^2}{2X_2}}$$

$R_2$ cancels completely. $T_{\max }$ depends only on $E_2$ (supply voltage) and $X_2$ (standstill reactance).

Conclusion:

• Changing $R_2$ shifts peak position ($s_{mT}=R_2/X_2$) but not peak value.
• At $R_2=X_2$: $s_{mT}=1$, so maximum torque coincides with starting torque. (Justified)

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Q2. Explain the blocked rotor test of an induction motor. [10 Marks]

Setup: Rotor is locked ($N=0$, $s=1$). Analogous to the short-circuit test of a transformer.

Circuit:

3-φ Supply → Variac → [V, A, W1+W2] → Stator terminals
                                        [Rotor: LOCKED]

Procedure:

1. Lock the rotor mechanically.
2. Short-circuit slip rings (wound rotor).
3. Start with zero voltage. Raise via variac until rated stator current flows.
4. Record: $V_{br}$, $I_{br}$, $P_{br}=W_1+W_2$.

Why low voltage? Core loss ∝ $V^2$. At 10–15% rated voltage, iron loss ≈ 0. So $P_{br}$ = copper losses only.

Parameters (star-connected): $$Z_{01}=\frac{V_{br}/\sqrt{3}}{I_{br}},R_{01}=\frac{P_{br}}{3I_{br}^2},X_{01}=\sqrt{Z_{01}^2-R_{01}^2}$$ $$R_2^{\prime }=R_{01}-R_1,X_1\approx X_2^{\prime }=\frac{X_{01}}{2}$$

Necessities:

1. Gives full-load copper loss → efficiency calculation.
2. Gives $R_{01}$, $X_{01}$ → series branch of equivalent circuit.
3. Starting current at rated voltage: $I_{sc}=I_{br}\times (V_{\text{rated}}/V_{br})$.
4. Estimates starting torque: $T_{st}\propto I_{sc}^2{R}_2$.
5. Needed for circle diagram construction (gives $\cos {\varphi }_{sc}$ and $I_{sc}$).

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CT-03 (26/08/2026): Transformers

Q1. Schematic of a 1-φ transformer with variables labeled. [10 Marks]

Primary variables:

Symbol	Meaning
$v_1(t)$, $V_1$	Applied terminal voltage (instantaneous / RMS)
$i_1(t)$, $I_1$	Primary current from AC source
$N_1$	Number of primary turns
$e_1(t)$, $E_1$	Self-induced counter-EMF (lags $\Phi$ by 90°)

Secondary variables:

Symbol	Meaning
$N_2$	Number of secondary turns
$e_2(t)$, $E_2$	Mutually induced EMF
$v_2(t)$, $V_2$	Secondary terminal voltage
$i_2(t)$, $I_2$	Load current
$Z_L=R_L+j{X}_L$	Load impedance

Core variables:

Symbol	Meaning
$\Phi (t)$	Instantaneous mutual flux (Wb)
${\Phi }_m$	Peak mutual flux (Wb)
$A$	Core cross-sectional area (m²)
$K=N_2/N_1$	Voltage transformation ratio

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Q2. Prove $E_1=4.44f{N}_1{\Phi }_m$ and $E_2=4.44f{N}_2{\Phi }_m$. [10 Marks]

Given: $\Phi (t)={\Phi }_m\sin (\omega t)$. Same flux links both windings (no leakage).

Derivation for $E_1$:

By Faraday's Law: $$e_1(t)=-N_1\frac{d\Phi }{dt}=-N_1\omega {\Phi }_m\cos (\omega t)=N_1\omega {\Phi }_m\sin (\omega t-90°)$$

Peak value: $$E_{m1}=N_1\omega {\Phi }_m=2\pi f{N}_1{\Phi }_m$$

RMS value: $$E_1=\frac{E_{m1}}{\sqrt{2}}=\frac{2\pi f{N}_1{\Phi }_m}{\sqrt{2}}=\sqrt{2}\pi f{N}_1{\Phi }_m$$

Since $\sqrt{2}\pi =4.44$:

$$\boxed{E_1=4.44f{N}_1{\Phi }_m}$$

For $E_2$: Same flux, same derivation with $N_2$:

$$\boxed{E_2=4.44f{N}_2{\Phi }_m}\text{(Proved)}$$

Alternative (average EMF method): In quarter-cycle $T/4=1/(4f)$, flux changes by ${\Phi }_m$: $$\text{Average EMF/turn}=4f{\Phi }_m$$ $$E_{\text{rms}}=K_f\times 4f{\Phi }_m=1.11\times 4f{\Phi }_m=4.44f{\Phi }_m$$ Multiply by turns: $E_1=4.44f{N}_1{\Phi }_m$, $E_2=4.44f{N}_2{\Phi }_m$ (Same result)

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CT-04 (09/09/2026): Transformers

Q1. Effect of leakage flux on transformer operation (with schematic). [10 Marks]

Three types of flux:

• ${\Phi }_m$: mutual flux, links both windings through core, transfers power.
• ${\Phi }_{l1}$: primary leakage flux, links only $N_1$ turns, path through air.
• ${\Phi }_{l2}$: secondary leakage flux, links only $N_2$ turns, path through air.

Schematic:

Effects:

1. Leakage reactances: Each leakage flux induces a self-EMF lagging by 90°. This behaves like a series reactance: $$X_1=2\pi f{L}_{l1},X_2=2\pi f{L}_{l2}$$

2. Voltage drop inside windings: $$V_1=-E_1+I_1{R}_1+j{I}_1{X}_1$$ $$V_2=E_2-I_2{R}_2-j{I}_2{X}_2$$

3. Worse voltage regulation: Under lagging pf loads, $j{I}_2{X}_2$ reduces $V_2$.

4. Fault current limiting (beneficial): During a short circuit, $X_1+X_2$ limits the fault current and protects the transformer.

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Q2. Phasor diagram of an R-L loaded ideal transformer: step by step. [10 Marks]

Ideal transformer assumptions: $R_1=R_2=X_1=X_2=0$, $I_0=0$, so $V_2=E_2$ and $V_1=-E_1$.

Step 1: Reference phasor: Draw ${\Phi }_m$ along the +X axis (horizontal reference).

Step 2: Induced EMFs: By Faraday's Law, induced EMF lags flux by 90°. Draw $E_1$ and $E_2$ vertically downward (−Y axis).

Step 3: Secondary terminal voltage: No internal drops in ideal transformer. $V_2=E_2$ (downward).

Step 4: Secondary load current: R-L load, so $I_2$ lags $V_2$ by ${\theta }_2={\tan }^{-1}(X_L/R_L)$. Draw $I_2$ clockwise from $V_2$.

Step 5: Primary current: Ampere-turn balance: $N_1{I}_1=N_2{I}_2$. Primary current is opposite to $I_2$: $$I_1=-\frac{N_2}{N_1}{I}_2\text{(180° reversed)}$$

Step 6: Primary voltage: No drops. $V_1=-E_1$ (vertically upward, +Y axis).

Step 7: Power factor angle: Angle between $V_1$ and $I_1$ equals ${\theta }_1={\theta }_2$. Same pf on both sides.

Phasor summary:

Phasor	Angle
${\Phi }_m$	0°
$E_1=E_2=V_2$	−90°
$V_1$	+90°
$I_2$	$-90°-{\theta }_2$
$I_1$	$90°-{\theta }_1$ (= $90°-{\theta }_2$)

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Source questions: CT_01.md · CT_02.md · CT_03.md · CT_04.md

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