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Department of Electrical & Computer Engineering (ECE), RUET

2nd Year Even Semester (Session 2023-24)

Class Test (CT) Questions — CT-03

Course Code: ECE 2207
Course Title: Electrical Machines-I
Date: 26/08/2026
Time: 20 minutes
Total Marks: 20 (10 marks per question)

Question Paper Scan

CT-03 Question Paper Scan

Question Paper Transcript

Sl.	Question	COs	POs	Marks
01.	Schematically representation a single-phase transformer assuming the primary winding is connected to a sinusoidal voltage source, while the secondary winding is connected to a load. Also identify and label the variables associated with the primary and secondary sides of the transformer in the schematic.	CO1	PO1	10
02.	Consider an ideal single-phase transformer operating under the following assumptions: • The permeability of the core is constant over the range of transformer operation, and hence the reluctance of the core is constant. • There is no leakage flux; therefore, the same sinusoidally varying flux links both the primary and secondary windings. If the core flux is sinusoidal with frequency $f$ and maximum value $Φ_{m}$ , prove that $E_{1} = 4.44 f N_{1} Φ_{m}$ and $E_{2} = 4.44 f N_{2} Φ_{m}$ Variables having their usual meanings.	CO1	PO1	10

(Note: "Schematically representation a single-phase transformer" appears verbatim from the original printed question sheet).

Comprehensive Solutions & Analysis

Question 01: Schematic Representation of a Single-Phase Transformer

Question: Schematically represent a single-phase transformer assuming the primary winding is connected to a sinusoidal voltage source, while the secondary winding is connected to a load. Also identify and label the variables associated with the primary and secondary sides of the transformer in the schematic. [Marks: 10, CO: 1, PO: 1]

1. Schematic Circuit Diagram

Schematic representation of a 1-phase transformer connected to a sinusoidal source on primary and load on secondary with all labeled variables

Core and Winding Schematic:

Schematic representation of mutual flux and primary/secondary leakage fluxes in a transformer

2. Identification and Labeling of Variables

Primary Side Variables:

$v_{1} (t)$ or $V_{1}$ : Instantaneous and RMS primary applied terminal voltage (V).
$i_{1} (t)$ or $I_{1}$ : Instantaneous and RMS primary current drawn from the sinusoidal AC source (A).
$N_{1}$ : Number of turns on the primary winding.
$e_{1} (t)$ or $E_{1}$ : Instantaneous and RMS counter electromotive force (self-induced EMF) in the primary winding (V).
Polarity Dots ( $∙$ ): Mark terminal polarity according to the right-hand rule.

Secondary Side Variables:

$N_{2}$ : Number of turns on the secondary winding.
$e_{2} (t)$ or $E_{2}$ : Instantaneous and RMS mutually induced electromotive force in the secondary winding (V).
$v_{2} (t)$ or $V_{2}$ : Instantaneous and RMS secondary terminal voltage across the connected load (V).
$i_{2} (t)$ or $I_{2}$ : Instantaneous and RMS secondary load current (A).
$Z_{L}$ : Load impedance connected across the secondary terminals ( $Ω$ ), where $Z_{L} = R_{L} + j X_{L}$ .

Core and Magnetic Variables:

$Φ (t)$ : Instantaneous mutual magnetic flux linking both primary and secondary windings (Wb).
$Φ_{m}$ : Maximum (peak) value of mutual magnetic flux in the core (Wb).
$B_{m}$ : Maximum flux density in the core material ( $B_{m} = Φ_{m} / A$ , in $Wb/m^{2}$ or Tesla).
$A$ : Effective cross-sectional area of the core limb ( $m^{2}$ ).
$f$ : Supply frequency (Hz), where $ω = 2 π f$ (rad/s).
$K$ : Voltage transformation ratio ( $K = \frac{N _{2}}{N _{1}} = \frac{E _{2}}{E _{1}}$ ).

Question 02: Derivation of the Transformer EMF Equation

Question: Under the assumptions of constant core permeability and zero leakage flux, prove that $E_{1} = 4.44 f N_{1} Φ_{m}$ and $E_{2} = 4.44 f N_{2} Φ_{m}$ . [Marks: 10, CO: 1, PO: 1]

1. Basis and Assumptions

Constant permeability ( $μ = constant$ ): The magnetic core does not saturate. Core reluctance $R = \frac{l}{μ A}$ remains constant.
Zero leakage flux: All magnetic flux remains confined to the iron core. Exactly the same flux $Φ (t)$ links all $N_{1}$ primary turns and all $N_{2}$ secondary turns.

Let the sinusoidal core flux be: $Φ (t) = Φ_{m} sin (ω t) = Φ_{m} sin (2 π f t)$

Here:

$Φ_{m}$ = Maximum flux in the core (Wb)
$f$ = Frequency of the AC supply (Hz)
$ω = 2 π f$ = Angular frequency (rad/s)

2. Derivation for Primary Induced EMF ( $E_{1}$ )

By Faraday's Law of Electromagnetic Induction, the instantaneous EMF induced in the primary winding is: $e_{1} (t) = - N_{1} \frac{d Φ}{d t}$

Differentiate the flux expression: $\frac{d Φ}{d t} = \frac{d}{d t} [Φ_{m} sin (ω t)] = ω Φ_{m} cos (ω t)$

Substitute into the EMF equation: $e_{1} (t) = - N_{1} ω Φ_{m} cos (ω t)$

Using trigonometric identity $- cos (θ) = sin (θ - 9 0^{\circ})$ : $e_{1} (t) = N_{1} ω Φ_{m} sin (ω t - \frac{π}{2})$

This shows that the induced EMF lags the core flux by $9 0^{\circ}$ .

The maximum value (peak amplitude) of the primary induced EMF is: $E_{m 1} = N_{1} ω Φ_{m} = 2 π f N_{1} Φ_{m}$

For a pure sine wave, the Root-Mean-Square (RMS) value is: $E_{1} = \frac{E _{m 1}}{2} = \frac{2 π f N _{1} Φ _{m}}{2} = 2 π f N_{1} Φ_{m}$

Substitute the numerical value of $2 π$ : $2 π \approx 1.4142 \times 3.1416 \approx 4.44288 \approx 4.44$

Therefore: $E_{1} = 4.44 f N_{1} Φ_{m}$

3. Derivation for Secondary Induced EMF ( $E_{2}$ )

Because there is no leakage flux, the identical flux $Φ (t)$ links all $N_{2}$ turns of the secondary winding.

By Faraday's Law: $e_{2} (t) = - N_{2} \frac{d Φ}{d t} = - N_{2} ω Φ_{m} cos (ω t) = N_{2} ω Φ_{m} sin (ω t - \frac{π}{2})$

The maximum secondary induced EMF is: $E_{m 2} = N_{2} ω Φ_{m} = 2 π f N_{2} Φ_{m}$

The RMS value of the secondary induced EMF is: $E_{2} = \frac{E _{m 2}}{2} = \frac{2 π f N _{2} Φ _{m}}{2} = 2 π f N_{2} Φ_{m}$ $E_{2} = 4.44 f N_{2} Φ_{m}$

(Both equations proved)

4. Alternative Method: Average EMF & Form Factor

In one quarter cycle of duration $Δ t = \frac{T}{4} = \frac{1}{4 f}$ seconds:

Core flux increases from $0$ to $Φ_{m}$ .
Change in flux $ΔΦ = Φ_{m} - 0 = Φ_{m}$ Wb.

The average rate of change of flux is: $\frac{ΔΦ}{Δ t} = \frac{Φ _{m}}{1/ ( 4 f )} = 4 f Φ_{m} Wb/s$

The average EMF induced per turn is: $Average EMF per turn = 4 f Φ_{m} Volts$

For a sine wave, the Form Factor $K_{f}$ is: $K_{f} = \frac{RMS Value}{Average Value} = 1.11$

Therefore: $RMS value of EMF per turn = 1.11 \times 4 f Φ_{m} = 4.44 f Φ_{m} Volts$

Multiplying by the number of turns:

For primary winding ( $N_{1}$ turns): $E_{1} = 4.44 f N_{1} Φ_{m}$
For secondary winding ( $N_{2}$ turns): $E_{2} = 4.44 f N_{2} Φ_{m}$

Cross-References & Study Vault Links

Classroom Board Notes: ClassNoteByRaidah/Class_13.md
Faculty Lecture Slides: SlidesByMaam/L-06_ECE-2207.md
Textbook Chapter: Books/B.L._Theraja/Ch-32_Transformer.md
Semester Final Repeats: PrevYearQuestions/2024.md, PrevYearQuestions/2023.md, PrevYearQuestions/2021.md, PrevYearQuestions/2019.md (EMF equation is a primary repeat topic).

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