Single phase transformer
What is Single phase transformer?
The phenomenon where a change in current in one coil induces an electromotive force (EMF) in an adjacent coil.
Key formula / rule: EMF Equation per winding
Key points
- Explain the construction and working principle of a single-phase transformer.
- Derive and apply the EMF equation and turns ratio relationships.
- Analyze the equivalent circuit of a practical transformer.
- Calculate efficiency and voltage regulation under various load conditions.
Common exam trap
Confusing ideal transformer assumptions with practical transformer behavior.
Definitions
- Term
Mutual Induction
- Meaning
The phenomenon where a change in current in one coil induces an electromotive force (EMF) in an adjacent coil.
- Term
Turns Ratio
- Meaning
The ratio of the number of turns in the primary winding to the number of turns in the secondary winding of a transformer.
- Term
Core Losses
- Meaning
Losses occurring in the magnetic core of a transformer, primarily due to hysteresis and eddy currents, which are largely independent of load.
- Term
Copper Losses
- Meaning
Losses occurring in the windings of a transformer due to the resistance of the conductors (I²R losses), which vary with the square of the load current.
- Term
Voltage Regulation
- Meaning
A measure of the change in the secondary terminal voltage from no-load to full-load conditions, expressed as a percentage of the full-load voltage.
- Term
Efficiency
- Meaning
The ratio of the output power to the input power of a transformer, indicating how effectively it converts electrical energy.
Learning objectives
Explain the construction and working principle of a single-phase transformer.
Derive and apply the EMF equation and turns ratio relationships.
Analyze the equivalent circuit of a practical transformer.
Calculate efficiency and voltage regulation under various load conditions.
Understand and differentiate between various types of losses in a transformer.
Formulae
- Name
EMF Equation per winding
- Note
E is RMS induced EMF, f is frequency, Φm is maximum flux in the core, N is number of turns.
- Expression
E = 4.44 f Φm N
- Name
Turns Ratio (Ideal Transformer)
- Note
N1, N2 are primary and secondary turns; V1, V2 are primary and secondary voltages; I1, I2 are primary and secondary currents.
- Expression
a = N1/N2 = V1/V2 = I2/I1
- Name
Total Losses
- Note
Pcore includes hysteresis and eddy current losses; Pcopper is I²R losses in windings.
- Expression
Plosses = Pcore + Pcopper
- Name
Efficiency
- Note
Pout is output power, Pin is input power. Often expressed as a percentage.
- Expression
η = (Pout / Pin) = (Pout / (Pout + Plosses))
- Name
Condition for Maximum Efficiency
- Note
Efficiency is maximum when constant losses (core) equal variable losses (copper).
- Expression
Pcore = Pcopper
- Name
Voltage Regulation
- Note
V2no-load is secondary voltage at no load, V2full-load is secondary voltage at full load. Often expressed as a percentage.
- Expression
VR = (V2no-load - V2full-load) / V2full-load
- Name
Approximate Voltage Regulation (referred to secondary)
- Note
I2FL is full-load secondary current, Req2 and Xeq2 are equivalent resistance and reactance referred to secondary, φ2 is load power factor angle.
- Expression
VR ≈ (I2FL * (Req2 cosφ2 + Xeq2 sinφ2)) / V2FL (for lagging power factor)
Prerequisites
Basic understanding of AC circuits (phasors, impedance, power factor).
Knowledge of Faraday's Law of Electromagnetic Induction.
Concepts of magnetic circuits (flux, MMF, reluctance).
Basic electrical engineering principles (Ohm's Law, Kirchhoff's Laws).
Common mistakes
Confusing ideal transformer assumptions with practical transformer behavior.
Incorrectly applying turns ratio for current transformation (I1/I2 = N2/N1).
Ignoring the impact of power factor on voltage regulation and efficiency calculations.
Miscalculating losses or efficiency, especially at partial loads.
Not understanding the significance of the equivalent circuit parameters.
Keywords
Transformer
Single Phase
Mutual Induction
EMF Equation
Turns Ratio
Core Loss
Copper Loss
Efficiency
Voltage Regulation
Equivalent Circuit
Step-up
Step-down
Practice preview
A 50 kVA single-phase transformer has full-load copper loss of 800 W and iron loss of 500 W. If the transformer operates at full load for 6 hours, half load for 10 hours, and no load for 8 hours in a day, calculate its a…
hard
What is the fundamental principle of operation of a single-phase transformer?…
easy
Which of the following is NOT a characteristic of an ideal single-phase transformer?…
easy