Module III: Single Phase Transformer - Solved Questions

Working Principle

Works on Faraday's law of electromagnetic induction and mutual induction.

1. AC voltage applied to primary → alternating current flows
2. Current produces alternating magnetic flux in core
3. Flux links both windings through iron core
4. Changing flux induces EMF in both windings (E₁ in primary, E₂ in secondary)
5. When load connected to secondary → power transfers

Transformation Ratio (K)

K = E₂/E₁ = N₂/N₁ = V₂/V₁ = I₁/I₂

Step-up: K > 1 (N₂ > N₁) | Step-down: K < 1 (N₂ < N₁)

Derivation

Let: N = turns, φ_m = max flux, f = frequency

Flux varies sinusoidally: φ = φ_m sin(2πft)

By Faraday's law: e = -N × dφ/dt

dφ/dt = φ_m × 2πf × cos(2πft)

Maximum dφ/dt = 2πf φ_m

Maximum EMF: E_m = N × 2πf φ_m

RMS value: E = E_m/√2 = (2π/√2) × f × N × φ_m

EMF Equation:

E₁ = 4.44 f N₁ φ_m

E₂ = 4.44 f N₂ φ_m

Definition

Efficiency = (Output Power / Input Power) × 100%

Formulas

η = Output / (Output + Losses) × 100%

η = (V₂I₂ cos φ) / (V₂I₂ cos φ + P_i + P_c) × 100%

At fraction 'x' of full load:

η = (x × VA × cos φ) / (x × VA × cos φ + P_i + x²P_c) × 100%

Max efficiency when: Iron losses = Copper losses (P_i = P_c)

1. Core Losses (Iron Losses) - Constant

Hysteresis Loss: Due to repeated magnetization/demagnetization

P_h = η × B_m^1.6 × f × V → Reduced by silicon steel

Eddy Current Loss: Due to circulating currents in core

P_e = K × B_m² × f² × t² → Reduced by laminated core

2. Copper Losses (I²R) - Variable

P_c = I₁²R₁ + I₂²R₂

Varies with square of load current

Loss	Nature	Reduction
Hysteresis	Constant	Silicon steel
Eddy current	Constant	Laminated core
Copper	Variable (∝I²)	Thick conductors

Parameter	Ideal	Practical
Winding resistance	Zero	Finite
Core losses	Zero	Has losses
Leakage flux	Zero	Some leakage
Magnetizing current	Zero	Required
Efficiency	100%	95-99%
Voltage regulation	Zero	2-5%
Power relation	P₁ = P₂	P₁ > P₂