A 25-kVA transformer has 500 turns on the primary and 50 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Find the full-load primary and secondary currents, the secondary e.m.f. and the maximum flux in the core. Neglect leakage drops and no-load primary current.

Given:

    
• Apparent power, S = 25 kVA = 25000 VA
    
• Primary turns, N₁ = 500
    
• Secondary turns, N₂ = 50
    
• Primary voltage, V₁ = 3000 V
    
• Frequency, f = 50 Hz

1. Full-load Primary Current (I₁):

The apparent power of a transformer is given by S = V₁I₁.

Therefore, I₁ = S / V₁

\[I_1 = \frac{25000 \text{ VA}}{3000 \text{ V}} = 8.333 \text{ A}\]

2. Secondary e.m.f. (E₂):

For an ideal transformer, the voltage ratio is equal to the turns ratio:

\[\frac{E_2}{E_1} = \frac{N_2}{N_1}\]

Since leakage drops are neglected, E₁ is approximately equal to V₁.

\[E_2 = V_1 \times \frac{N_2}{N_1}\]

\[E_2 = 3000 \text{ V} \times \frac{50}{500} = 3000 \text{ V} \times 0.1 = 300 \text{ V}\]

3. Full-load Secondary Current (I₂):

The apparent power is also given by S = E₂I₂.

Therefore, I₂ = S / E₂

\[I_2 = \frac{25000 \text{ VA}}{300 \text{ V}} = 83.333 \text{ A}\]

4. Maximum Flux in the Core (Φ_m):

The EMF equation for a transformer primary winding is:

\[E_1 = 4.44 f N_1 \Phi_m\]

Again, assuming E₁ ≈ V₁:

\[\Phi_m = \frac{V_1}{4.44 f N_1}\]

\[\Phi_m = \frac{3000 \text{ V}}{4.44 \times 50 \text{ Hz} \times 500 \text{ turns}}\]

\[\Phi_m = \frac{3000}{111000} \approx 0.02703 \text{ Weber (Wb)}\]