Question

A one-phase, 50 Hz core type transformer has core of cross-section 400 cm². The permissible maximum B = 1 T, Find the number of turns on high and low voltage sides for a 3000 V/220 V ratio.
1. N_lv = 338 turns, N_hv = 25 turns
2. N_lv = 25 turns, N_hv = 338 turns
3. N_lv = 48 turns, N_hv = 654 turns
4. N_lv = 35 turns, N_hv = 477 turns

Answer 1

Correct Answer - Option 2 : N_lv = 25 turns, N_hv = 338 turns

Concept:

In a transformer, an alternating current is applied to the primary winding, a current in the primary winding (magnetizing current) produces alternating flux in the core of the transformer. This alternating flux gets linked with the secondary winding, by mutual induction, so an emf gets induced in the secondary winding. This induced emf is given by the EMF equation of the transformer.

 E₁ = 4.44f N₁ ϕ_m     ……….(1)

 E₂ = 4.44f N₂ ϕ_m     ……….(2)

Where, N₁ = Number of turns in primary winding (high voltage side)

N₂ = Number of turns in secondary winding (low voltage side)

ϕ_m  = Maximum flux in the core (Wb)

Φ_m = B_m ×  A

B_m = Max flux density (T)

A = Area of core (m²)

f = Frequency of the AC supply (Hz)

E₁ = Induced emf on the primary side (high voltage side) (V)

E₂ = Induced emf on the secondary side (low voltage side) (V)

Calculation:

Given: f = 50 hz

B_m = 1 T

A = 400 cm² = 0.04 m²

E₁ = 3000 V

E₂ = 220 V

From equation (1)

E₁ = 4.44 B_m × A × f × N₁

_{\( {N_1} = \;\frac{{{E_1}}}{{4.44 \times {B_m} \times A \times f}}\)}

\( {N_1} = \;\frac{{3000}}{{4.44 \times 1 \times 0.04 \times 50}} = 337.84\)

N₁ ≈  338 turns

Similarly from equation (2)

\({N_2} = \;\frac{{{E_2}}}{{4.44 \times \;{B_m} \times A \times f\;}} = \;\frac{{220}}{{4.44 \times 1 \times 0.04 \times 50}} = 24.77\)

N₂ ≈  25 turns

So,Low voltage side turn N_lv = N₂ = 25 turns

High voltage side turn N_hv = N₁  = 338 turns