Voltage Transformation Ratio (K)

VOLTAGE TRANSFORMATION RATIO (K)

From equation (1) and (2) we get

E₂/E₁ = N₂/N₁ = K

This constant K is known as voltage transformation ratio.

(i) If N₂>N₁ then transformer is called as step-up transformer.

(ii) If N₂ < N₁ then transformer is called as step-down transformer.

For an ideal transformer input V_A = output V_A

V₁ I₁ = V₂ I₂ (or)

I₂/ I₁ = V₁/V₂ = 1/K

Example: 13

The maximum flux density in the core of a 250/3200 volts, 50 Hz single- phase transformer is 1.2 wb/m². It the emf per turn is 8 volts. Calculate (i) primary and secondary turns (ii) area of the core.

Solution:

E₁ =250 volts, E₂ = 3200 volts, ƒ = 50 Hz, B_m = 1.2 wb/m², emf/turn = 8 volts

(i) E₁ = N₁ × emf induced/turn

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 (ii) We know that

E₂ = 4.44 ƒN₂ B_m A

3200 = 4.44 × 50 × 400 × 1.2 × A

A = 0.03

Example: 14

A single-phase transformer has 400 primary and 1000 secondary turns. The net cross-sectional area of the core is 60 cm². If the primary winding be connected to a 50 Hz supply at 520 V. Calculate

(i) peak value of flux density in the core

(ii) voltage induced in the secondary windings

Solution:

N₁ = 400, N₂ = 1000, A = 60 cm², ƒ = 50 Hz, E₁ = 520 V.

(i) K = N₂/N₁

= 1000/400 = 2.5

E₂/E₁ = K

E₂ = K E₁

E₂ = 2.5 × 520

E₂ = 1300 volts

(ii) E₁ = 4.44 ƒ N₁ B_m A

520 = 4.44 × 50 × 400 × B_m × 60 × 10^-4

B_m = 0.976 wb/m²

Example: 15

The primary and secondary windings of a single-phase transformer are 350 and 35 respectively. If the primary is connected to a 2.2 KV, 50 Hz supply, determine the secondary voltage on no-load.

Solution:

N₁ = 350, N₂ = 35, E₁ = 2.2 KV, ƒ = 50 Hz

K = N₂/N₁ = 35/350 = 0.1

K = 0.1

E₂/E₁ = K

E₂ = K E₁

E₂ = 0.1 × 202 × 10³

E₂ = 220 volts

Example: 16

A 3000/200 V, 50 Hz, single-phase transformer is built on a core having an effective cross-sectional area of 150 cm2 and has 80 turns in the low-voltage windings. Determine

(i) the value of the maximum flux density

(ii) number of turns in high voltage windings

Solution:

E₂ = 3000 volts

E₁ = 200 volts

ƒ = 50 Hz

A = 150 cm²

N₁ = 80

E₁ = 4.44 ƒN₁ B_m A

200 = 4.44 × 50 × 80 × B_m × 150 × 10^-4

B_m = 0.75 wb/m²

E₂ = 4.44ƒN₂ B_m A

3000 = 4.44 × 50 × N₂ × 0.75 × 150 × 10^-4

N₂ = 1200

Example: 17

A 3300/230 V, 50 Hz, 1-phase transformer is to be worked at a maximum flux density of 1.2 wb/m² in the core. The effective cross-sectional area of the transformer core is 150 cm². Calculate the suitable values of the primary and secondary turns.

Solution:

E₁ = 230 volts

E₂ = 3300 volts

ƒ = 50 Hz

B_m = 1.2 wb/m²

A = 150 cm²

E₁ = 4.44 ƒN₁ B_m A

230 = 4.44 × 50 × N₁ × 1.2 × 150 × 10^-4

N₁ = 58

E₂ = 4.44 ƒN₂ B_m A

3300 = 4.44 × 50 × N₂ × 1.2 × 150 × 10^-4

N₂ = 825