Question

20kVA, 300/240 V, 50Hz singal phase transformer has 80 turns on secondary winding. calculate number of primary winding turns, full load primary & secondary currents & maximum value of flux in the core.

Thank uh !​

Answer 1

The number of primary winding turns is 100 turns.

The full load primary & secondary currents are 66.66 A and 53.33 A.

The maximum value of flux in the core is 23.88 Weber.

What is transformer?

A transformer transfers electric energy from one AC circuit to one or more other circuits, either stepping up or stepping down the voltage.

a) Turns ratio N₁/N₂ = V₁/V₂ where N₁ = number of turns of secondary coil = 80 turns, N₂ = number of turns of secondary coil , V₁ = voltage in primary circuit = 300 V and , V₂ = voltage in secondary circuit = 240 V

Substitute the values and we get

N₁  =V₁/V₂ x  N₂

N₁= 300/240 x 80

N₁= 100 turns

Thus, the number of primary winding turns is 100 turns.

b) Power = V₁ I₁

20 x1000 = 300 x I₁

I₁ = 66.66 A

Turn ratio also represented as N₁/N₂ = I₂/I₁

Put the values, we have current in secondary circuit is

I₂ = N₂/N₁ x I₁

I₂= 80/100 x 66.66

I₂= 53.33 A

Thus, the current in the primary and secondary circuit is  66.66 A and 53.33 A.

c) Maximum flux equals (√2Vrms) / (N₁ x ω).

The rms voltage is Vrms = V₁ /√2

Vrms = 300/√2 = 212.1 V

The angular frequency ω = 2π /f

ω = 2π /50= 0.1256 rad/s

Substitute the values into the expression, we get

Φmax = (√2Vrms) / (N₁ x ω).

          = (√2 x 212.1)  / (100 x 0.1256)

          =23.88 Weber

Thus, the maximum flux is 23.88 Weber.

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