Answer to Question #124256 in Electric Circuits for victoria

Question

A three–phase transformer has its primary winding delta-connected and its secondary winding starconnected. The number of turns per phase on the primary is four times that on a secondary, and the secondary line voltage is 400V. A balanced load of 20kW, at power factor 0.8, is connected across the secondary terminals. Assuming an ideal transformer, calculate the primary voltage and the phase and line currents on the secondary and primary sides. Sketch a circuit diagram and indicate the values of the voltages and currents on the diagram.

Answer 1

There are two circuit connection stated in the problem first is the Delta connection and from the name itself "delta" means the shape of this connection is triangular and delta's elements are connected in series, and are 120 degrees apart.Second, is the Star connection wherein there is a neutral point that connects the elements or the coils of the circuit that are 120 degrees apart.

In this problem we are asked to find the primary voltage, phase and line currents on the secondary and primary sides.

Given:

V₂(secondary/load) = 440V

load = 20kW at pf(power factor) = 0.8

N₁(turns per phase on the primary) = 4

N₂(turns per phase on the secondary) = 1

Since the secondary transformer is Star connection. Therefore, secondary phase current is equal to the secondary line current.

Required:

V₁(primary) = ?

I_{primary ph}(phase current) = ?

I_{primary l-l}(line current) = ?

I_{sec l-l}(line current) = I_{sec ph}(phase current) = I₂=?

Formulas:

$V_1 = \frac{N_1*V_2}{N_2}* \frac{1}{\sqrt3}$ ,we multiply it by $\frac{1}{\sqrt3}$ because primary transformer is

connected in Delta.

$I_2= \frac{load}{\sqrt3*V_2*pf}$

I_{primary l-l}(line current) = $\frac{\sqrt3*I_2}{N_1}$

I_{primary ph}(phase current) = I_{primary l-l}(line current)* $\frac{1}{\sqrt3}$

Substitute the given:

$V_1 = \frac{4*440V}{1}* \frac{1}{\sqrt3} = 1016.136V$

$I_2= \frac{20kW}{\sqrt3*440V*0.8} = 32.804 A$

I_{primary l-l(line current)} = $\frac{\sqrt3*32.804}{4} = 14.205A$

I_{primary ph}(phase current) = $\frac{14.205A}{\sqrt3} = 8.201A$

Sketch:

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