Question

A single phase, 100 KVA, 2000/200V, 50 Hz transformer has animpendence drop of 10% and resistance 5%. Calculate the (1) Regulation at
full load 0.8 PE. lagging (2) The value of the P.f at current regulation is zero​

Answer 1

Regulation at full load with 0.8 lagging is 9.2 and the value of power factor at current regulation is zero, is 0.866.

A single phase, 100 KVA, 2000/200V, 50 Hz transformer has an impedance drop of 10% and resistance 5%.

We have to calculate ;

1. Regulation at full load with 0.8 lagging.
2. The value of the power factor at current regulation is zero.

Voltage regulation at full load = %R cosФ + %X sinФ

Where

• Ф is power factor.
• %R is voltage drop due to resistance.
• and %X is voltage drop due to reactance.

We know,

$Z=\sqrt{R^2+X^2}\\\\\implies10=\sqrt{5^2+X^2}\\\\\implies X=5\sqrt{3}$

power factor, cosФ = 0.8 ⇒ sinФ = 0.6

Now, voltage regulation = 5 × (0.8) + 5√3 × (0.6) ≈ 9.2

Therefore the voltage regulation at full load with 0.8 lagging is 9.2

2. From above, it is clear that percentage drop due to reactance, %X = 5√3

For zero regulation, tanФ = IR/IX = %R/%X = 5/5√3 = 1/√3 = tan30°

⇒ Ф = 30°

Now the value of power factor = cosФ = 30° = √3/2 = 0.866

Therefore the value of power factor at current regulation is zero, is 0.866.

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