Question

A single-phase, 300 kVA, 11 kV/2.2 kV, 60 Hz transformer has the following equivalent circuit parameters referred to the high-voltage side: RC,p=57.60 kΩ XM,p=16.34 kΩ Req,p=2.78 Ω Xeq,p=8.45 Ω Answer the following questions:

(a) What is the no-load current as a percentage of full-load current (you need to find both currents referred to the primary side, no-load current: apply rated primary voltage to the primary, full-load current: apply rated secondary voltage to the load)?

Answer 1

The full-load primary current for the given transformer is 27.27 A, and the no-load current is approximately 0.182 A. The no-load current as a percentage of the full-load current is about 0.67%.

Step-by-step explanation:

To calculate the no-load current (Io) as a percentage of the full-load current (Ifl) for the given transformer, we first need to determine the full-load current on the primary side, since the transformer is rated at 300 kVA and 11 kV. The full-load primary current can be calculated using the formula:

Ifl = S / Vp

Where S is the apparent power (300 kVA) and Vp is the primary voltage (11 kV).

Using the formula, we get:

Ifl = 300,000 VA / 11,000 V = 27.27 A (Full-load current on the primary side)

The no-load current is found when the primary is energized without any load connected to the secondary. The magnetizing branch represented by Rc,p and Xm,p will draw the no-load current. Since we're dealing with a no-load condition, and the coil resistance and reactance are high, we can approximate that no-load current to be largely reactive and is calculated by:

Io = Vp / √(Rc,p^2 + Xm,p^2)

Plugging in the given values:

Io = 11,000 V / √(57,600,000^2 + 16,340,000^2) = 0.182 A

Finally, to find the no-load current as a percentage of the full-load current:

Percentage = (Io / Ifl) × 100 = (0.182 A / 27.27 A) × 100 ≈ 0.67%