# secondary voltage of practical transformer question

I am working on a practical transformer question as seen below. My thought process to solve this question was:

• 1) Calculate the total current supplied by dividing the apparent power by the voltage.
• 2) Find the current in the excitation branch and subtract it from the total current supplied.
• 3) Use this value to multiply the series impedance to find the voltage drop across it.
• 4) subtract this value from the input voltage and use this value with the turns ratio to find the output voltage of the transformer.

however there seems to be missing information as I cannot move forward with step 2. How do I find the excitation current without the reactive power? There is another question in my assignment with the same information supplied so what am I missing here? Thank you.

I believe that 230/6600 implies that this is a step-up transformer. Since you are given the impedance referred to the high-voltage side, you probably should assume that the no-load losses consist entirely of iron losses on the input side. Assume the input is 6600 V across a resistor representing the iron losses.

I cannot move forward with step 2. How do I find the excitation current without the reactive power?

I think the question is suggesting that there are no core saturation losses; only core iron losses and that amounts to 700 watts. These losses are purely due to eddy currents. Without significantly more information in the question regarding the core hysteresis curve, I don't think you can make any assumptions about saturation losses. So assume these to be zero.

At 230 volts and 700 watts, the primary unloaded current is 3.043 amps.

But the input winding does have series components and you can assume that they are roughly equally shared between primary and secondary. So, the primary losses (referred to the secondary) are half of 5.34 + j16.76 i.e. 2.67 + j8.38 and, when returned to the primary (by dividing by the turns ratio squared = 823) you get (3.24 + j10.2) milli ohms.

I think that gives the full picture now although not knowing the magnetization current makes it somewhat of a theoretical exercise.

In a good design, efficiency of a transformer will be maximum when copper loss and iron losses are equal. But efficiency losses cannot be minimized then raising the cost of short circuit protection gear to become prohibitive.

Units are also designed to give rated voltage out at full MVA load with resistive load. Thus an apparent full load with pf=0.8 may have a slightly higher load impedance.

The no load excitation and core losses may start at 4% but at full load become linear asymptotic thru the origin around 10% due Zo and its inverse, 1/Zo, which defines the short cct./max load current ratio.

The efficiency thus rises from 0% vs 100% max load current with an asymptote to 90% effic. @ 10% max current. So the no load power of 700W is irrelevant here for computing output voltage given the output source impedance.

# Analysis

If pf=1 Given data.
150kVA, 220V 50Hz primary to 6600V secondary

• minimum load is $$\V^2/Pd=R=6600^2/150kVA=290.4 Ohms\$$

• I rated= 150kVA/6600V=

• source impedance, Zs=5.34 +j16.76
• next use pf=0.8 to define complex load impedance or current
• then compute output voltage in Vpu or actual.

There are 2 standard methods. Compute load loss or voltage rise with no load. Take your pick.