# Transformers question HELP PLEASE

#### brown_dynamite

Joined Feb 4, 2012
6
I would appreciate any help with this question:

A 100kVA transformer has 400 turns on the primary and 80 turns on the secondary. The primary
and secondary resistances are 0.3ohms and 0.01ohm respectively, and the corresponding leakage
reactances are 1.1ohms and 0.035ohm respectively. The supply voltage is 2.2kV. Calculate equivalent
impedance of the transformer referred to the primary side, and the magnitude of the secondary
terminal voltage for full load having a power factor of 0.8 lagging. Repeat this for full load with a
power factor of 0.8 leading. Assume as an approximation that the power factors seen on the
primary and secondary side of the transformer are the same and Sin=Sload
I can get the first part which is the equivalent impedance of 2.05 ohms angle
77.4 degrees.

But the next part the answer should be 425V and 447.2V

I tried working out the primary which is 100KVA/2.2KV = 45.45 A
hen using the turns ratio Is = 227.25
hence on the secondary side it is 100KVA/227.25 = 440V

#### brown_dynamite

Joined Feb 4, 2012
6
if anyone needs clarification or more info I can post it. Thanks

#### t_n_k

Joined Mar 6, 2009
5,455
if anyone needs clarification or more info I can post it. Thanks
Yes you should post your detailed working. Then it might be possible to spot where you are having difficulty.

The answer of 425V for 0.8pf lagging load is correct - as is 447V for 0.8pf leading.

The 'exact' values I have are 425.18V and 447.67V

Last edited:

#### brown_dynamite

Joined Feb 4, 2012
6
Hi first of all thank you for helping me. I think i have got the answer and here is my working
Ip= 100KVA / 2200 = 45.45<-36.87
then i work out the power including losses
Ip^2(Zeq) + S
(45.45<-36.87)^2(2.05<74.4) + 100E3<36.87
=103447<35.484

Ip'= 103447<35.484/2200
Ip'47<35.84
Is= 5* Ip'
Is=235.1<35.484

on secondary side
|Vs|=100KVA/235.1A = 425.35 V

of course it will differ slightly because of the amount of decimal places I used

#### brown_dynamite

Joined Feb 4, 2012
6
can anyone advise if the method i used is correct

#### t_n_k

Joined Mar 6, 2009
5,455
Firstly, there's a (mathematically) small but important error.

For a lagging power factor the apparent power angle is negative.

So S=100E3<-36.87 for the lagging pf case.

And S=100E3<+36.87 for the leading pf case.

Secondly, if the question allows you to assume the source apparent power equals the load apparent power then presumably you may simply find the primary referred load voltage for the lagging case as

which gives a secondary side load voltage of 425.37<-1.53° at a secondary current of Is=(100e3<-36.87)/(425.37<-1.53°)

In reality the approach of assuming a constant apparent power from input to output is theoretically & practically 'flawed' but the difference in the outcome is insignificant in this example.

#### brown_dynamite

Joined Feb 4, 2012
6
yes i think the question does say to assume Sin=Sload
but does go on to say that assuming Sin does not equal Sload is more accurate
The method i used overall is it still valid.

#### t_n_k

Joined Mar 6, 2009
5,455
yes i think the question does say to assume Sin=Sload
but does go on to say that assuming Sin does not equal Sload is more accurate
The method i used overall is it still valid.
I guess I'm making the point if you use the assumption that Sin=Sload why then go to the bother of including the step which re-estimates the value of Sin - contrary to the initial convenience of making assumption.

Why do this ....

Ip^2(Zeq) + S
(45.45<-36.87)^2(2.05<74.4) + 100E3<36.87
=103447<35.484

• brown_dynamite