# Transformer Question

#### spikes2020

##### Member
I am studdying for the PE exam next week and have a question on one of the practice problems. I have the answer but it is twice as big as i thought it was... Is that because I^2 * R = P so (2)^2 * R = P so its 4 times not twice?

A power transformer has the following test data at rated voltage:
% of Name Plate kVA || Total Losses (W)
0 || 460
50 || 2370

The total transformer losses (W) at 100% of nameplate kVA are most nearly:

A 5200
B 8100
C 9480
D 11320

Solution

Poc = 460 W

P50% = 2370 - 460 = 1910 W
P100% = 4 * 1910 = 7640 W

Pt = 7640 + 460 = 8100W

#### GoldDigger

##### Moderator
Staff member
I am studdying for the PE exam next week and have a question on one of the practice problems. I have the answer but it is twice as big as i thought it was... Is that because I^2 * R = P so (2)^2 * R = P so its 4 times not twice?

A power transformer has the following test data at rated voltage:
% of Name Plate kVA || Total Losses (W)
0 || 460
50 || 2370

The total transformer losses (W) at 100% of nameplate kVA are most nearly:

A 5200
B 8100
Ci320

Solution

Poc = 460 W

P50% = 2370 - 460 = 1910 W
P100% = 4 * 1910 = 7640 W

Pt = 7640 + 460 = 8100W
Yes. At half power the transformer will be carrying half of the full load current, since the output voltage can be considered constant.
But the internal resistance is constant (ignoring temperate coefficient) so the internal losses will increase as the square of the load current.
There will be a very small correction because the magnetizing current in the primary will cause the full load current to be less than twice the 50% load current, but I am sure the question ignores this. Which is fine if the load has a high PF or that the resistive losses are all in the secondary only.

Last edited: