Transformer Efficiency Calculations


by MathsDude69
Tags: calculations, efficiency, transformer
MathsDude69
MathsDude69 is offline
#1
May13-10, 02:30 PM
P: 27
Hey. Im currently doing a bit of revision for an exam and I am struggling on the following question:

Calculate the efficiency of a 500W transformer, which reduces the 230 AC voltage to 24 V. For the calculations, assume the resistances of the primary and secondary windings as 0.05 Ohm and 0.01 Ohm, respectively. The losses in the transformer core Pc = 70mW.
I understand the whole voltage/current/power/impedance ratio stuff, but cant seem to find any information on calculating efficiencies using the resistances or the transformer core losses. Any Suggestions?
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MathsDude69
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#2
May13-10, 03:26 PM
P: 27
Hey. Ive found a bit more information regarding copper losses in transformers. It states:

Copper Loss = Ip2Rp + Is2Rs
I would think given the question that the asnwer would be:

P = VI ∴ I = P/V

For the primary:
500/230 = 2.27 Amps

For the secondary:
500/24 = 20.83 Amps

Copper loss for primary:
2.272 x 0.05 = 0.258 W

Copper loss for secondary:
20.832 x 0.01 = 4.34 W

Total Copper Loss:
0.258 + 4.34 = 4.598 W

Total loss including core loss:

4.598 + 0.07 = 4.668 W

Total efficiency =
(500-4.668)/500 = 0.9907 = 99.07% efficient.

The problem with this is if we were only getting an efficiency of 99.07% the secondary wouldnt be developing 20.83 Amps! Does anyone know how to actually solve this question?
Averagesupernova
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#3
May13-10, 04:30 PM
P: 2,450
@ Maths:
I skimmed through this pretty fast but I would say that you should not have 500 watts for both the primary and secondary. With any efficiency of less than 100% you will naturally get you a lower secondary current than 20.83.

Studiot
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#4
May13-10, 04:39 PM
P: 5,462

Transformer Efficiency Calculations


You are nearly there except for the last bit.

The 500W refers to the output not the input for transformers so

input power = output power + total losses = 500 + 4.668 = 504.668.

efficiency = 1 - losses/input power = 1 - 4.668/504.668 = 99.07% expressed as a percentage.
MathsDude69
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#5
May13-10, 05:46 PM
P: 27
See this is where I am getting confused. If the input power is 504.668W instead of 500W wouldnt that make the current and thus the losses higher on the primary winding??
Averagesupernova
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#6
May13-10, 09:09 PM
P: 2,450
More power is lost, but more is put in too. The percentage of loss should be the same.


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