
#1
May1310, 02:30 PM

P: 27

Hey. Im currently doing a bit of revision for an exam and I am struggling on the following question:




#2
May1310, 03:26 PM

P: 27

Hey. Ive found a bit more information regarding copper losses in transformers. It states:
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: 2.27^{2} x 0.05 = 0.258 W Copper loss for secondary: 20.83^{2} 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 = (5004.668)/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? 



#3
May1310, 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. 



#4
May1310, 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. 



#5
May1310, 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??




#6
May1310, 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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