According to my Hughes Electrical book,
eff=output power/(output power+losses) = 9.8/(9.8+0.53) = 94.87%
or more accurately...
eff= (input power-losses)/input power = (10-0.53)/10 = 94.7%
so I don't know.
Where's Tony?
The two equations are derived from first principles and are equivalent ways of calculating the efficiency of a transformer, neither is more accurate than the other.
The first answer is in error because the copper lossses are proportional to the load VI[SUP]2[/SUP], the copper loss at 9.8kW is therefore [0.33kW/(10/9.8)[SUP]2[/SUP]]= 0.317kW, the iron losses are approximately constant over the working load range, giving total losses at 9.8kW of 0.517kW.
So plugging these into equation 1:
eff = [9.8/(9.8+0.517)] x 100/1
eff = 95.0% (to 1DP)
The second answer is in error because the rated power 10kW is not the input power at 9.8kW.
From above the Input Power = Output Power + Total Losses
Input Power = 9.8kW + 0.517kW
Input Power = 10.317kW
Checking using equation 2
eff = (10.317-0.517)/10.317
eff= 95.0% (to 1DP) as was expected from equation 1
The above workings are based on the data provided in the OPs original question and as can be seen does not yield any of the multiple choice answers.
To achieve answer c) 90.4% the total losses would obviously need to be higher.
Total Losses = [(Output Power/eff)- output power]
Total Losses = [(9.8/0.904)-9.8]kW
Total Losses = 1.04kW
If we assume iron losses in the original question were correct at 0.2kW
Copper Losses = Total Losses-Iron Losses
Copper Losses = 1.04-0.2
Copper Losses = 0.84kW
Suggesting that the Copper Losses in the original question of 0.33kW might be a transcription error of the intended 0.83kW which would yield a rounded answer of c) 90.4% efficiency as proposed by Moses.
