Hey folks i am answering this question for a class i do, do i add fixed losses into the total amount of losses calculated like i have or would fixed losses be taken out of the total amount?

Easy enough question just not worded great, 30000/30000+1250* 100 = 97.4% efficiency that is by adding all losses together.

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`Tafe `

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It looks like there are 4 cases that they want you to consider, the 4 different power levels. So the fixed loss is added to the variable loss to get the total loss at each level. For the 25% case it would be 500W of total losses operating at 7500W power.

Where did you come up with that formula? it makes no sense.

% efficiency = (1- (fixed + variable power losses)/output power) * 100%

@25%, 93.33% efficient

Now i used to think that i was cool,

drivin' around on fossil fuel,

until i saw what i was doin',

was drivin' down the road to ruin. --JT

My math is

N= P out/ P out + losses *100

Output power plus losses is equal to input power, we got told to it that way.

Thanks for the help.

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Okay that makes sense, do it the way the instructor told you.

Now i used to think that i was cool,

drivin' around on fossil fuel,

until i saw what i was doin',

was drivin' down the road to ruin. --JT

That is correct and implicit in the question because it states ‘output power’ of the motor, i.e. that is the input power minus the losses. The losses are in the table, so for a given output power you calculate the required input power, rather than the other case which is from a given input power with losses at such power, calculating the actual output power that the motor then provides.