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Yes, I meant you got it right, it is the output, not the input. Nice scenario and indeed interesting..

I did some research and generally in transformer efficiency problems they give you the LOAD at which you are required to find the efficiency at. Thus, it is the output not the input as said by @rg1. One more thing to consider, sometimes they specify the VA rating of the load and the PF. in this...

yes, I think very difficult to notice the 25% after using it to calculate the losses and you are ready to calculate for the efficiency. sometimes it is better to slow down and not rushing fast into solving problems familiar to you. unfortunately, I do. Ok, for the next part, I need you to...

interestingly, you did the same two mistakes I did. can you figure them out? I think you will..

@rg1 yes, calculation is straight, however sometimes I miss it in the math part. can you try it, I just want to see your solution before I tell my confusion.

I am posting this problem since I usually get it wrong first time. this post will help me remember not to make the same mistake always. Try it, A single phase 100 kVA, 4.6/0.480 kV transformer no-load loss is measured to be 300 W. The total loss at 75% loading is measured at 1200 W...

Look at this topic, it does have some good graphs also http://peguru.com/2011/06/per-unit-system-practice-problem-solved-for-easy-understanding/

just to mention, in the per unit system, voltage across the transformer is 1 Pu, both in primary and secondary, however actual voltages are different , that mean bases are different, right? this problem is all about getting the new base voltage in the motor side while given base voltage on the...

interesting, I think you should get both answers same. I will concentrate on your first approach since, I think, it is the one required by the question. Base MVA is defined as 200, no problem with this one and it is fixed across the whole system. base voltage is defined as 22 Kv on the...

As I said, just look at it intuitively and don't concentrate too much on the (ratio or the reverse). moving from LV side to the HV side then your multiplier (ratio) is bigger than 1, in this case 220/25 for T1. moving from HV side to the LV side then your multiplier (ratio) is less than 1, in...

I, usually think of it intuitively. suppose the base voltage of T1 on the primary side is 25 Kv same as its primary rating, then the base on the secondary is 220 Kv same as its secondary rating. thus, to transfer 25 Kv to secondary you multiply by 220/25 (ratio of transformer). in your case...

Thanks @rg1 and @TNPE for elaborating on the subject. As you mentioned rg1, the formulation of the question could have been better.

Ok, that mean capacitance required is different depending on the connection. will assume for this specific question star as it is usually the case. this will result in capacitance of 9 micro farad per phase. what confused me from the beginning was the solution provided. I think they are...

if we assumed delta connection, as you suggest power per phase is 50VA, however voltage across the capacitor is 208 V line voltage compared to 208/sqrt(3) across capacitors in wye connection. therefore, answer will be different. Is it so?

here we go

Thanks rg1, this is what I thought, but I faced one problem where they used the sqrt(3) formula. I will post later. Anyway, Xc we get, is it total or per phase?

Hi all, just a quick one, for a 3 phase system, we have V line and total reactive power consumed by the system Q. then capacitive reactance Xc = V line**2/Q  or sqrt(3)*V line**2/Q ? I am confused.