# Single Phase/Three Phase Transformer problem



## baddriver (Dec 20, 2010)

Here's the problem:

A 480v to 480/277v three phase transformer is connected delta-wye. A single phase load drawing 10A is connected to one line of the secondary. There are no other loads on the transformer.

What are the primary winding currents?

What are the primary line currents?


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## jv21 (Dec 20, 2010)

thinking:

1. 277v

2. 480

but i really have no idea. Is this multiple guess?

ahhh.... helps to read the question..... current... of coarse. Then in that case - no clue


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## Flyer_PE (Dec 20, 2010)

I'm assuming the single phase load is a 277 Volt load.

If you connect a 10 amp single phase load to the wye-winding, you will have one primary winding with (10 * 277/480 = 5.77) 5.77 Amps. The other two windings will have no current.

For the lines supplying the delta winding, there will be current in only two of the phases. The magnitude of each will be 5.77 Amps.


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## baddriver (Dec 20, 2010)

Flyer_PE said:


> I'm assuming the single phase load is a 277 Volt load.
> If you connect a 10 amp single phase load to the wye-winding, you will have one primary winding with (10 * 277/480 = 5.77) 5.77 Amps. The other two windings will have no current.
> 
> For the lines supplying the delta winding, there will be current in only two of the phases. The magnitude of each will be 5.77 Amps.


Flyer, that's exactly how I solved the problem. The issue was a disagreement with a co-worker over the solution. His position is there must be some current on the third primary line that I say is zero. His reasoning is because of the 'mutual inductance' of the transformer magnetics. My position is this can be solved as a single 480/277 single phase transformer coil and the answer is the same.


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## Kahrlo (Dec 20, 2010)

baddriver said:


> Flyer_PE said:
> 
> 
> > I'm assuming the single phase load is a 277 Volt load.
> ...



Since only one phase of the Y winding has a load, the corresponding delta winding is the only one with current. The line currents therefore is equal to the phase current but in opposite direction (going to/ away) from the winding. But take note of the phase shift when considering delta-wye transformer. 30 deg. leading on the higher voltage side. (positive sequence)


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## baddriver (Dec 30, 2010)

I like to check problems like this with the simple "Power in = Power out" adage.

We are single phase, so P = VI

On the secondary side we have Psec = (277v)(10A) = 2770 VA

On the primary side we have Ppri = (480v)(5.77A) = 2770 VA

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Unfortunately I couldn't stop there.

Repeat the problem for the case where there are now loads on two of the secondary phases, connected line to neutral, drawing 10A each. Now what are the primary line currents and primary winding currents?


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## Kahrlo (Dec 30, 2010)

baddriver said:


> I like to check problems like this with the simple "Power in = Power out" adage.
> We are single phase, so P = VI
> 
> On the secondary side we have Psec = (277v)(10A) = 2770 VA
> ...



For a balanced load, defined in phasors, positive sequence:

the line current magnitudes are the same, only the phase angle for the HV side leads LV side by 30 deg.

the phase current for the delta winding is reduced by sqrt(3) and lags the line current phase angle by 30 deg.

So on the primary side, the line current magnitudes are 10A, the phase current magnitudes are 5.77A.


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## Flyer_PE (Dec 31, 2010)

^For a balanced system, that is correct. With just two phases in use, the line currents will not be balanced. What you will end up with in this condition is two of the primary windings will have currents of 5.77 Amps. The third primary winding current will be 0.

If you do the vector math, there will be two line currents at 5.77 Amps. The third will be 10 Amps.


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