# Having trouble understanding current in Delta load.

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##### Senior Member
I want to thank you for this! Not only have you answered my original question but have also helped me understand other problems I have had, mostly concerning open - delta connections. :thumbsup:

I'm sorry for the the other readers who had problems understanding the question.

#### Smart \$

##### Esteemed Member
If you could disconnect the interconnection of the 3 coils (that were connected in a delta), what would the voltages be

across any of the 3 coils (I assume they will all be the same)?

between the two coils where they used to be connected?

Thanks to anyone that can answer this (remember, you are talking to someone of limited education, thanks).
If you disconnected at the interconnection, the voltages would be zero

Primary - assuming the windings remain connected to the supply lines, yes voltage would remain same, but still zero across separated winding connections as each pair is still connected in some way to the same supply line.

Secondary - same as primary, but if not connected to common load lines the winding to winding voltage would be dependent on the mutual coupling effect (i.e. inductive coupling).

#### hardworkingstiff

##### Senior Member
If you disconnected at the interconnection, the voltages would be zero

Primary - assuming the windings remain connected to the supply lines, yes voltage would remain same, but still zero across separated winding connections as each pair is still connected in some way to the same supply line.

Secondary - same as primary, but if not connected to common load lines the winding to winding voltage would be dependent on the mutual coupling effect (i.e. inductive coupling).

1st, thanks for attempting to decipher my question and answer it.

2nd, your answer has me even more confused. I can see too that I ask questions that don't have enough info. So, here I go again.

I was under the understanding that we were mostly just talking about the secondary of a transformer (in this thread) that had the secondary connected as a delta. So that would mean (if I understand this correctly) we have 3 coils in which the + of coil 1 is connected to the - of coil 2 and the + of coil 2 is connected to the - of coil 3 and the + of coil 3 is connected to the - of coil 1 (or something similar). The 3 points in which have the +/- connections from the coils would be the line voltage output from the transformer (with or without a load).

So, am I to understand that if those 3 connections were taken apart, the primary would no longer induce a voltage on the 3 coils of the secondary? I think that's what you said, but I'm not sure. If I indeed did understand correctly, then I need some education (if someone is willing) as to why disconnecting the interconnection of those 3 connections will disable all of the secondary coils. It does not make sense to me. (sorry I'm so dense here).

#### Smart \$

##### Esteemed Member
...
So, am I to understand that if those 3 connections were taken apart, the primary would no longer induce a voltage on the 3 coils of the secondary? ...
Each winding will still have the same voltage.

#### kwired

##### Electron manager
1st, thanks for attempting to decipher my question and answer it.

2nd, your answer has me even more confused. I can see too that I ask questions that don't have enough info. So, here I go again.

I was under the understanding that we were mostly just talking about the secondary of a transformer (in this thread) that had the secondary connected as a delta. So that would mean (if I understand this correctly) we have 3 coils in which the + of coil 1 is connected to the - of coil 2 and the + of coil 2 is connected to the - of coil 3 and the + of coil 3 is connected to the - of coil 1 (or something similar). The 3 points in which have the +/- connections from the coils would be the line voltage output from the transformer (with or without a load).

So, am I to understand that if those 3 connections were taken apart, the primary would no longer induce a voltage on the 3 coils of the secondary? I think that's what you said, but I'm not sure. If I indeed did understand correctly, then I need some education (if someone is willing) as to why disconnecting the interconnection of those 3 connections will disable all of the secondary coils. It does not make sense to me. (sorry I'm so dense here).

If you take the primary connections apart you have open circuit in portions you disconnected and no current will flow.

Now don't forget you can disconnect one coil of a delta primary and it will still operate as an open delta system, it will not be able to carry same maximum load without overloading though.

#### hardworkingstiff

##### Senior Member
View attachment delta current question.PDF

If you will indulge one more question. Let's say we have a delta secondary, and we only have 1 single-phase 10-amp load connected. How do the currents divide between the 3 coils?

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#### Smart \$

##### Esteemed Member
...

If you will indulge one more question. Let's say we have a delta secondary, and we only have 1 single-phase 10-amp load connected. How do the currents divide between the 3 coils?
Each "coil" has an impedance of "z" and you have two paths for current. One path has one "coil" (1z), the other path two "coils"(2z). So the path with one "coil" will carry 2/3 the total current, while the remaining 1/3 by the other two "coil" path.

#### hardworkingstiff

##### Senior Member
Each "coil" has an impedance of "z" and you have two paths for current. One path has one "coil" (1z), the other path two "coils"(2z). So the path with one "coil" will carry 2/3 the total current, while the remaining 1/3 by the other two "coil" path.

Thanks Smart money. That makes sense.

Is the correct way to figure the current through the coils 10A*.33 and 10A*.67 (in the example I posted)? So we have 3.33 amps through 2 coils and 6.67 amps through the 3rd coil? I didn't know if the sqrt3 would come into play or not. I assumed not because it's a single-phase load, but it's a 3-phase supply so I'm unsure.

I know this is remedial stuff for you, thanks for your help.

#### hardworkingstiff

##### Senior Member
I'm thinking some more, (rut roh).

Assuming now a 3-phase load of 10-amps on the line current. The phase current would be 10/1.732=5.77-amps.

I would assume that the same amount of power being consumed by the load would go through the transformer (with the addition of transformer losses, but for this discussion, I am going to ignore the losses unless y'all say I can't).

So the load is using 10-amps, 3-phase @240-volts. The phase current is 5.77 amps.

The load (assuming unity PF) power is 10*(240*1.732)= 4.156kva

Ignoring transformer losses, wouldn't that same power be transferred through the transformer? If yes, then the secondary side of the transformer would be 4.156kva/(5.77A)= 720/1.732=416V internal to the secondary of the transformer? Did that even make sense? What am looking at the wrong way? I'm not sure I even know how to ask what I think my question is.

#### david luchini

##### Moderator
Staff member
I'm thinking some more, (rut roh).

Assuming now a 3-phase load of 10-amps on the line current. The phase current would be 10/1.732=5.77-amps.

I would assume that the same amount of power being consumed by the load would go through the transformer (with the addition of transformer losses, but for this discussion, I am going to ignore the losses unless y'all say I can't).

So the load is using 10-amps, 3-phase @240-volts. The phase current is 5.77 amps.

The load (assuming unity PF) power is 10*(240*1.732)= 4.156kva

Ignoring transformer losses, wouldn't that same power be transferred through the transformer? If yes, then the secondary side of the transformer would be 4.156kva/(5.77A)= 720/1.732=416V internal to the secondary of the transformer? Did that even make sense? What am looking at the wrong way? I'm not sure I even know how to ask what I think my question is.

The load current seen by the 3 phase transformer is 10A, not 5.77A. 4.156kVA/10A= 415.6/1.732 = 240V.

If you wanted to address each coil of the transformer separately, then you would have 4.156kVA/3 = 1.385kVA per coil. 1.385kVA/5.77A= 240V.

#### hardworkingstiff

##### Senior Member
The load current seen by the 3 phase transformer is 10A, not 5.77A. 4.156kVA/10A= 415.6/1.732 = 240V.

If you wanted to address each coil of the transformer separately, then you would have 4.156kVA/3 = 1.385kVA per coil. 1.385kVA/5.77A= 240V.

Thanks. Can you help me understand better what this means?

ILine = IPhase x square Root of 3
I thought the "phase" current was internal to the transformer (which would be the coils, right?), am I wrong in that assumption?

#### Smart \$

##### Esteemed Member
...

Is the correct way to figure the current through the coils 10A*.33 and 10A*.67 (in the example I posted)? So we have 3.33 amps through 2 coils and 6.67 amps through the 3rd coil? I didn't know if the sqrt3 would come into play or not. I assumed not because it's a single-phase load, but it's a 3-phase supply so I'm unsure.

...
Correct.

#### Smart \$

##### Esteemed Member
Thanks. Can you help me understand better what this means?
ILine = IPhase x square Root of 3
I thought the "phase" current was internal to the transformer (which would be the coils, right?), am I wrong in that assumption?
No. Thats what it means... or it can apply to a balanced delta load.

#### david luchini

##### Moderator
Staff member
I thought the "phase" current was internal to the transformer (which would be the coils, right?), am I wrong in that assumption?

You got that right, but you were incorrectly applying the internal current in each coil as 3 phase which gave you the wrong voltage.

It might be easier to think of the current at the load rather than in the transformer. If you had a 4.156kVA, 3phase delta connected load, you would see 10A line currents, and 5.77A in each leg of the load.

Now disconnect the 3 phase load, and connect three individual 1.385kVA, 240V, single phase loads. Each load current will be 5.77A. Each conductor supplying each individual load will also see 5.77A (If load 1 is between A and B, the A and B conductors supplying load 1 will see 5.77A, and if load 2 is between B and C, the B and C conductors supplying load 2 will see 5.77A, etc.)

But if you clamped an ammeter around both of the B conductors, you will read 10A.

#### kwired

##### Electron manager
I'm thinking some more, (rut roh).

Assuming now a 3-phase load of 10-amps on the line current. The phase current would be 10/1.732=5.77-amps.

I would assume that the same amount of power being consumed by the load would go through the transformer (with the addition of transformer losses, but for this discussion, I am going to ignore the losses unless y'all say I can't).

So the load is using 10-amps, 3-phase @240-volts. The phase current is 5.77 amps.

The load (assuming unity PF) power is 10*(240*1.732)= 4.156kva

Ignoring transformer losses, wouldn't that same power be transferred through the transformer? If yes, then the secondary side of the transformer would be 4.156kva/(5.77A)= 720/1.732=416V internal to the secondary of the transformer? Did that even make sense? What am looking at the wrong way? I'm not sure I even know how to ask what I think my question is.

Disregarding any losses for a moment, power in equals power out. You can transform or convert voltage, number of phases, AC to DC, DC to AC, etc. Changing any of those characteristics should directly change current. Total power remains the same though.

Does that help?

#### hardworkingstiff

##### Senior Member
You got that right, but you were incorrectly applying the internal current in each coil as 3 phase which gave you the wrong voltage.

It might be easier to think of the current at the load rather than in the transformer. If you had a 4.156kVA, 3phase delta connected load, you would see 10A line currents, and 5.77A in each leg of the load.

Now disconnect the 3 phase load, and connect three individual 1.385kVA, 240V, single phase loads. Each load current will be 5.77A. Each conductor supplying each individual load will also see 5.77A (If load 1 is between A and B, the A and B conductors supplying load 1 will see 5.77A, and if load 2 is between B and C, the B and C conductors supplying load 2 will see 5.77A, etc.)

But if you clamped an ammeter around both of the B conductors, you will read 10A.
Thanks David, this helped. (I think, :lol: )

Actually, thanks to everyone that helped me understand this better, you guys are great!

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