3 phase circuit
- Thread starter kimrichi
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kwired
Electron manager
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- NE Nebraska
Exactly what flows where depends on if you have a wye or delta configured load, but for either you have current flowing in on one phase and returning on the other two phases, but you also have this for all three of the phases. If you have resistive loads connected in delta configuration and each load draws 5 amps as a single phase load the net on each phase is 5 x 1.732 (square root of 3) or 8.66 amps. Each resistive load will still see 5 amps. I can draw this out if you wish and post it, but will have to be later on.
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Through the circuit conductors and the transformer windings. It takes some time to get your head around it, but you don't need a neutral to complete a circuit.
What kwired said, to begin with, but where I highlighted "phase" in red, you (plus kwired et al) should substitute "line" or "line conductor", such as L1, L2, and L3. Phases are actually Line-to-Line, i.e. voltage across the pairs L1-L2, L2-L3, and L3-L1 and are the phasing references. In kwire's example, each 5A load constitutes a "phase" load, and the net currrrent expressed is the line current (8.66A).
Line current is out-of-phase with the phase voltage because it is the sum of the connected phase loads. Taking L1 line current as an [extended] example, it has a 30? shift in phase angle because it sums the phase currents which are 0? and -60? (not absolutely technical but will suffice). The current through the phase loads at each instant in time are not equal. This is why you only get 8.66A of line current, rather than double (or zero). The same thing happens at the nodes of L2 and L3 with the phasing angle reference shifted to 120? and 240? respectively. Note the preceding phase load having the -60? phase current is the same phase load connected to L2 having a load current phase angle of 120?... the 180? complement of -60?, and this is just because we have used the opposite polarity to previous node (i.e. we flopped our positive and negative probes), while the phasing of the L2 to L3 connected load would be +60? at the L2 end and 240? at the L3 end. At the L3 node, the L3 to L1 connected load would be 180? at the L3 end and 0? at the L1 end. As you can see, we are back where we started.
Now if you were to plot the phase load and line currents at each node with respect to time, and draw a vertical line through any instance in time, the instantaneous vaues would sum to zero... i.e. whatever current enters the node also leaves the node. This is true of all three nodes, occurring simultaneously.
If you were to plot all three line currents and draw vertical lines as above, those instantaneous values would also sum to zero.
I'm short on time to elaborate more, so I'll just ask if you have questions, or invite you to say I'm giving a reply which is too complex
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kwired
Electron manager
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- NE Nebraska
Any single phase circuit has that same characteristic. Three phase is more complex - as Smart$ tells us:
What kwired said, to begin with, but where I highlighted "phase" in red, you (plus kwired et al) should substitute "line" or "line conductor", such as L1, L2, and L3. Phases are actually Line-to-Line, i.e. voltage across the pairs L1-L2, L2-L3, and L3-L1 and are the phasing references. In kwire's example, each 5A load constitutes a "phase" load, and the net currrrent expressed is the line current (8.66A).
Line current is out-of-phase with the phase voltage because it is the sum of the connected phase loads. Taking L1 line current as an [extended] example, it has a 30? shift in phase angle because it sums the phase currents which are 0? and -60? (not absolutely technical but will suffice). The current through the phase loads at each instant in time are not equal. This is why you only get 8.66A of line current, rather than double (or zero). The same thing happens at the nodes of L2 and L3 with the phasing angle reference shifted to 120? and 240? respectively. Note the preceding phase load having the -60? phase current is the same phase load connected to L2 having a load current phase angle of 120?... the 180? complement of -60?, and this is just because we have used the opposite polarity to previous node (i.e. we flopped our positive and negative probes), while the phasing of the L2 to L3 connected load would be +60? at the L2 end and 240? at the L3 end. At the L3 node, the L3 to L1 connected load would be 180? at the L3 end and 0? at the L1 end. As you can see, we are back where we started.
Now if you were to plot the phase load and line currents at each node with respect to time, and draw a vertical line through any instance in time, the instantaneous vaues would sum to zero... i.e. whatever current enters the node also leaves the node. This is true of all three nodes, occurring simultaneously.
If you were to plot all three line currents and draw vertical lines as above, those instantaneous values would also sum to zero.
I'm short on time to elaborate more, so I'll just ask if you have questions, or invite you to say I'm giving a reply which is too complex
Very well may be too complex for as simple as the question was, as well as inexperienced inquiring minds. Exact phase angles and vector math may scare someone new from ever wanting anything to do with three phase. That knowledge helps you understand more complex things but is not needed for many basics. You really just need to understand that The current that is supplied by Line A is going to return on B, C or N for a single phase load. For a three phase load the current supplied by line A is going to return partially on B and partially on C. Line A is also going to have some current from B as well as C returning to A.
I know it may be too complex... that is why I elicited a comment to that effect from kimrichi. Additionally, a simple question does not always mean only basic information is desired.
Also, many are confused by using terms such as "supply" and "return", which are fine when learning DC theory. But with AC power, each line acts as both supply AND return.. while all lines together are said to supply the load.
kwired
Electron manager
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- NE Nebraska
There is a supply and a return for 1/2 cycle, then you flip flop them for the other 1/2 cycle
Supply and return is still somewhat usable terms with single phase circuits, as your RMS values are interchangeable with DC values in most voltage and current calculations.I still think it was a fairly good explanation without getting too technical, and for someone that does not understand higher level mathematics (myself somewhat included) it gives a person enough of an idea of what is happening. Not knowing the knowledge of this person it is fair to give him a simple as well as a complex answer - he can use one or both answers if he wants. Now that anyone that reads it knows the simple answer is not entirely fact they may still get something from it if they don't fully understand the more complex answer.
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I think of it short and simple, right or wrong: half of the current leaving A is seeking B, and the other half leaving A is seeking C. A neutral is just another path to get to those destinations. All phases are similar, half the current from each seeking the other two.
kwired
Electron manager
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- NE Nebraska
Me too. And then you have to rememer that at same time the same thing is happening on both B and C. You don't necessarily have to know or remember all the technical details for many basic things, you just need to know that the net is 1.73 times what it would be for single phase in most of your calculations.
I understand there is a little more to it, but is it not still similer? The current returns through the other phase conductors, that is why you don't need a neutral conductor.
Thanks
Mike
Giving it just a tad bit of thought makes it quite obvious.
3? source. 3? load. Three wires. Each with current.
Where else could the current go under normal operating conditions? ... :slaphead:
(kwired: How's that for basic?
)
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kwired
Electron manager
- Location
- NE Nebraska
Giving it just a tad bit of thought makes it quite obvious.
3? source. 3? load. Three wires. Each with current.
Where else could the current go under normal operating conditions? ... :slaphead:
(kwired: How's that for basic?
:thumbsup::thumbsup:
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