Single phase understanding

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jwhit

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I calculate 3 phase power and 120v single phase circuits for various applications all the time. I have trouble understanding how single phase 208 works (2 hots and a ground). I typically figure each leg is pulling the same amount of current which is calculated using the 208volts as E. The confusion comes when I try to understand the phase relationship and why it's considered single phase.

Thanks,
 
Re: Single phase understanding

It is not single phase it is two phases of a three phase system. The phases are still 120 degrees out of phase, not 180 degrees like in a single phase system.

Chris
 
Re: Single phase understanding

Does that mean one would use the square root of 2 in the calculations?
 
Re: Single phase understanding

Originally posted by raider1: It is not single phase it is two phases of a three phase system.
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Perhaps it would be more accurate to say that it is single phase, having been created from two phases of a three phase system. It can be called single phase because the current that goes out on one wire, and that then goes through the load, has no alternative but to return via the other wire. In addition, despite their having started from two legs that are 120 degrees apart from each other, the current going out is in phase with the current returning to the source. (SIDE NOTE - Depending on how you label the two currents and what you define as "positive current," you might have to describe the two as being 180 degrees out of phase with each other).

By contrast, when a three phase system supplies a three phase load, the current leaving on (let us say) Phase A will pass through the load, and will find three paths it can take to return to the source. It can return on Phase B, or on Phase C, or on the neutral. In addition, the current going out on Phase A will be 120 degree out of phase with the current on Phase B, and 240 degrees out of phase with the current on Phase C.
 
Re: Single phase understanding

Originally posted by jwhit: Does that mean one would use the square root of 2 in the calculations?
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No. Not a bad question, but no. You had it right. In a single phase 208 volt load, power equals 208 times the current.
 
Re: Single phase understanding

I am still not clear if a leg of a 3 phase circuit has 3 choices doesn't it stand to reason that the one path that is choosen would become the return (Just as in a single phase)
 
Re: Single phase understanding

One leg of a three phase load will use both legs as its return path.

I try to think of single phase as using one coil of a transformer and three phase as using three coils.

However, this would not be true for a single phase 208 V load on a grounded center tapped delta system.

N/M Charlie's explanation was better.
 
Re: Single phase understanding

Originally posted by jwhit: . . . doesn't it stand to reason that the one path that is choosen would become the return . . .
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My point is that there will not be "one path that is chosen." Current leaving on one leg will return on both the other legs and on the neutral (if the circuit has a neutral and if the loads are not perfectly balanced). How much current will take which path depends on the degree to which the loads are balanced.
 
Another single phase question

Another single phase question

I experimented and now I have another question concerning single phase 208. How come when I measure the current with a clamp on amp meter I read the same current thru both legs before and after the load.
 
Re: Another single phase question

Re: Another single phase question

jwhit said:
I experimented and now I have another question concerning single phase 208. How come when I measure the current with a clamp on amp meter I read the same current thru both legs before and after the load.
Click to expand...


Think of it as a 2 wire circuit. The current is the same at every point on the entire circuit.
 
Yes, of course. Thanks for the eye opening. So when calculating the current each leg is pulling how do I know what leg of a single phase circuit is the current coming from. For example L1 and L2 are connected accross a single phase load. I assume L1 leads L2 so for current calculations the current is leaving L1 and returning on L2. Is this correct?
 
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