I thought I understood how phase applies to adding sine waves but apparently not...

I was reading about 220V house wiring, and how each of the 110V hot wires are 180 degress out of phase with each other - and that makes sense, alright. But then how do we achieve 220 volts if the waves are out of phase with each other? Seems to me they should cancel.

I'm being told they add, when they're out of phase and that if they were 'in phase' then all that would do is double the duty cycle. This seems completely backward from what I've been taught.

Any help?

 

If you combine them they would make zero, but since you're pulling power from both waves, you have -110V on one line, and +110V on the other if I'm not mistaken.

 

You likely only have a single phase 240v transformer that is center tapped. This gives you two opposite polarity legs @120v, not two phases. Google split phase. To get 240, the outer connections are used.

http://www.allaboutcircuits.com/vol_2/chpt_10/1.html

 

Awesome read. I needed to take the time to read that whole section. That totally makes sense now.

You guys are outstanding in this forum. Thanks a bunch.

 

You likely only have a single phase 240v transformer that is center tapped. This gives you two opposite polarity legs @120v, not two phases. Google split phase. To get 240, the outer connections are used.

http://www.allaboutcircuits.com/vol_2/chpt_10/1.html

I have a follow up question on the neutral wire / grounded center tap...

The astute observer will note that the neutral wire only has to carry the difference of current between the two loads back to the source. In the above case, with perfectly “balanced” loads consuming equal amounts of power, the neutral wire carries zero current.

Sadly, I am not an astute observer. I can "see" the split condition here, partioning the loads into two 120 loops - but I don't get this "difference of current" between the two loads.

I read, and re-read, and could not find where this is expanded upon or explained anywhere. How does the neutral wire *not* carry some current here? In case I haven't advertised my ignorance enough...it would seem the positive cycle would ground out, for each source, rather than power the load.

 

My understanding and explanation would probably be fuzzy at best, and incorrect at worst. Perhaps someone like TNK or RGPATTER would chime in and help us both.... I have read their posts and have recently grasped the calcs for 3 phase neutral current. I'm sure they will be more help than I.

 

Referring to the drawing in post #5

At some particular moment in time, the top line is at +9 volts compared to the middle line. At that same moment, the bottom line is at -9 volts compared to the middle line. (The sources are synchronous.)

Substitute a pair of imaginary, perfect, 9 volt batteries into the drawing where the AC sources are and think in those terms. This will demonstrate that, if you remove the middle wire, there is no difference in current flow.

deriving 1.44 ohms per load, the current at 9 volts will be 6.25 amps per leg.

very soon thereafter, a pair of 12v batteries can be substituted to represent a tiny moment in time.

The entire sine wave can be reconstructed, a microsecond at a time, by substituting the correct battery size at each microsecond.

Is your head tracking this?

 

I have a follow up question on the neutral wire / grounded center tap...

Sadly, I am not an astute observer. I can "see" the split condition here, partioning the loads into two 120 loops - but I don't get this "difference of current" between the two loads.

I read, and re-read, and could not find where this is expanded upon or explained anywhere. How does the neutral wire *not* carry some current here? In case I haven't advertised my ignorance enough...it would seem the positive cycle would ground out, for each source, rather than power the load.

There is a (very) basic principle in electricity that the sum of the currents flowing into any junction in a circuit must be zero. This is called Kirchhoff's Current law. http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.KCL.html

Given that this is the case, because the currents flowing in the two outer branches of your example circuit sum to zero, there is no resultant current left over to flow in the neutral return. If the currents in the outer branches were not equal, the difference would flow in the neutral.

This is something which was established a long time ago, and is absolutely key to understanding circuit operation. Perhaps it is something to be taken on trust?

 

Referring to the drawing in post #5

At some particular moment in time, the top line is at +9 volts compared to the middle line. At that same moment, the bottom line is at -9 volts compared to the middle line. (The sources are synchronous.)

Substitute a pair of imaginary, perfect, 9 volt batteries into the drawing where the AC sources are and think in those terms. This will demonstrate that, if you remove the middle wire, there is no difference in current flow.

deriving 1.44 ohms per load, the current at 9 volts will be 6.25 amps per leg.

very soon thereafter, a pair of 12v batteries can be substituted to represent a tiny moment in time.

The entire sine wave can be reconstructed, a microsecond at a time, by substituting the correct battery size at each microsecond.

Is your head tracking this?

Ok, yes that does make more sense. I envisioned clockwise loop currents, which presents equally opposing current along that neutral wire if both loads are the same. Not sure why that didn't dawn on me before. Do I have that right?

That would make sense then that "difference" current would show up on the line. Sticking with this moment in time...If one load had double the resistance, then I'm going to see current along that line because the opposing currents are no longer equal due to one losing half of its current - using KCL I'm thinking I'll see 3.125 Amps on that neutral line, the other half working on the 2.88 Ohm load. (Not sure I'm doing KCL correctly...)

Now, in real life, I can't say that loads are perfectly balanced like this in my home, right? I mean, in my panel I have two busses with the same number of breakers, but the loads they serve are as diverse as can be.

There is a (very) basic principle in electricity that the sum of the currents flowing into any junction in a circuit must be zero. This is called Kirchhoff's Current law. http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.KCL.html

Given that this is the case, because the currents flowing in the two outer branches of your example circuit sum to zero, there is no resultant current left over to flow in the neutral return. If the currents in the outer branches were not equal, the difference would flow in the neutral.

This is something which was established a long time, and is absolutely key to understanding circuit operation. Perhaps it is something to be taken on trust?

Thanks Adjuster. I was working KCL at the same time you posted - nice to see I was at least on the right track. The key was #12 pointing out the polarity and thinking of this as a DC circuit for that moment in time.

I'm really not that strong on KCL though, that's for sure. After doubling one of those load resistances to 2.88 Ohms to determine what's going on across the neutral line, I assumed that current in 1.44 Ohm load would remain at 6.25 Amps. After that assumption, KCL was fine.

 

It's always wonderful to get a "thank you" and the validation that my kind of really primitive thinking sometimes unlocks it for other people.

Ps, I am not responding to questions in post #15 at this time because you are getting them right by yourself.

 

Well your primitive thinking clicks with me, apparently. Definitely thank you for the help.

I am curious about the load balancing at my residence and if the neutral in my house is ever carrying any current. But I will continue to dig on my own, as I get time.

 

If you are in the USA, the loads are random, as you said. Your neutral wire is almost never without current.

 

And as I think about this more (this is where I usually get into trouble) that schematic only makes sense from the transformer to my panel.

Once I start looking at outlets in my house, it would seem neutral would return every bit of the current from the hot wire to the panel, upon use. Because the load is wired to one hot wire and one neutral - the other hot wire is not connected to this load in any way that I can tell (except through the neutral being a center tap way back at the transformer).

Furthermore, since the neutral wire in the panel is terminated on the same bar with bare ground, and that is wired to a grounding rod, I'm not sure how my inside home wiring cooperates with the arrangement in that schematic.

 

And as I think about this more (this is where I usually get into trouble) that schematic only makes sense from the transformer to my panel.

Once I start looking at outlets in my house, it would seem neutral would return every bit of the current from the hot wire to the panel, upon use. Because the load is wired to one hot wire and one neutral - the other hot wire is not connected to this load in any way that I can tell (except through the neutral being a center tap way back at the transformer).

Since all the neutral wires are electrically common, the schematic still holds, you're just spreading it out across the house.

Furthermore, since the neutral wire in the panel is terminated on the same bar with bare ground, and that is wired to a grounding rod, I'm not sure how my inside home wiring cooperates with the arrangement in that schematic.

Check on that, I don't think neutral wires are wired to the ground.

 

OK...when you consider only the circuit breakers that are connected to the top line in the drawing, they are connected to outlets, light bulbs, kitchen appliances. When only those things are using current, all the current is returned to the transformer through the neutral line. Let's pretend that it amounts to 20 amps for this moment.

As soon as you turn on a load that is power by the bottom line, its current meets the other current at the neutral line in the breaker box and, being out of phase, they cancel. If you turned on a 12 amp toaster oven on the lower line, the neutral current to the transformer will now be 8 amps.

You can duplicate this by setting up a circuit with (2) 9v batteries and using milliamps instead of amps. Put a 470 ohm resistor in the top line and measure 19.14 ma in the neutral line. Put a 750 ohm resistor from the bottom line to the neutral line and measure 7.14 ma going back to the junction of the batteries.

There's nothing like hooking it up and measuring to become a believer.
You could also "simulate" it, but I am so old that simulating doesn't prove much to me. You can make a mistake in simulating and never know it. If you use real electrons and measure them, you simply have to believe the results because electrons never lie.

 

Check on that, I don't think neutral wires are wired to the ground.

In the US, earth and neutral share the same bus in the distribution panel. The current carriers (white and bare copper) are run separately.

 

ps, the neutral wire in Indiana is connected to the ground wire in the breaker box, then both of them are connected to the Earth ground rod and the neutral of the transformer.

I don't know about the house that zippoinc lives in because he isn't telling where he lives, but I do know about Indiana. I lived there for 19 years.

 

Thank-you as well from me, When I get time I will look it over and see if I get it as well. We did DC analysis in school, but AC theory wasn't really stressed.

 

OK...when you consider only the circuit breakers that are connected to the top line in the drawing, they are connected to outlets, light bulbs, kitchen appliances. When only those things are using current, all the current is returned to the transformer through the neutral line. Let's pretend that it amounts to 20 amps for this moment.

As soon as you turn on a load that is power by the bottom line, its current meets the other current at the neutral line in the breaker box and, being out of phase, they cancel. If you turned on a 12 amp toaster oven on the lower line, the neutral current to the transformer will now be 8 amps.

You can duplicate this by setting up a circuit with (2) 9v batteries and using milliamps instead of amps. Put a 470 ohm resistor in the top line and measure 19.14 ma in the neutral line. Put a 750 ohm resistor from the bottom line to the neutral line and measure 7.14 ma going back to the junction of the batteries.

There's nothing like hooking it up and measuring to become a believer.
You could also "simulate" it, but I am so old that simulating doesn't prove much to me. You can make a mistake in simulating and never know it. If you use real electrons and measure them, you simply have to believe the results because electrons never lie.

I like the way you explained it. And, you're right I should do exactly that. I'll hit radioshack at lunch tomorrow.

My main quandary was about the current on the neutral line tied from the receptacle back to the breaker panel. That bit of length would seem to carry all of the current its paired hot wire delivered to the receptacle through whatever load.

And you actually confirmed that for me when you said "its current meets the other current at the neutral line in the breaker box".

Much thanks again. I like how you think.

Check on that, I don't think neutral wires are wired to the ground.

In the US, earth and neutral share the same bus in the distribution panel. The current carriers (white and bare copper) are run separately.

ps, the neutral wire in Indiana is connected to the ground wire in the breaker box, then both of them are connected to the Earth ground rod and the neutral of the transformer.

I don't know about the house that zippoinc lives in because he isn't telling where he lives, but I do know about Indiana. I lived there for 19 years.

I live in Missouri for now. Here it's like beenthere and #12 said.

 

I think I got it now as well, I was trying to picture "A moment in time" and envisioning the position of the sine wave at that moment. It became clear when I replaced the source w/DC, and used superposition theory. Perhaps that is why we didn't spend a lot of time w/AC, The RMS calcs can be treated like DC. Thanks again!