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Current flow in a home power installation

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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

Current flow in a home power installation

In a typical home power delivery scenario, there is distribution transformer with two oppositely phased hot conductors and a center tap neutral. The home's 110v loads are connected across either hot conductor and the neutral. When (simultaneously) one of the hot conductors is in the positive state and the other is in the negative state, how can the current flow both out of the neutral connector into the more negatively charged hot conductor and into the neutral from the more positively charged hot conductor, at the same instant? What is actually going on?

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Posted: ‎2021-06-29 06:21 PM . Last Modified: ‎2024-03-21 03:47 AM

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Posted: ‎2021-06-29 06:21 PM . Last Modified: ‎2024-03-21 03:47 AM

In a typical home power delivery scenario, there is distribution transformer with two oppositely phased hot conductors and a center tap neutral. The home's 110v loads are connected across either hot conductor and the neutral. When (simultaneously) one of the hot conductors is in the positive state and the other is in the negative state, how can the current flow both out of the neutral connector into the more negatively charged hot conductor and into the neutral from the more positively charged hot conductor, at the same instant? What is actually going on?

See Answer In Context

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BillP
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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

If you've got 2 hots180 degrees out of phase and a neutral, the current will only flow one way on the neutral. If for instance you have 10 amps on 1 circuit and 5 on the other, the neutral will only have 5 amps on it. Let's say that current flows from positive to negetive (some people say it is negative to positive or hole flow). Let's also say that you have circuits A and B and for this 120th of a second, the hot leg feeding B is positive, they share a neutral, and they're 180 degrees out of phase. The load on B is drawing 10 amps. The load on A is drawing 5 amps. With respect to B's hot leg, neutral is negative. 10 amps of curent are flowing on B's hot and 5 are returning down the neutral. The other 5 amps are flowing from that neutral point to A's hot leg as it's load is using 5 amps.

If both circuits were drawing 10 amps, there would actually be no current on the neutral.

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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

I am confused about the last sentence. You said if both circuits were drawing 10 amps, there was no current on the neutral. Let's say you took the neutral away and the two individual ten amp loads were then in series, they wouldn't draw 10 amps. So how are they each drawing 10 amps if the neutral is there but not really doing anything?

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BillP
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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

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Posted: ‎2021-06-29 06:20 PM . Last Modified: ‎2024-03-21 03:47 AM

Let's say you have 2 loads. Each one is drawing 1200 watts exactly (12 ohms). One of these loads on a 120 volt circuit will draw exactly 10 amps. 2 of these loads in series (2400 watts) on a 240 volt circuit will still draw 10 amps.

(2) 12 ohm loads in series = 24 ohms.

P = E 2 / R
or
wattage = voltage squared divided by resistance.
P = 240 squared / 24.
P = 57,600 / 24.
P = 2400.

I = P / E
or
Current = Wattage / Voltage
! = 2400 / 240
I = 10

You still have 10 amps.
I found a nice little ohms law calculator if you'd like:
http://www.the12volt.com/ohm/page2.asp#resistance

I found another link today that may help with both this question and your grounding question.
http://www.tkk.fi/Misc/Electronics/wire_mains.html
This has quite a bit of information for you. One of the sections within this document also references the shared neutral aspect of this question:
http://dotznize.com/electric/?a=sh
Similar to what I mentioned in my last response, this document says:
The answer is that the neutral does not carry the sum of the two currents; it carries the difference. If both loads draw the same amount of current, the neutral will carry no current

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BillP
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Posted: ‎2021-06-29 06:21 PM . Last Modified: ‎2024-03-21 03:47 AM

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Posted: ‎2021-06-29 06:21 PM . Last Modified: ‎2024-03-21 03:47 AM

With or without a Neutral, the voltage drop in the middle of the 2 equal loads will be 1/2 of the total voltage across the 2. When you have multiple light bulbs plugged into the same circuit, they are actually connected in parallel and each will have the same voltage across it. if you took 2 120 volt light bulbs, connected them in series, and put 240 volts across the 2 of them, you would get a 120 volt measurement from the wire in the middle to either of the 2 hot legs.

Using calculations from before, we know that there is 10 amps total. We also know that there was 12 ohms per load. (2) 12 amp loads in series adds their resistance for a total of 24 ohms. Across 240 volts, that was 2400 watts and 10 amps. Using that same ohm's law that states 10 amps across a 24 ohm load is 240 volts, by that same math, 10 amps across a 12 amp load is 120 volts.

http://www.techlearner.com/DCPages/DCReseries.htm

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Posted: ‎2021-06-29 06:21 PM . Last Modified: ‎2024-03-21 03:47 AM

In a typical home power delivery scenario, there is distribution transformer with two oppositely phased hot conductors and a center tap neutral. The home's 110v loads are connected across either hot conductor and the neutral. When (simultaneously) one of the hot conductors is in the positive state and the other is in the negative state, how can the current flow both out of the neutral connector into the more negatively charged hot conductor and into the neutral from the more positively charged hot conductor, at the same instant? What is actually going on?

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