# 3-wire circuit Theorm

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**3-wire circuit Theorm**

3-wire circuits are often of interest to participants in this Section of the Forum. Electricians know empirically that the current in the Neutral conductor which is "common" to the 2 connected loads is the difference of the current-values of the 2 loads.

If the current thru Load "A" is 10 amps and the current thru load "B" is 5 amps, the Neutral-current is 10-5 = 5 amps. Also, the Neutral current could de determined by ammeter readings. But how could the Neutral current be derived by a precise calculation? This question and the concept of an "Open-Neutral", a condition that un-balances the normally equal voltage-division across the 2 loads, leads to a ingenious circuit-theorm that is useful when Ohm's Law cannot be applied.It also introduces some unique circuit concepts.

We have a 3-wire circuit consisting of two 100-volt batteries suppying currents to a 30-Ohm load and a 70 Ohm load, the Neutral "common" to both loads.The current thru the 30-Ohm load is 100/30 = 3.33 amps, and the current thru the 70-Ohm load is 100/70 = 1.43 amps. The Neutral-current is 3.33 - 1.43 = 1.9 amps.

We have a removable link in the Neutral conductor, fixed to 2 terminals, and we remove the link which opens the Neutral and un-balances the former equal voltage divsion across the 2 loads. The circuit is now comprised of two 100-volts batteries in series with the 2 loads which are in series.

The circuit current is now 100 + 100 / 30 + 70 = 200/100 = 2 amps. The voltage-drop across the 30-Ohm load is 2 X 30 = 60 volts and the voltage-drop across the 70-Ohm load is 2 X 70 = 140 volts.

We now apply the theorm by calculating an "equivilent voltage", Veq. Veq. is the voltage across the 2 terminals in the "open-Neutral" connection, which is 140 - 100 = 100 - 60 = 40 volts = Veq.

Next we calculate an "equivilant resistance", Req., which is the resistance across the 2 terminals with Zero Ohms across both batteries. The equivilent circuit is 70 Ohms in parallel with 30 Ohms so Req. = 70 X 30 / 70 + 30 = 2100 / 100 = 21 Ohms = Req.

We next connect Veq. and Req. is series across the 2 terminals, and re-connect the link to the 2 terminals, and calculate the current which is the Neutral current of the 3-wire connection.

With a value of Zero Ohms for the link we have I = Veq. / Req. + 0 = 40/ 21 + 0 = 1.9 amps. If the 3-wire circuit was complicated by an un-wanted resistance of say 6 Ohms in the Neutral conductor, we would presume a 6-Ohm resistor, instead of a "link", was connected to the 2 terminals and apply the theorm as described.

If the current thru Load "A" is 10 amps and the current thru load "B" is 5 amps, the Neutral-current is 10-5 = 5 amps. Also, the Neutral current could de determined by ammeter readings. But how could the Neutral current be derived by a precise calculation? This question and the concept of an "Open-Neutral", a condition that un-balances the normally equal voltage-division across the 2 loads, leads to a ingenious circuit-theorm that is useful when Ohm's Law cannot be applied.It also introduces some unique circuit concepts.

We have a 3-wire circuit consisting of two 100-volt batteries suppying currents to a 30-Ohm load and a 70 Ohm load, the Neutral "common" to both loads.The current thru the 30-Ohm load is 100/30 = 3.33 amps, and the current thru the 70-Ohm load is 100/70 = 1.43 amps. The Neutral-current is 3.33 - 1.43 = 1.9 amps.

We have a removable link in the Neutral conductor, fixed to 2 terminals, and we remove the link which opens the Neutral and un-balances the former equal voltage divsion across the 2 loads. The circuit is now comprised of two 100-volts batteries in series with the 2 loads which are in series.

The circuit current is now 100 + 100 / 30 + 70 = 200/100 = 2 amps. The voltage-drop across the 30-Ohm load is 2 X 30 = 60 volts and the voltage-drop across the 70-Ohm load is 2 X 70 = 140 volts.

We now apply the theorm by calculating an "equivilent voltage", Veq. Veq. is the voltage across the 2 terminals in the "open-Neutral" connection, which is 140 - 100 = 100 - 60 = 40 volts = Veq.

Next we calculate an "equivilant resistance", Req., which is the resistance across the 2 terminals with Zero Ohms across both batteries. The equivilent circuit is 70 Ohms in parallel with 30 Ohms so Req. = 70 X 30 / 70 + 30 = 2100 / 100 = 21 Ohms = Req.

We next connect Veq. and Req. is series across the 2 terminals, and re-connect the link to the 2 terminals, and calculate the current which is the Neutral current of the 3-wire connection.

With a value of Zero Ohms for the link we have I = Veq. / Req. + 0 = 40/ 21 + 0 = 1.9 amps. If the 3-wire circuit was complicated by an un-wanted resistance of say 6 Ohms in the Neutral conductor, we would presume a 6-Ohm resistor, instead of a "link", was connected to the 2 terminals and apply the theorm as described.

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**3**
Nice calculations. But I have a couple of notes.

First of all, the reference to "batteries" may be a bit misleading. Your calculations apply to two 100-volt AC power supplies which are of the same frequency and 180 degrees out of phase. I'm not sure I've ever seen an AC battery.

Second, I don't see the relevance of your calculation that yielded 1.9 amps. Under what circumstances would we see 1.9 amps, and where would we see it?

First of all, the reference to "batteries" may be a bit misleading. Your calculations apply to two 100-volt AC power supplies which are of the same frequency and 180 degrees out of phase. I'm not sure I've ever seen an AC battery.

Second, I don't see the relevance of your calculation that yielded 1.9 amps. Under what circumstances would we see 1.9 amps, and where would we see it?

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**4**
Remember that multiwire branch circuits were originally called edison circuits and they were devised as a way of decreasing the voltage drop in DC distribution.

240/120 volt, center tapped, battery banks can still be found in older buildings as a supply for incandescent emergency lighting. This permits the emergency lighting panel to be a single phase dual voltage panel that can supply three wire circuits for lighting and still take it's emergency power from the battery bank directly without the need for an expensive inverter.

--

Tom

240/120 volt, center tapped, battery banks can still be found in older buildings as a supply for incandescent emergency lighting. This permits the emergency lighting panel to be a single phase dual voltage panel that can supply three wire circuits for lighting and still take it's emergency power from the battery bank directly without the need for an expensive inverter.

--

Tom

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**5***Originally posted by hotarc*

Good info. Here's a general rule of thumb:

Multiwires and DIYer's usually don't mix real well!

**Any DIY that does not take the time to understand multiwire branch circuits should forswear doing there own electrical work.**

I'm not saying that electrical work is beyond a concientious DIY but I am saying that taking the time to learn what you are doing is esential to the safe execution of the work.

--

Tom Horne

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**7**
The above is a rehash of the work of Thevenin and/or Norton that was done a long time ago. Today such knowledge is taught in electrial engineering classes all over the country. Those engineers who stayed awake in class know it, some probably forgot it, but, I believe, most learned it at least once. Some may even have the opportunity to learn it again here a second time. Remember, leave it to an engineer to make the simple, complex.

*Last edited by jughead; 01-05-04 at 12:16 AM.*

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**8***Originally posted by hornetd*

I'm not saying that electrical work is beyond a concientious DIY but I am saying that taking the time to learn what you are doing is esential to the safe execution of the work.

--

Tom Horne

**Any DIY that does not take the time to understand multiwire branch circuits should forswear doing there own electrical work.**I'm not saying that electrical work is beyond a concientious DIY but I am saying that taking the time to learn what you are doing is esential to the safe execution of the work.

--

Tom Horne

I agree with you 100%, but from what I've seen, I have to stand by my statement. Many electrical DIYers do not take the time to read up on the codes and procedures involved with multiwires. They just think, "two circuits for the price of one!" This is definitely not right, but unfortunately that's the way it is sometimes.

A lot of the "handyman" types I've seen can do a decent job running wires and mounting boxes, etc., but they don't have a clue as far as theory is concerned.

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*Originally posted by jughead*

**The above is a rehash of the work of Thevenin and/or Norton that was done a long time ago. Today such knowledge is taught in electrial engineering classes all over the country. Those engineers who stayed awake in class know it, some probably forgot it, but, I believe, most learned it at least once. Some may even have the opportunity to learn it again here a second time. Remember, leave it to an engineer to make the simple, complex.**

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"Leave it to an engineer to make the simple complex"--------

The purpose of the theorm is to reduce a complex circuit to a simple one. By "removing" the Neutral connection, a 3-wire circuit is reduced to the most simple of circuits- a series-circuit with only one current-path.

We can make the following observations on 3-wire circuit;

Load "A" ( 10 amps) in connected in parallel with "Power-Source A" (120 volts), Load "B" ( 5 amps) is connected in parallel with "Power-Source B" ( 120 volts), and Loads "A" and "B" are connected together in series across Power-Sources "A" and "B" in series ( 240 volts).

The current thru Load "A", 10 amps, has a 120-volt parallel-component", 5 amps. This is the Neutral current. The Load "A" current also has a 240-volt "series-component", 5 amps, which is the current thru "A" and "B" in series. Although Load "B" is connected directly across "Power-Source B" via the Neutral, the Neutral current thru Load "B" is Zero. Although connected together at the Neutral, both circuits are essentaily "independent" circuits. Dis-connecting Load "A" has no effect on the operation of Load "B".

I thought it would be of interest to know how the current-divisions of such a combination of circuits could be calculated with a theorem.

J.N. mentions an "out-of-phase" connection, but I believe this is mis-leading. Two 120-volt single-phase voltages, with a 180 degree phase-difference, can be connected in parallel for a 2-wire, 120 volt power-source,or in series for a 120/240 volt 3-wire power-source. One the connection is made, the "out-of-phase" reference is essentialy irrelevant in determing circuit values.

J.N.--- what if I were to describe the 2 power-sources as not "Batteries" but as "Zero-Frequency A.C. Power-Sources"????????? (ha ha )

The purpose of the theorm is to reduce a complex circuit to a simple one. By "removing" the Neutral connection, a 3-wire circuit is reduced to the most simple of circuits- a series-circuit with only one current-path.

We can make the following observations on 3-wire circuit;

Load "A" ( 10 amps) in connected in parallel with "Power-Source A" (120 volts), Load "B" ( 5 amps) is connected in parallel with "Power-Source B" ( 120 volts), and Loads "A" and "B" are connected together in series across Power-Sources "A" and "B" in series ( 240 volts).

The current thru Load "A", 10 amps, has a 120-volt parallel-component", 5 amps. This is the Neutral current. The Load "A" current also has a 240-volt "series-component", 5 amps, which is the current thru "A" and "B" in series. Although Load "B" is connected directly across "Power-Source B" via the Neutral, the Neutral current thru Load "B" is Zero. Although connected together at the Neutral, both circuits are essentaily "independent" circuits. Dis-connecting Load "A" has no effect on the operation of Load "B".

I thought it would be of interest to know how the current-divisions of such a combination of circuits could be calculated with a theorem.

J.N. mentions an "out-of-phase" connection, but I believe this is mis-leading. Two 120-volt single-phase voltages, with a 180 degree phase-difference, can be connected in parallel for a 2-wire, 120 volt power-source,or in series for a 120/240 volt 3-wire power-source. One the connection is made, the "out-of-phase" reference is essentialy irrelevant in determing circuit values.

J.N.--- what if I were to describe the 2 power-sources as not "Batteries" but as "Zero-Frequency A.C. Power-Sources"????????? (ha ha )

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**11**
Try to find an example where Ohm's law would be harder to use than a theorem and you'll get an engineer excited. In your given example the only 'simplification' necessary is Ohm's law. There is no need make the simple, complex when simple simply will do.

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**12****DIYer here using the multi-wire configuration.**

I've read the theory, I've read the code, now I'm putting it into practice.

I ran two cables of romex 12-3 into my attic. Let's call them A and B. On the A cable, I'm pushing 7 ceiling lights and one fan on the red hot (about 950 watts) and on the black hot I'm also pushing 7 ceiling lights and one fan (950 watts).

On the B cable, I'm pushing three lights and a doorbell transformer on the black hot (about 310 watts) and 10 lights and one fan on the red hot (1250 watts).

BTW: Once I got the 12-3 romex into my attic, I spliced each hot wire into a 12-2 romex circuit for ease of use.

Do any of you geniuses see any problems with my DIY configuration?

Thanks.

I ran two cables of romex 12-3 into my attic. Let's call them A and B. On the A cable, I'm pushing 7 ceiling lights and one fan on the red hot (about 950 watts) and on the black hot I'm also pushing 7 ceiling lights and one fan (950 watts).

On the B cable, I'm pushing three lights and a doorbell transformer on the black hot (about 310 watts) and 10 lights and one fan on the red hot (1250 watts).

BTW: Once I got the 12-3 romex into my attic, I spliced each hot wire into a 12-2 romex circuit for ease of use.

Do any of you geniuses see any problems with my DIY configuration?

Thanks.

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**13**
It would have been better to have run a neutral and a ground along with the two hots (a total of four wires) on cables A and B. That would cut down on the shock hazzard.

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**14****I am running the neutral and grouds everywhere.**

I'm a little confused here. When I say that I'm running lights off of the red or the black hots, I am also running the neutral and grounds to each of these fixtures. Everything is to code. I'm not quite sure how it would work if I didn't.

BTW: The 12-3 romex has 4 wires in it. They don't count the ground wire for some inexplicable reason.

Thanks.

George

BTW: The 12-3 romex has 4 wires in it. They don't count the ground wire for some inexplicable reason.

Thanks.

George

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jughead------ We have a 240/120 volt 3-wire circuit suppyling power to Loads A and B. Load A is a resistive load of 30 Ohms and Load B is a resistive load of 70 Ohms.

At the point where the Neutral of the 3-wire Branch-Circuit cable connects to the 2 Neutrals of the 2-wire cables that connect to Loads A and B, someone incorrectly connects a 20 Ohm resistive load in the Neutral circuit with the result that there is now 20 Ohms of resistance in the Neutral circuit between the point where the Neutral of the 3-wire cable connects at the panel ,and the point where the Neutral of the 3-wire cable connects to the 2 Neutral wires of Loads A and B.

What is the Neutral current with a resistance of 20 Ohms "inserted" in the Neutral conductor of the 3 wire Branch-Circuit cable?

At the point where the Neutral of the 3-wire Branch-Circuit cable connects to the 2 Neutrals of the 2-wire cables that connect to Loads A and B, someone incorrectly connects a 20 Ohm resistive load in the Neutral circuit with the result that there is now 20 Ohms of resistance in the Neutral circuit between the point where the Neutral of the 3-wire cable connects at the panel ,and the point where the Neutral of the 3-wire cable connects to the 2 Neutral wires of Loads A and B.

What is the Neutral current with a resistance of 20 Ohms "inserted" in the Neutral conductor of the 3 wire Branch-Circuit cable?

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**17**
Everything is to code. I'm not quite sure how it would work if I didn't.

George, what you did is fine as long as the two hot wires are connected to breakers on opposite legs of the service (not the same thing as opposite sides of the panel). If this is not true, then it will work but present a significant fire hazard.

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**18**
Providing the simple circuit I scribbled matches the description you gave in your post the neutral current would be as follows:

I = (120/(30+20)) - (120/(70+20)) = 1.067 amps

Ohm's law was just applied twice. Since the currents flowing in the two loops are always in opposite directions I just subtracted the current in one loop from the other.

If I misunderstood the question the answer could be different.

I = (120/(30+20)) - (120/(70+20)) = 1.067 amps

Ohm's law was just applied twice. Since the currents flowing in the two loops are always in opposite directions I just subtracted the current in one loop from the other.

If I misunderstood the question the answer could be different.

*Last edited by jughead; 01-17-04 at 10:35 AM.*

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Step 1; "remove" the 20 Ohm resistance in the Neutral conductor.

This reduces the circuit to 100 Ohms ( R30 + R70) across 240 volts.

The current is 240/100 = 2.4 . The V-D across R30 is 30 X 2.4 = 72 volts. The V-D across R70 is 70 X 2.4 = 168 volts.

"pretending" that the 20 Ohm resistance was connected to the Neutral via 2 terminals, we calculate the Equivilent Voltage, Veq., across the 2 terminals;

168-120 = 120 -72 = 48 volts = Veq.

The Equivilent Resistance, Req., across the terminals is R30 in parallel with R70. Req. = 30 X 70 / 30 + 70 = 2100/100 = 21 Ohms.

We now re-connect to 20 Ohm Neutral resistance to the terminals and calculate the Neutral current thru the 20 Ohm resistance, which is in series with Veq. and Req.

In = 48 / 21 + 20 = 48 / 41 = 1.17 amps.

The V-D across the 20 Ohm resistance = 20 X 1.17 = 23.4 volts.

"Presuming" the 1.17 Neutral current is a "component" of the current thru R30, the V-D across R30 = 120 - 23.4 = 96.54 volts.

The current thru R30 is 96.54 / 30 = 3.22 amps.

The current thru R70 is 3.22 - 1.17 = 2.05 amps, and the V-D across R70 = 70 x 2.05 = 143.5 volts.

The V-D's across R30 and R70 in series equals the line voltage; 96.54 + 143.5 = 240 volts.

This reduces the circuit to 100 Ohms ( R30 + R70) across 240 volts.

The current is 240/100 = 2.4 . The V-D across R30 is 30 X 2.4 = 72 volts. The V-D across R70 is 70 X 2.4 = 168 volts.

"pretending" that the 20 Ohm resistance was connected to the Neutral via 2 terminals, we calculate the Equivilent Voltage, Veq., across the 2 terminals;

168-120 = 120 -72 = 48 volts = Veq.

The Equivilent Resistance, Req., across the terminals is R30 in parallel with R70. Req. = 30 X 70 / 30 + 70 = 2100/100 = 21 Ohms.

We now re-connect to 20 Ohm Neutral resistance to the terminals and calculate the Neutral current thru the 20 Ohm resistance, which is in series with Veq. and Req.

In = 48 / 21 + 20 = 48 / 41 = 1.17 amps.

The V-D across the 20 Ohm resistance = 20 X 1.17 = 23.4 volts.

"Presuming" the 1.17 Neutral current is a "component" of the current thru R30, the V-D across R30 = 120 - 23.4 = 96.54 volts.

The current thru R30 is 96.54 / 30 = 3.22 amps.

The current thru R70 is 3.22 - 1.17 = 2.05 amps, and the V-D across R70 = 70 x 2.05 = 143.5 volts.

The V-D's across R30 and R70 in series equals the line voltage; 96.54 + 143.5 = 240 volts.

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**20**
See the link to my diagram and calculations below. These were done using the results of the work of

and commonly known as Kirchhoff's voltage laws. I also used Kirchhoff's current law as well.

What you are doing is closer to the work of Thevenin, but all are just derived from Kirchhoff's laws.

**Physicist Gustav R. Kirchhoff in 1847**and commonly known as Kirchhoff's voltage laws. I also used Kirchhoff's current law as well.

What you are doing is closer to the work of Thevenin, but all are just derived from Kirchhoff's laws.

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**21**
There is some evidence that the Chinese, before the time of Christ, made use of bamboo rods to develop methods of solving simultaneous equations. A more modern approach took form in 1683 in Japan and about 1693 in Germany. In Japan see the work of Seki Kowa and in Germany look at the work of G.W. Leibniz. Works of these men were used by G. Cramer of Switzerland in 1750 to produce the methods used here:

These days I can just use a HP48 calculator instead of bamboo rods.

These days I can just use a HP48 calculator instead of bamboo rods.