Page 4 of 9

Wrong. The neutral is "effectively disconnected"***only*** if the loads on the
two legs are exactly the same. The two legs function as two parallel circuits
with respect to 120V loads. Obviously they are indeed in series WRT 240V
loads.

[snip]

Which they are (either in the 200A+200A example or the 1A+1A one).

In a parallel circuit BOTH ends of the loads are connected together (or at least to identical voltages). Neither is true here.

Strangely, I get the idea that you actually know this stuff.

In this 200A service there are THREE current-carrying conductors. Each of these conductors is of the proper size to carry 200A. OK so far?

You say (when this service is fully loaded) that two of these conductors is carrying 200A (for a total of 400A, as you say).

Then where is that 400A going? The only remaining conductor is the neutral, a big enough conductor for 200A (yes, this 400A was at 120V but current is still current and voltage doesn't change the conductor's current capacity).

Somehow I'm imagining a bridge that can handle 200 cars per minute, but that can be 400 if half the cars are blue :-)

And that almost never happens in real life, either....

Wrong -- both are true.

OK

400A @ 120V, or 200A @ 240V, yes.

Cute. Just answer these questions; assume a 240V 200A service.

What is the maximum power that service can provide?

If all the loads supplied by that service are 120V loads (e.g. blender, toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get when you divide that maximum power by 120V?

Exactly so.

200A each. Total of 400A of 120V loads -- as you said.

Agree.

If you also put a 120V load across Line 2 and the neutral, then

Don't agree with this. If the second load on line 2 is equal to the load already on line 1, then the current flow is in on line 1 and back out on line 2. No current flows in the neutral.

If the second load on line 2 were half the size of the load on line 1, then half the current from line 1 would flow back out on line 2 and half the current from line 1 would flow back out the neutral.

The key here is look at that service cable coming from the transformer and you have a circuit running a max of 200 amps. Agree?

No, it not. Once again, the physical current in the service cable is 200 amps. You have 200 amps flowing IN on one hot, 199 flowing OUT on the other hot and 1 amp flowing OUT on the neutral. The current flowing in that circuit is 200 amps. You don't count current twice.

How much current is flowing in a simple 120W light bulb plugged into a 120Volt outlet? 1 amp. Now simply add another single wire in parallel with one of those supplying the light bulb. The current will be split now, with some part of it going via one wire, some via the other. For sake of argument, let;s assume it just splits evenly. So, now you have 1 amp coming in on one wire, 1/2 amp leaving via one wire, 1/2 amp leaving via the second wire. How much current is flowing in this "service" circuit to the light bulb? There is no confusion, it's 1 amp.

It's a simple matter of applying Kirchoffs law.

Balanced or unbalanced matters not a wit. There is still a current of 200 amps coming and going either way. The only difference balance makes is which PATH that 200 amp current takes.

No, never said that.

You're going to have 200A flowing from Line 1 to neutral. Again, that is 200 amps flowing in the service cable. As I said, if it's fully loaded, you will have a max of 200 amps flowing in the service cable. That's 200 amps coming in, 200 amps going out. The only difference is whether some of the current flows on the neutral or not. If the neutral has some current, then one of the hot lines must have less than 200 amps or you'd violate Kirchoff's law.

There will be 0 amps in the neutral. The loads form two legs of a balanced bridge, the windings in the transformer form the other two legs, When they are balancedas you described therer will be ZERO current in the neutral. Dont believe me check it for yourself. Make a test circuit out of two 100 watt light bulbs connected in series, put them across a 240VAC connection with no neutral connection then connect the neutral between them. Measure the current in the neutral. It will be ZERO. measure the current in each hot leg. The neutral is only there to carry the difference between the two legs when there is an imbalance. If the current in one leg was 100 amps and the the other had 50 amps the neutral current would be 100 -50 + 50 amps.. When the load is balanced and you measure the current in each leg you are just measuring the same current twice.

Jimmie

Doesnt matter if there is no current flow in the neutral, Quit trying to reason what you dont understand and build a mock up circuit if this really concerns you that much. I am growing very bored of talking to people who are more interested in defending their argument than actually trying to learn something. Any decent text that explains Kerchov's(sp?) Law as applied to AC circuits will give you the answer.

Jimmie

Yes. You have a 3 wire service cable coming into the house. Use Kirchoff's law and add up the current coming into the house at any point in time under any conditions through that cable any you have a max of 200 amps. Add up all the current leaving the house at the same point in time and you have a max of 200 amps. In my world, that's a 200 amp service. You don't measure 300, or 400 by counting electrons twice.

And also, to support a max of 400 amps of 120volt load in the house which is the hot topic, you have to have the special case where the loads are perfectly balanced so that 200 amps is on each side. And then you in fact have a SINGLE circuit with 200 amps flowing in on one hot leg, throught the loads in series, and back out the other hot leg. Nothing flows in the neutral. Again, in my world, that's 200 amps flowing in the service, not 400.

#### Site Timeline

- posted on October 25, 2009, 6:47 pm

Wrong. The neutral is "effectively disconnected"

- posted on October 26, 2009, 1:50 am

Which they are (either in the 200A+200A example or the 1A+1A one).

In a parallel circuit BOTH ends of the loads are connected together (or at least to identical voltages). Neither is true here.

Strangely, I get the idea that you actually know this stuff.

In this 200A service there are THREE current-carrying conductors. Each of these conductors is of the proper size to carry 200A. OK so far?

You say (when this service is fully loaded) that two of these conductors is carrying 200A (for a total of 400A, as you say).

Then where is that 400A going? The only remaining conductor is the neutral, a big enough conductor for 200A (yes, this 400A was at 120V but current is still current and voltage doesn't change the conductor's current capacity).

Somehow I'm imagining a bridge that can handle 200 cars per minute, but that can be 400 if half the cars are blue :-)

- posted on October 26, 2009, 3:08 am

And that almost never happens in real life, either....

Wrong -- both are true.

OK

400A @ 120V, or 200A @ 240V, yes.

Cute. Just answer these questions; assume a 240V 200A service.

What is the maximum power that service can provide?

If all the loads supplied by that service are 120V loads (e.g. blender, toaster, light bulbs, range hood, stereo, TV, computer, etc.) what do you get when you divide that maximum power by 120V?

- posted on October 26, 2009, 2:49 pm

On Oct 25, 11:08 pm, snipped-for-privacy@milmac.com (Doug Miller) wrote:

*all.invalid> wrote:*

But a balanced load is exactly what was shown in the simple circuit example above that he understood and is discussing.

Which matters not a wit. Unless of course you are trying to get close to the maximum capacity of the service. If it's totally unbalanced, guess what? You get 200 amps at 120V, or exactly half the power capacity of the service. Gee, I wonder why? Could it be that it's because the service can only handle 200AMPS? And that with a 200 amp unbalanced load at 120V, 200 amps is coming in on one hot and it's all going back on the neutral?

Wow, it's getting really strange here. Of course, by definition, a parallel circuit is one where the ends of the individual elements are connected together. A series circuit is one where elements are connected one after the other, in series.

And there you go again, inserting voltage into a question of amperage. Amperage is a measure of the charge, ie electrons passing through the conductor and IS NOT LINKED TO VOLTAGE.

You;ve asked that question multiple times and it's always been answered the same: 48KVA

Now answer his question that you avoided. Apply Kirchoff's law and tell us where current is flowing in a 200 amp service cable that totals up to 400 amps. All of us here agree and can account for 200amps. So explaing the missing 200.

If it's a balanced load, you get 400 amps because half the load is in SERIES with the other half. As I've outlined about 6 times now, you have 200 amps coming in on one hot, going through the loads in series and then out the other hot. 200 amps is flowing in the service. If you say it's 400, then why isn't it 2 amps that flows in a 120watt light bulb plugged into an outlet? 1 amp comes in one wire, 1 amp goes out the other wire. Yet the world agrees that only 1 amp is flowing, not 2.

If it's a totally unbalanced 120V load, then you can't just divide the power by 120 as youu imply, because you have 200 amps flowing in on one hot, and 200 amps flowing out on the neutral. So you have a 120V, 200 amp load and only a power of 24KVA.

No matter how you slice and dice it, there is a max of 200 amps flowing in the service. Since you believe otherwise, outline the current flows as I have here and how it adds up to greater than 200 amps flowing in the service conductors.

But a balanced load is exactly what was shown in the simple circuit example above that he understood and is discussing.

Which matters not a wit. Unless of course you are trying to get close to the maximum capacity of the service. If it's totally unbalanced, guess what? You get 200 amps at 120V, or exactly half the power capacity of the service. Gee, I wonder why? Could it be that it's because the service can only handle 200AMPS? And that with a 200 amp unbalanced load at 120V, 200 amps is coming in on one hot and it's all going back on the neutral?

Wow, it's getting really strange here. Of course, by definition, a parallel circuit is one where the ends of the individual elements are connected together. A series circuit is one where elements are connected one after the other, in series.

And there you go again, inserting voltage into a question of amperage. Amperage is a measure of the charge, ie electrons passing through the conductor and IS NOT LINKED TO VOLTAGE.

You;ve asked that question multiple times and it's always been answered the same: 48KVA

Now answer his question that you avoided. Apply Kirchoff's law and tell us where current is flowing in a 200 amp service cable that totals up to 400 amps. All of us here agree and can account for 200amps. So explaing the missing 200.

If it's a balanced load, you get 400 amps because half the load is in SERIES with the other half. As I've outlined about 6 times now, you have 200 amps coming in on one hot, going through the loads in series and then out the other hot. 200 amps is flowing in the service. If you say it's 400, then why isn't it 2 amps that flows in a 120watt light bulb plugged into an outlet? 1 amp comes in one wire, 1 amp goes out the other wire. Yet the world agrees that only 1 amp is flowing, not 2.

If it's a totally unbalanced 120V load, then you can't just divide the power by 120 as youu imply, because you have 200 amps flowing in on one hot, and 200 amps flowing out on the neutral. So you have a 120V, 200 amp load and only a power of 24KVA.

No matter how you slice and dice it, there is a max of 200 amps flowing in the service. Since you believe otherwise, outline the current flows as I have here and how it adds up to greater than 200 amps flowing in the service conductors.

- posted on October 26, 2009, 11:21 pm

snipped-for-privacy@optonline.net wrote:

Now divide 48kVA by 120V and tell me what you get.

Which is exactly what I've been telling you for the last three days. Glad you finally figured it out.

Now divide 48kVA by 120V and tell me what you get.

Which is exactly what I've been telling you for the last three days. Glad you finally figured it out.

- posted on October 27, 2009, 12:12 am

On Mon, 26 Oct 2009 03:08:49 GMT, snipped-for-privacy@milmac.com (Doug Miller)
wrote:

True, although it happens in examples, such as the ones used here.

IF those 2 legs have identical voltages, the difference between them is 0 (that's what "identical" means). In that case, 120V loads should be OK but 240V loads would get nothing.

That's some strange reality. In this one voltage and current are different things, and you can't change one into another with arithmetic.

There is still no 400A at any voltage.

I notice you ignored mine. You have yet to show any non-imaginary location of that 400A.

48KW. Of course we were talking about CURRENT.

That would be 400A. Of course that's only in your imagination since the math is invalid (120V is obtained by splitting the service into 2 separate halves, each of which is only 24KW).

True, although it happens in examples, such as the ones used here.

IF those 2 legs have identical voltages, the difference between them is 0 (that's what "identical" means). In that case, 120V loads should be OK but 240V loads would get nothing.

That's some strange reality. In this one voltage and current are different things, and you can't change one into another with arithmetic.

There is still no 400A at any voltage.

I notice you ignored mine. You have yet to show any non-imaginary location of that 400A.

48KW. Of course we were talking about CURRENT.

That would be 400A. Of course that's only in your imagination since the math is invalid (120V is obtained by splitting the service into 2 separate halves, each of which is only 24KW).

- posted on October 27, 2009, 2:37 am

Exactly so.

200A each. Total of 400A of 120V loads -- as you said.

- posted on October 27, 2009, 4:00 am

On Oct 26, 10:37 pm, snipped-for-privacy@milmac.com (Doug Miller) wrote:

*all.invalid> wrote:*

Where in the box can you measure 400 amps? If the panel is controlling 48KW there will be no current on the neutral because the currents will be balanced. The current that flows through one half of the breaker is the same current that flows through the other half of the breaker. In this case what you have is two 200 amp breakers in series. Doug you have more current coming into the box than going out and that shouldnt happen.

Where in the box can you measure 400 amps? If the panel is controlling 48KW there will be no current on the neutral because the currents will be balanced. The current that flows through one half of the breaker is the same current that flows through the other half of the breaker. In this case what you have is two 200 amp breakers in series. Doug you have more current coming into the box than going out and that shouldnt happen.

- posted on October 27, 2009, 5:35 am

wrote:

The power is coming in from a transformer secondary winding that is center-tapped. Let's call the 3 wires Line 1, the neutral & Line 2 (seee the link below that shows a transformer secondary at the bottom of the page). When you put 120V loads across Line 1 & neutral, they are independent of Line 2. In effect, you're only using half of the transformer secondary, so you're only going thru the Line 1 half of the main breaker. The current path is from the Line 1 side of the secondary winding, thru the Line 1 side of the main breaker, thru the load, and back thru the neutral to the Line 1 half of the secondary winding. If you also put a 120V load across Line 2 and the neutral, then the current path is from the Line 2 side of the secondary winding thru the Line 2 side of the main breaker, thru the load, and back thru the neutral (in the opposite direction of current flow of the Line 1 current thru the neutral) and back to the Line 2 side of the secondary winding. Both loads form their own circular loops that are independent of each other, except for sharing the neutral (in opposite directions) to complete their separate circuits. Here is a great explanation of the transformer secondary, using the battery analogy which the author (not me) originally designed to show balanced loads, but is also useful in showing how 120V loads form independent circuits on each side of the secondary. You can even close the various switchs and see the effect.

http://home.comcast.net/~ronaldrc/wsb/ax.htm

The power is coming in from a transformer secondary winding that is center-tapped. Let's call the 3 wires Line 1, the neutral & Line 2 (seee the link below that shows a transformer secondary at the bottom of the page). When you put 120V loads across Line 1 & neutral, they are independent of Line 2. In effect, you're only using half of the transformer secondary, so you're only going thru the Line 1 half of the main breaker. The current path is from the Line 1 side of the secondary winding, thru the Line 1 side of the main breaker, thru the load, and back thru the neutral to the Line 1 half of the secondary winding. If you also put a 120V load across Line 2 and the neutral, then the current path is from the Line 2 side of the secondary winding thru the Line 2 side of the main breaker, thru the load, and back thru the neutral (in the opposite direction of current flow of the Line 1 current thru the neutral) and back to the Line 2 side of the secondary winding. Both loads form their own circular loops that are independent of each other, except for sharing the neutral (in opposite directions) to complete their separate circuits. Here is a great explanation of the transformer secondary, using the battery analogy which the author (not me) originally designed to show balanced loads, but is also useful in showing how 120V loads form independent circuits on each side of the secondary. You can even close the various switchs and see the effect.

http://home.comcast.net/~ronaldrc/wsb/ax.htm

- posted on October 27, 2009, 6:37 am

wrote:

I need to make a correction - the Line 2 current flow would be down the neutral, thru the load, up thru the Line 2 half of the breaker, back to the Line 2 side of the transformer. The currents flow in the same direction thru the transformer secondary "halves", and in opposite directions thru the neutral.

I need to make a correction - the Line 2 current flow would be down the neutral, thru the load, up thru the Line 2 half of the breaker, back to the Line 2 side of the transformer. The currents flow in the same direction thru the transformer secondary "halves", and in opposite directions thru the neutral.

- posted on October 27, 2009, 3:45 pm

Agree.

If you also put a 120V load across Line 2 and the neutral, then

Don't agree with this. If the second load on line 2 is equal to the load already on line 1, then the current flow is in on line 1 and back out on line 2. No current flows in the neutral.

If the second load on line 2 were half the size of the load on line 1, then half the current from line 1 would flow back out on line 2 and half the current from line 1 would flow back out the neutral.

The key here is look at that service cable coming from the transformer and you have a circuit running a max of 200 amps. Agree?

- posted on October 27, 2009, 7:16 pm

wrote:

Agree.

If you also put a 120V load across Line 2 and the neutral, then

OK, you're right. If they're perfectly balanced. So lets make the Line 1-to-neutral load a 200A load (0.6 ohms), and the Line 2-to-neutral load 199A (0.603 ohms). Now we have an unbalanced current of 1 amp thru the neutral.So we have 200A thru one side of the breaker, and 199A thru the other side. So what's the total amperage now?? Is it 200A or 399A, or? It's all in how you look at it. In the end ,it's a matter of total energy or kW. 200A x 120V$000W & 199A x 120V= 23880W for a total of 47880W or 47.88kW

With a perfectly balanced load - yes, one circuit. With a slightly unbalanced load - then, two circuits.

Agree.

If you also put a 120V load across Line 2 and the neutral, then

OK, you're right. If they're perfectly balanced. So lets make the Line 1-to-neutral load a 200A load (0.6 ohms), and the Line 2-to-neutral load 199A (0.603 ohms). Now we have an unbalanced current of 1 amp thru the neutral.So we have 200A thru one side of the breaker, and 199A thru the other side. So what's the total amperage now?? Is it 200A or 399A, or? It's all in how you look at it. In the end ,it's a matter of total energy or kW. 200A x 120V$000W & 199A x 120V= 23880W for a total of 47880W or 47.88kW

With a perfectly balanced load - yes, one circuit. With a slightly unbalanced load - then, two circuits.

- posted on October 27, 2009, 9:47 pm

No, it not. Once again, the physical current in the service cable is 200 amps. You have 200 amps flowing IN on one hot, 199 flowing OUT on the other hot and 1 amp flowing OUT on the neutral. The current flowing in that circuit is 200 amps. You don't count current twice.

How much current is flowing in a simple 120W light bulb plugged into a 120Volt outlet? 1 amp. Now simply add another single wire in parallel with one of those supplying the light bulb. The current will be split now, with some part of it going via one wire, some via the other. For sake of argument, let;s assume it just splits evenly. So, now you have 1 amp coming in on one wire, 1/2 amp leaving via one wire, 1/2 amp leaving via the second wire. How much current is flowing in this "service" circuit to the light bulb? There is no confusion, it's 1 amp.

It's a simple matter of applying Kirchoffs law.

Balanced or unbalanced matters not a wit. There is still a current of 200 amps coming and going either way. The only difference balance makes is which PATH that 200 amp current takes.

- posted on October 27, 2009, 10:13 pm

wrote:

No, it not. Once again, the physical current in the service cable is 200 amps. You have 200 amps flowing IN on one hot, 199 flowing OUT on the other hot and 1 amp flowing OUT on the neutral. The current flowing in that circuit is 200 amps. You don't count current twice.

How much current is flowing in a simple 120W light bulb plugged into a 120Volt outlet? 1 amp. Now simply add another single wire in parallel with one of those supplying the light bulb. The current will be split now, with some part of it going via one wire, some via the other. For sake of argument, let;s assume it just splits evenly. So, now you have 1 amp coming in on one wire, 1/2 amp leaving via one wire, 1/2 amp leaving via the second wire. How much current is flowing in this "service" circuit to the light bulb? There is no confusion, it's 1 amp.

It's a simple matter of applying Kirchoffs law.

Umm...so you're saying that if you have a 200A load on Line1to neutral that you're going to have 200A going thru both sides of the 200A breaker regardless of the load on the Line 2 to neutral side??

What if you have a 200A load on the Line 1 to neutral side, and nothing loading the Line 2 to neutral side? Are you still going to have 200A going thru each "leg" of the main breaker?

No, it not. Once again, the physical current in the service cable is 200 amps. You have 200 amps flowing IN on one hot, 199 flowing OUT on the other hot and 1 amp flowing OUT on the neutral. The current flowing in that circuit is 200 amps. You don't count current twice.

How much current is flowing in a simple 120W light bulb plugged into a 120Volt outlet? 1 amp. Now simply add another single wire in parallel with one of those supplying the light bulb. The current will be split now, with some part of it going via one wire, some via the other. For sake of argument, let;s assume it just splits evenly. So, now you have 1 amp coming in on one wire, 1/2 amp leaving via one wire, 1/2 amp leaving via the second wire. How much current is flowing in this "service" circuit to the light bulb? There is no confusion, it's 1 amp.

It's a simple matter of applying Kirchoffs law.

Umm...so you're saying that if you have a 200A load on Line1to neutral that you're going to have 200A going thru both sides of the 200A breaker regardless of the load on the Line 2 to neutral side??

What if you have a 200A load on the Line 1 to neutral side, and nothing loading the Line 2 to neutral side? Are you still going to have 200A going thru each "leg" of the main breaker?

- posted on October 28, 2009, 9:49 am

No, never said that.

You're going to have 200A flowing from Line 1 to neutral. Again, that is 200 amps flowing in the service cable. As I said, if it's fully loaded, you will have a max of 200 amps flowing in the service cable. That's 200 amps coming in, 200 amps going out. The only difference is whether some of the current flows on the neutral or not. If the neutral has some current, then one of the hot lines must have less than 200 amps or you'd violate Kirchoff's law.

- posted on October 28, 2009, 4:49 pm

wrote:

No, never said that.

But you still didn't answer the question about the current flow path. If you have a 120V 200A load between Line 1 & neutral,and no load between Line 2 & neutral, how much current flows thru the Line 1 side of the main breaker, how much current flows thru the Line 2 side of the main breaker, and how much thru the neutral?

No, never said that.

But you still didn't answer the question about the current flow path. If you have a 120V 200A load between Line 1 & neutral,and no load between Line 2 & neutral, how much current flows thru the Line 1 side of the main breaker, how much current flows thru the Line 2 side of the main breaker, and how much thru the neutral?

- posted on October 28, 2009, 5:22 pm

There will be 0 amps in the neutral. The loads form two legs of a balanced bridge, the windings in the transformer form the other two legs, When they are balancedas you described therer will be ZERO current in the neutral. Dont believe me check it for yourself. Make a test circuit out of two 100 watt light bulbs connected in series, put them across a 240VAC connection with no neutral connection then connect the neutral between them. Measure the current in the neutral. It will be ZERO. measure the current in each hot leg. The neutral is only there to carry the difference between the two legs when there is an imbalance. If the current in one leg was 100 amps and the the other had 50 amps the neutral current would be 100 -50 + 50 amps.. When the load is balanced and you measure the current in each leg you are just measuring the same current twice.

Jimmie

- posted on October 28, 2009, 5:37 pm

wrote:

Re-read the question. I'm asking about a SINGLE 120V 200A load between Line 1 & neutral, not 2 loads in series across 240V.

Re-read the question. I'm asking about a SINGLE 120V 200A load between Line 1 & neutral, not 2 loads in series across 240V.

- posted on October 29, 2009, 8:25 pm

Doesnt matter if there is no current flow in the neutral, Quit trying to reason what you dont understand and build a mock up circuit if this really concerns you that much. I am growing very bored of talking to people who are more interested in defending their argument than actually trying to learn something. Any decent text that explains Kerchov's(sp?) Law as applied to AC circuits will give you the answer.

Jimmie

- posted on October 28, 2009, 9:40 am

Yes. You have a 3 wire service cable coming into the house. Use Kirchoff's law and add up the current coming into the house at any point in time under any conditions through that cable any you have a max of 200 amps. Add up all the current leaving the house at the same point in time and you have a max of 200 amps. In my world, that's a 200 amp service. You don't measure 300, or 400 by counting electrons twice.

And also, to support a max of 400 amps of 120volt load in the house which is the hot topic, you have to have the special case where the loads are perfectly balanced so that 200 amps is on each side. And then you in fact have a SINGLE circuit with 200 amps flowing in on one hot leg, throught the loads in series, and back out the other hot leg. Nothing flows in the neutral. Again, in my world, that's 200 amps flowing in the service, not 400.

- Kitchen faucet chatter
- - next thread in Home Repair

- What would you do with this shower?
- - previous thread in Home Repair

- OT: How many days?
- - newest thread in Home Repair

- Expected Life Of a Pedestal Sump Pump ?
- - last updated thread in Home Repair

- OT: How many days?
- - the site's newest thread. Posted in Home Repair

- Expected Life Of a Pedestal Sump Pump ?
- - the site's last updated thread. Posted in Home Repair