Theory question again

[zappy]

If you have a parallel 120v circuit with one,two or 100 loads, how is there current on the neutral, if from what I've read a load always uses up all the voltage. I'm guessing it has something to do with DC and not AC. Am I correct? Thank you for your help.

 

[dkarst]

It is possible I am misinterpreting your question so let's take a simple example with three 100 W light bulbs plugged into three receptacles on the same circuit. The current through each bulb would = P/V or a bit over ~ 1 amp. Each of these one amp currents would add on the neutral conductor and flow back to the panel. The ~ 3 amps flow out through the breaker, ~ 1 amp through each bulb, combine to be ~ 3 amps in the neutral, and back to panel. Your comment "the load uses up all the voltage" would be better stated "the voltage is dropped across the load"... in this case, the 115V is dropped across our bulbs so there is 0 volts on the neutral (ignoring supply voltage wire drop) but that DOES NOT mean there is zero amps flowing on neutral. Hope I helped.

 

[LarryFine]

If you have a parallel 120v circuit with one,two or 100 loads, how is there current on the neutral, if from what I've read a load always uses up all the voltage. I'm guessing it has something to do with DC and not AC.

Presuming that, by 'parallel', you mean MWBC's (more than one hot sharing a neutral):

First, it has nothing to do with AC vs. DC. Remember that the voltage is normally all "used up" by the load because it's forced to by the low impedance of the rest of the circuit.

A resistive section of the circuit will allow some voltage to be dropped there, leaving less to be dropped across the load. The relative amounts depend on the relative resistances.

In the case of a genuinely-balanced MWBC, there's no neutral current because the voltage at the loads' neutral point already happens to equal the voltage at the supply's neutral point.

In all other cases, the neutral does carry current in its effort to keep the supply and load neutral points as close to the same voltage as possible, just as with line-conductor voltages.

 

[LarryFine]

After reading DK's post, if you mean a single 2-wire circuit, the current in the grounded conductor is always equal to the current in the hot conductor, except during a shock or other leakage to earth, which a GFCI can detect.

The voltage to earth from each is dependent on which one is grounded, but not that from wire to wire, which is what the load sees. It does not matter to the load which conductor is grounded, only the voltage between them.

 

[zappy]

It is possible I am misinterpreting your question so let's take a simple example with three 100 W light bulbs plugged into three receptacles on the same circuit. The current through each bulb would = P/V or a bit over ~ 1 amp. Each of these one amp currents would add on the neutral conductor and flow back to the panel. The ~ 3 amps flow out through the breaker, ~ 1 amp through each bulb, combine to be ~ 3 amps in the neutral, and back to panel. Your comment "the load uses up all the voltage" would be better stated "the voltage is dropped across the load"... in this case, the 115V is dropped across our bulbs so there is 0 volts on the neutral (ignoring supply voltage wire drop) but that DOES NOT mean there is zero amps flowing on neutral. Hope I helped.

Yes this is what I was wondering. So you can have current without voltage? I thought if there's no voltage there can't be current? Interesting.

 

[mivey]

Yes this is what I was wondering. So you can have current without voltage? I thought if there's no voltage there can't be current? Interesting.

There is a slight voltage, but He said to ignore the supply voltage wire drop.

 

[mivey]

So you can have current without voltage? I thought if there's no voltage there can't be current? Interesting.

You can have current with no voltage in a super conductor. In our world, there will be a voltage.

 

[mivey]

If you have a parallel 120v circuit with one,two or 100 loads, how is there current on the neutral, if from what I've read a load always uses up all the voltage. I'm guessing it has something to do with DC and not AC. Am I correct? Thank you for your help.

The resistance in the neutral is a load. Just a relatively small load.

Add: The ungrounded conductor resistance is a small load as well.

 

[Volta]

Yes this is what I was wondering. So you can have current without voltage? I thought if there's no voltage there can't be current? Interesting.

Remember that to observe a voltage you need two points. There will be 120 volts between the two conductors the whole time. 'Zero' volts is a relative term here. The neutral has 120 volts on it just as the 'hot' wire does, in this two-wire example. In fact, in a two-wire example, we don't need to consider it 'neutral' anymore, because we don't have the third wire present. It is just another conductor carrying the same current.

You can have current with no voltage in a super conductor.

How's that? Super conductors can do some amazing things, even counter-intuitive things, but they won't cause current to flow with no voltage present.

 

[mivey]

Remember that to observe a voltage you need two points. There will be 120 volts between the two conductors the whole time. 'Zero' volts is a relative term here. The neutral has 120 volts on it just as the 'hot' wire does, in this two-wire example. In fact, in a two-wire example, we don't need to consider it 'neutral' anymore, because we don't have the third wire present. It is just another conductor carrying the same current.

But there is not 120 volts across the neutral wire. The two points under consideration are the neutral point at the source and the neutral point at the load. There is a small voltage different when current flows. Just remove the picture of a wire in your mind and replace all of the wires with resistors.

How's that? Super conductors can do some amazing things, even counter-intuitive things, but they won't cause current to flow with no voltage present.

The super conductor does not cause the current to flow. A regular conductor does not cause current to flow either.

They inject the current into a super conducting loop. Some of these have been running for years now with no drop in the current running around the loop.

 

[Volta]

But there is not 120 volts across the neutral wire. The two points under consideration are the neutral point at the source and the neutral point at the load. There is a small voltage different when current flows. Just remove the picture of a wire in your mind and replace all of the wires with resistors.

I'm just saying that in a two-wire example that the 'neutral' wire exhibits exactly the same results as the 'hot' conductor. If the circuit is drawn symmetrically, with either wires or resistors, the voltages and reference points can be flipped and the readings would be the same.

Neither the fact that the identified conductor is grounded nor that it can be considered a neutral to the larger system matter in the two-wire example.

The super conductor does not cause the current to flow. A regular conductor does not cause current to flow either.

True.

They inject the current into a super conducting loop. Some of these have been running for years now with no drop in the current running around the loop.

Interesting. So as this is a closed loop, do we need to consider that there are precisely zero volts present in the system now (no matter how the current was initially injected)?
This circuit seems short. My simple mind at least needs to assume that voltage was present to start this flow.
Do you know who is running this experiment?

 

[gar]

100213-0904 EST

zappy:

Try to draw this circuit on paper. At the left is a voltage source. Draw this as a circle with a single cycle sine wave in its center. Put a 120 V label beside the circle. This will be a theoretical idea voltage source. That means no matter how much load current you draw from it the voltage will remain at 120 V.

From the top of the circle draw a line upward a ways, then a right angle to the right. Horizontally at this point put a 0.1 ohm resistor (the resistance of your hot wire from the voltage source to your load), and continue with a line to the right.

Do the same coming from the bottom of the voltage source for your neutral wire. Again an 0.1 ohm resistor. Remember 1.6 ohms/1000 ft for #12 copper. Thus our length from the voltage source to the load is 62.5 ft.

For our load consider a 100 W bulb in parallel with a 250 W. At 120 V these resistances are about 14400/100 = 144 ohms and 14400/250 = 57.6 ohms. These two resistances in parallel = 144*57.6/(144+57.6) = 41.14 ohms.

Vertically place a 41.14 ohm resistor between the right hand ends of the two 0.1 ohm supply line resistances.

The result is a series circuit of 0.1 + 41.14 + 0.1 = 41.34 ohms as the total load on the 120 V supply. This series circuit has a current flow of 120/41.34 = 2.903 A.

The voltage drop across each supply line is 0.1 * 2.903 or 0.29 V (290 MV). This is what you would read from the neutral buss in the main panel to the neutral terminal at the load point. Same voltage drop occurs from the output of the breaker to the hot terminal of the load.

The voltage drop across the two parallel lamps is 120 - 0.29 -0.29 = 119.42, or another way 2.903 * 41.14 = 119.43. The difference is in my rounding of numbers.

The current splits between the two loads as 119.43/144 = 0.8294 and 119.43/57.6 = 2.073 . Note 2.073 + 0.829 = 2.903 . A check on the calculations.

When you run experiments at your home that we have talked about you will find similar results, or maybe you will find some bad connections resulting in unexpected voltage drops on either the neutral or hot or both.

.

 

[ibew501ed]

nice job GAR
very clearly explained

 

[mivey]

I'm just saying that in a two-wire example that the 'neutral' wire exhibits exactly the same results as the 'hot' conductor. If the circuit is drawn symmetrically, with either wires or resistors, the voltages and reference points can be flipped and the readings would be the same.

Neither the fact that the identified conductor is grounded nor that it can be considered a neutral to the larger system matter in the two-wire example.

Correct

Interesting. So as this is a closed loop, do we need to consider that there are precisely zero volts present in the system now (no matter how the current was initially injected)?

Correct

Do you know who is running this experiment?

Kamerlingh Omnes was the first to do this. He let it run for a year and found no significant current loss. He won the Nobel Prize in 1913 for this discovery.

Link about a new measuring technique:
http://www.sciencedaily.com/releases/2009/10/091011071349.htm

Many people measure it. Google "persistent current" and "ring".

Even we can get in on the fun:

http://www.futurescience.com/manual/sc1000.html

or kit K18 - Superconducting Energy Storage Kit from here: http://www.users.qwest.net/~csconductor/

 

[Volta]

Cool stuff!

 

[zappy]

Thank you everyone

Thank you everyone

100213-0904 EST

zappy:

Try to draw this circuit on paper. At the left is a voltage source. Draw this as a circle with a single cycle sine wave in its center. Put a 120 V label beside the circle. This will be a theoretical idea voltage source. That means no matter how much load current you draw from it the voltage will remain at 120 V.

From the top of the circle draw a line upward a ways, then a right angle to the right. Horizontally at this point put a 0.1 ohm resistor (the resistance of your hot wire from the voltage source to your load), and continue with a line to the right.

Do the same coming from the bottom of the voltage source for your neutral wire. Again an 0.1 ohm resistor. Remember 1.6 ohms/1000 ft for #12 copper. Thus our length from the voltage source to the load is 62.5 ft.

For our load consider a 100 W bulb in parallel with a 250 W. At 120 V these resistances are about 14400/100 = 144 ohms and 14400/250 = 57.6 ohms. These two resistances in parallel = 144*57.6/(144+57.6) = 41.14 ohms.

Vertically place a 41.14 ohm resistor between the right hand ends of the two 0.1 ohm supply line resistances.

The result is a series circuit of 0.1 + 41.14 + 0.1 = 41.34 ohms as the total load on the 120 V supply. This series circuit has a current flow of 120/41.34 = 2.903 A.

The voltage drop across each supply line is 0.1 * 2.903 or 0.29 V (290 MV). This is what you would read from the neutral buss in the main panel to the neutral terminal at the load point. Same voltage drop occurs from the output of the breaker to the hot terminal of the load.

The voltage drop across the two parallel lamps is 120 - 0.29 -0.29 = 119.42, or another way 2.903 * 41.14 = 119.43. The difference is in my rounding of numbers.

The current splits between the two loads as 119.43/144 = 0.8294 and 119.43/57.6 = 2.073 . Note 2.073 + 0.829 = 2.903 . A check on the calculations.

When you run experiments at your home that we have talked about you will find similar results, or maybe you will find some bad connections resulting in unexpected voltage drops on either the neutral or hot or both.

.

I took a 60w. 120v. bulb, used my ohmmeter and the ohms was 12. Now when I do ohms law based on 60w. 120v. it says the ohms are 250. That's why I have no idea how to use my ohmmeter.

 

[gar]

100213-1231 EST

Good question zappy.

A standard incandescent lamp today is made with a tungsten filament. As you probably know most metals have a positive temperature coefficient of resistance. Meaning the resistance increases with temperature.

When you measure the lamp resistance with your ohmmeter you are measuring the filament resistance at about room temperature. When 120 V is applied to the lamp the filament gets very hot.

There is a lot of good information at
http://en.wikipedia.org/wiki/Incandescent_light_bulb
and the following quote is from near the end in a section titled "Voltage, light output, and lifetime"

In flood lamps used for photographic lighting, the tradeoff is made in the other direction. Compared to general-service bulbs, for the same power, these bulbs produce far more light, and (more importantly) light at a higher color temperature, at the expense of greatly reduced life (which may be as short as 2 hours for a type P1 lamp). The upper limit to the temperature at which metal incandescent bulbs can operate is the melting point of the metal. Tungsten is the metal with the highest melting point, 3695 K (6192?F). A 50-hour-life projection bulb, for instance, is designed to operate only 50 ?C (90 ?F) below that melting point. Such a lamp may achieve up to 22 lumens per watt, compared with 17.5 for a 750-hour general service lamp.[71]

A normal 1000 hour life time lamp has a hot to cold resistance ratio of about 14 to 1.

You can run your own experiment. A 50 or 60 W tungsten filament bulb, a Variac with 1 A capability, an ammeter, a voltmeter, pencil, paper, and a calculator are the tools needed to measure resistance vs voltage.

Another experiment. On a very sunshiny day connect your ohmmeter to a 250 W or smaller reflector flood lamp. Go outside and point in different directions noting the resistance. Now point directly at the sun. What happens?

.

 

[mivey]

...Another experiment. On a very sunshiny day connect your ohmmeter to a 250 W or smaller reflector flood lamp. Go outside and point in different directions noting the resistance. Now point directly at the sun. What happens?

.

I can't see the meter and I have a splitting headache.

 

[gar]

100213-2334 EST

mivey:

Sorry I was incomplete in my statement. Point the reflector in various directions, and then point the reflector flood at the sun.

I know you were joking, but my suggested experiment was not very clear. Although I believe I thought it was obvious what to do.

On a cloudy day with a 250 W bulb and my Fluke 27 I can not see a change, but with a higher resistance detector and/or a more sensitive instrument I should be able to find the sun behind the clouds.

But the experiment of changing the radiant energy directed at an incandescent lamp and seeing the lamp resistance change is something most people do not think about. Actually I can take a second 250 W flood and point it at the one of which I am measuring and see the resistance change. I have to be fairly close to see a change.

These are sort of kitchen sink experiments. Simple things to illustrate a point.

.

 

[ibew501ed]

GAR
these are great little teaching tool KEEP THEM COMING apprentices (and others) love hands on things to learn. Science, ohms law, meter use, all wrapped up in one