Residential Voltage drops on circuits

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suemarkp

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Kent, WA
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Retired Engineer
But in a MWBC you have 2 voltage sources out of phase with each other using a common return path. If the circuit under test is from Leg A, and Leg B has an existing 5A load in the neutral, you could see the voltage drop in the neutral go negative (relative to testing with no load on A) when you add a 5A load on Leg A. In this example, the neutral currents cancel so there is no current on the neutral (so no voltage drop) and 5A in each hot leg.

This effect was obvious in my detached workshop when I used a saw on leg 1 and the lights on leg 2 got a bit brighter. There was nothing wrong with the neutral (it was not high resistance or failed and running current through other loads), it is just that the neutral current went close to 0 so that lighting circuit had a reduced voltage drop from "normal".

Because none of us know how much current this meter actually sends don the wires, and whether it is extrpolating or not, and if there is even a MWBC involved and whether the other leg was loaded or there are high resistance connections, I'm not sure anyone can really know what is going on here without tracing out this circuit and monitoring currents at the panel.
 

GoldDigger

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Retired PV System Designer
But in a MWBC you have 2 voltage sources out of phase with each other using a common return path. If the circuit under test is from Leg A, and Leg B has an existing 5A load in the neutral, you could see the voltage drop in the neutral go negative (relative to testing with no load on A) when you add a 5A load on Leg A. In this example, the neutral currents cancel so there is no current on the neutral (so no voltage drop) and 5A in each hot leg.

This effect was obvious in my detached workshop when I used a saw on leg 1 and the lights on leg 2 got a bit brighter. There was nothing wrong with the neutral (it was not high resistance or failed and running current through other loads), it is just that the neutral current went close to 0 so that lighting circuit had a reduced voltage drop from "normal".

Yes, this is an absolutely normal occurrence and is yet another reason that MWBCs cause homeowner complaints even when wired in a code-compliant manner. A 3% voltage change will be easily visible in an incandescent light if it happens suddenly.

But in the OPs case, the voltage drop in the neutral changing polarity would still correspond to a drop in line to neutral voltage seen by the meter. If turning on the saw caused lights on the same phase to get brighter, then you would have something interesting!

One interesting interaction on an MWBC is that if the load on one side is primarily reactive and there is a high resistance neutral, the effects of changing the load on the other side is less obvious. And in the situation of an open neutral with the right values of inductive load on one side and capacitive load on the other side, you can end up with a resonant circuit with near zero reactance and an effective short circuit.
So the visible experiment is that with the neutral connected, both sides draw a reactive current which could be, for example, 1/10 of the breaker trip point. But when you open the neutral, the breaker(s) trip. Seeing a breaker trip when you disconnect a wire is a very counterintuitive experience!
 

Strathead

Senior Member
Location
Ocala, Florida, USA
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Electrician/Estimator/Project Manager/Superintendent
But in a MWBC you have 2 voltage sources out of phase with each other using a common return path. If the circuit under test is from Leg A, and Leg B has an existing 5A load in the neutral, you could see the voltage drop in the neutral go negative (relative to testing with no load on A) when you add a 5A load on Leg A. In this example, the neutral currents cancel so there is no current on the neutral (so no voltage drop) and 5A in each hot leg.

This effect was obvious in my detached workshop when I used a saw on leg 1 and the lights on leg 2 got a bit brighter. There was nothing wrong with the neutral (it was not high resistance or failed and running current through other loads), it is just that the neutral current went close to 0 so that lighting circuit had a reduced voltage drop from "normal".

Because none of us know how much current this meter actually sends don the wires, and whether it is extrpolating or not, and if there is even a MWBC involved and whether the other leg was loaded or there are high resistance connections, I'm not sure anyone can really know what is going on here without tracing out this circuit and monitoring currents at the panel.

This doesn't make sense to me. I am going to tell why, not to question you per se, but to have someone here who is smarter than me explain. First, I know assume that you misspoke when you stated that the "2 voltage sources are out of phase." I am assuming that this is a typical single phase 120/240 house feed. Of course you could be right if this is a far less typical 2 legs of a 120/208 utility transformer. That said my though process proceeds as follows:
, if the neutral connection is good, then any unbalanced current from your saw will take the path of least resistance, which is from the source, through the saw back on the neutral. Only the unbalanced portion of the current should travel through the lamp. I can wrap my head around the possibility that the resistance of the neutral may cause an extremely minor change in the total resistance of the lamp circuit, but without drawing it out and smoking my brain I am not sure. However, the change would have to be so minuscule that no standard human could detect it, it seems to me.
 

Strathead

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Location
Ocala, Florida, USA
Occupation
Electrician/Estimator/Project Manager/Superintendent
Sorry if I missed someone else pointing this out, but the OP mentioned the breaker tripping if more than one item was plugged in. I would just like to point out that all of the discussion about bad connections etc. is not likely to cause this unless possibly the connection is at the breaker and the heat is transferring to the thermal trip element in the breaker.
 

GoldDigger

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Staff member
Location
Placerville, CA, USA
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Retired PV System Designer
Sorry if I missed someone else pointing this out, but the OP mentioned the breaker tripping if more than one item was plugged in. I would just like to point out that all of the discussion about bad connections etc. is not likely to cause this unless possibly the connection is at the breaker and the heat is transferring to the thermal trip element in the breaker.

It was mentioned that the breaker trip was inconsistent with a simple high resistance neutral, but not that a heat generating high resistance phase termination at the breaker terminal itself could indeed cause this. That would take care of half of the observed anomalies and leave just the meter reading to be explained.
 

ELA

Senior Member
Occupation
Electrical Test Engineer
35.5% of 120.3 = 42.7 round up to 43 and mistakenly add a 1 in front of it?

Nice meter :happysad:
 
This doesn't make sense to me. I am going to tell why, not to question you per se, but to have someone here who is smarter than me explain. First, I know assume that you misspoke when you stated that the "2 voltage sources are out of phase." I am assuming that this is a typical single phase 120/240 house feed. Of course you could be right if this is a far less typical 2 legs of a 120/208 utility transformer. That said my though process proceeds as follows:
, if the neutral connection is good, then any unbalanced current from your saw will take the path of least resistance, which is from the source, through the saw back on the neutral. Only the unbalanced portion of the current should travel through the lamp. I can wrap my head around the possibility that the resistance of the neutral may cause an extremely minor change in the total resistance of the lamp circuit, but without drawing it out and smoking my brain I am not sure. However, the change would have to be so minuscule that no standard human could detect it, it seems to me.

I have not heard back from AMPROBE yet. You are correct in that the tenant living in this home mentioned if they plugged multiple devices into the up stairs outlets the breaker would trip.
 

GoldDigger

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Placerville, CA, USA
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Retired PV System Designer
I have not heard back from AMPROBE yet. You are correct in that the tenant living in this home mentioned if they plugged multiple devices into the up stairs outlets the breaker would trip.
How much load is "multiple devices" relative to the breaker size? One toaster and one electric heater, for example? If the whole upstairs is on one 15 amp circuit, it would not take much.
Is it an immediate trip or a delayed (thermal) trip? If delayed, roughly how long? (I realize that you may not have the answers to this, but raising them anyway as diagnostic questions.)
 

suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
This doesn't make sense to me. I am going to tell why, not to question you per se, but to have someone here who is smarter than me explain. First, I know assume that you misspoke when you stated that the "2 voltage sources are out of phase." I am assuming that this is a typical single phase 120/240 house feed. Of course you could be right if this is a far less typical 2 legs of a 120/208 utility transformer. That said my though process proceeds as follows:
, if the neutral connection is good, then any unbalanced current from your saw will take the path of least resistance, which is from the source, through the saw back on the neutral. Only the unbalanced portion of the current should travel through the lamp. I can wrap my head around the possibility that the resistance of the neutral may cause an extremely minor change in the total resistance of the lamp circuit, but without drawing it out and smoking my brain I am not sure. However, the change would have to be so minuscule that no standard human could detect it, it seems to me.

In a 120/240V system, the 120V "phases" are 180 degrees out of phase with one another. That is how you get the 240V -- one 120V waveform is at max +, while the other is at max -.

There were a lot of lights in my example (about 10A worth). So with just the lights on, there is 10A going through the neutral. When the saw kicks on (powered from the opposite power leg), if it draws 10A, then the neutral current will become 0 because its 10A is current in the opposite direction of the 10A from the lights. If the saw drew 12A, then the neutral would have 2A going through it (12A - 10A = 2A). If the saw drew 8A, it would be the same thing (8A - 10A = -2A, direction {plus or minus} doesn't matter for voltage drop). Your eyes can notice a sudden change of 2 or 3 volts (the voltage drop on the neutral feeder or MWBC) on an incandescent light.
 

GoldDigger

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Location
Placerville, CA, USA
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Retired PV System Designer
If the saw drew 12A, then the neutral would have 2A going through it (12A - 10A = 2A). If the saw drew 8A, it would be the same thing (8A - 10A = -2A, direction {plus or minus} doesn't matter for voltage drop).

Yes and no. In one case above the voltage "drop" in the neutral will be in phase with the voltage to the lights, in the other case it would be out of phase with the voltage to the lights. What is important is that the 12A or 8A load from the saw will in either case an increase the voltage applied to the lights.
When we talk about voltage drop we are used to talking about the drop caused by the load itself (increase in neutral voltage, decrease in phase voltage) rather than a change caused by a load on the other side of the MWBC. That leads to lots of potential confusion.
 

suemarkp

Senior Member
Location
Kent, WA
Occupation
Retired Engineer
Yes. And I wonder if that confusion is what is confusing the test equipment. It would probably be a lot less confused if there was an equipment gorund at the test receptacle. Perhaps he should retest with a grounding wire strung along the floor back to the panel to get a better picture of what the meter is seeing.
 

kwired

Electron manager
Location
NE Nebraska
With all due respect to a Senior Member, what you are proposing violated Kirchoff's Laws and several other basic principles of "equivalent circuits". Regardless of the complexity of the circuitry on the other side of the receptacle, it can be modelled as a voltage source and a series resistance. The only ways that the voltage at the interface could go up when the current draw increased would be:
1. The equivalent circuit has a negative resistance, or
2. The act of drawing the current caused some other device to switch on or off, thus changing the equivalent circuit.

The voltage between neutral and ground could certainly rise, but that would reduce the measured line to neutral voltage, not increase it! And the line to ground voltage could not increase at all.

One thing that occurred to me was that if the tester applies a current pulse rather than a linear load, the presence of a resonant circuit on the other side of the receptacle somewhere could cause ringing which could be misinterpreted as an increase in voltage.

No violation of Kirchoff's laws. With the MWBC and a open neutral the thing that changes is applied voltage. Loads that were 120 each with a good neutral are now in series and 240 volts is applied to the series. The load with the higher impedance will have more voltage across it. Input voltage will be 240 and sum of all voltage drops in the circuit will equal applied voltage.

If the neutral is not completely open things are more complex to calculate, as the resistance of the bad neutral likely changes according to conditions, but you are still going to have more drop across the side of the circuit that has the higher impedance.
 

GoldDigger

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No violation of Kirchoff's laws. With the MWBC and a open neutral the thing that changes is applied voltage. Loads that were 120 each with a good neutral are now in series and 240 volts is applied to the series. The load with the higher impedance will have more voltage across it. Input voltage will be 240 and sum of all voltage drops in the circuit will equal applied voltage.

If the neutral is not completely open things are more complex to calculate, as the resistance of the bad neutral likely changes according to conditions, but you are still going to have more drop across the side of the circuit that has the higher impedance.

Exactly. And when you decrease the impedance of the tester, it will see a lower voltage. The equivalent circuit is a 240 volt source with a series impedance which is the impedance of the load on the other side of the circuit. Which we have to assume is fixed for the duration of the test. This is exactly the opposite of what the tester is reporting. Either there is a bug in its program or something very strange is happening in the circuit.

Since the tester (and any real load for a more definitive test) is connected only to two terminals, the entire rest of the system, all the way back through the distribution system if you wish, can be modelled by an equivalent circuit which is either a single voltage source with a series resistor or a single current source with a parallel resistor. Either way, there is no way that drawing more current from those two terminals can cause the voltage at the same two terminals to rise unless the series resistance of the equivalent circuit is a negative resistance. Other than devices like fluorescent tubes, zener diodes and under some conditions motors, you will not find negative resistance in purely passive circuits.

There are some rather esoteric situations which could lead to a condition like the OP's meter claimed it saw, but they depend on non-linear loads somewhere on the other side of the receptacle and are for the most part not consistent with a perfectly normal nominal voltage when no load is applied.

Here is one carefully constructed example of how a current pulse drawn from a DC circuit could cause the terminal voltage to rise:

Imagine a 24 volt battery, with an internal resistance of 1 ohm.
Connected to the + and - of the battery is a voltage divider of two 1200 ohm resistors in series. The load terminals of our network are the top and bottom of the top `1200 ohm resistor. With no load applied, the voltage seen is 12 volts. If you start to draw current, the voltage will decrease. Eventually with a near zero ohm load applied, the voltage will drop to zero and the full 24 volts will appear across the bottom resistor. So far, so good.
This behavior can be modelled by a 12 volt battery with a 600 ohm series resistor.
Now put in parallel with the bottom resistor a circuit composed of an an SCR, zener and resistor such that the SCR will fire and latch when the applied voltage excedes 18 volts. Initially the SCR is not conducting, so the model matches the earlier case exactly.

Look what happens now as you start to draw increasing current from the load points: the voltage decreases toward 6 volts. At that point the voltage on the other resistor reaches 18 volts and the SCR fires. Suddenly the voltage across the lower resistor drops to near zero and the voltage at our load rises to nearly 24 volts.
From this point on, the model of the two terminal network becomes a 24 volt battery in series with a ~1 ohm resistor. Very different.

If all you did was apply a large enough current pulse to the terminals, it would appear to increase the terminal voltage, but that is because of the presence of non-linear active elements which change the equivalent circuit.
I do not think there is any credible equivalent condition on an AC circuit, MWBC or otherwise, which will cause a similar condition.

My current best guess leans toward the presence of a reactive load somewhere in the house which is causing the meter to see a ringing voltage waveform when a step current pulse is applied. Or a programming error in the tester.
 
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