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How could a damaged wire in split-phase power delivery create these voltages?


Recently, my home suffered a partial power outage, and due to curiosity and a desire to learn more about residential AC power, I'm trying to understand how the event that took place resulted in the symptoms I observed.

My home is served by split-phase power from a nearby transformer, single-phase power transformed down to 240V with a center tap. For the sake of explanation I refer to the wires coming from the transformer to the main junction box on my house as L1, L2 and G. L1 and L2 are the "hot" wires, the voltage between which should ideally be 240V. G is the center tap wire, which is grounded and should ideally have a voltage difference of 120V to either of L1 or L2.

When the outage occurred, I observed the following:

  • The L1/G voltage difference was 120V (good).
  • The G/L2 voltage difference was 90V (bad).
  • The L1/L2 voltage difference was 30V (very bad!).

Despite my taking these measurements with a generic digital multimeter with a 10 MΩ input impedance (which I'm told is not reliable for this case), I believe them to be accurate as the power company repairmen also made measurements and confirmed that theirs matched my own.

The power company repairman investigated and determined that someone had been digging nearby and nicked the insulation to the L2 cable going from the transformer to my house. He said that this damage resulted in corrosion and build-up of mineral residue from electrolysis, and that the result was that impedance of the L2 cable increased drastically. (This cable was actually a cluster of several thick wires, and he said that sometimes in cases like this the corrosion is so bad that only 2 of the wires in the cluster remain fully intact.)

What I don't understand is how that can result in those voltage measurements. I can understand how high impedance on L2 could cause a drop in the L1/L2 voltage, and how L1/G could remain 120V since the high impedance of L2 would not be a part of the L1/G circuit. But I'm thoroughly puzzled by how the L1/L2 voltage of 30V can be lower than the G/L2 voltage of 90V; it seems like both voltages to the center tap, namely L1/G and G/L2, should sum up to L1/L2... unless some kind of really weird voltage phase shift is going on.

Any answers that could educate me in the aspects of AC electrical power that might explain this would be greatly appreciated!

Why should this post be closed?


3 answers


Here is a basic diagram of the power feed to your house, according to what the power company said they found:


R1 represents the additional series resistance due to the cable being corroded. R2 represents the leakage current to ground due to the broken insulation. Both these can vary over time, whether it recently rained, and even recent usage. Significant usage will heat up R1, and various chemical effects can make R2 not even look like a resistor.

However, this is still the basic concept. This alone explains nicely why the L2-G voltage is low, but not why L2-L1 could be lower than L2-G. For that we have to consider what is connected in the house.

Here is the circuit considering some of what is in the house:


R3 shows the small but inevitable leakage and "phantom loads", and C1 capacitance between the wires and inside 240 V appliances.

Note that both R3 and C1 can allow substantially more current than might be apparent at first glance. Modern appliances often use a little standby power, even when "off". There may be a clock, for example. Most modern appliances aren't turned on by actually switching main power directly by a button you touch. That would require way too large and expensive user buttons. Something has to always be on to sense when you press one of those little membrane panel buttons, for example.

C1 isn't just the stray capacitance between wires in the walls. The large appliances in particular probably have hefty line filters to reduce their conducted emissions to just below the legal limit. There could easily be some deliberate capacitance across the power input for that purpose alone.

The symptoms you found can be explained by the right relative values of the various components in the above circuit. To prove this, consider the extreme case where R1 and R2 are completely open:


With an infinite impedance voltmeter, you'd measure 120 V L1-G, 120 V L2-G, and 0 V L2-L1. Your actual case is somewhere between the top circuit and this last one. Hopefully you can see that with the various ratios between, R1, R2, R3, and C1, you can construct lots of different scenarios, including the one you actually had.

  • The L1/G voltage difference was 120V (good).
  • The G/L2 voltage difference was 90V (bad).
  • The L1/L2 voltage difference was 30V (very bad!).

Note that your numbers add up nicely if you assume the L2 connection from the street is completely open. In other words, all the L2 voltage you see is coming from L1. The 30 V L1-L2 is because phantom loads and leakage to ground on L2 pull down on what is getting dumped onto L2 from L1. That leakage is causing a 30 V drop, hence L2 is at 120 V - 30 V = 90 V. When you measure L2 to L1, you get that 30 V drop directly.

In the context of the second schematic:

  • R1 = ∞
  • R2 = 3 * R3
  • C1 = 0

This is all quite plausible, and doesn't require phase shifts to explain.


This simple scenario would exactly cause your measurements:


R2 is shown outside the house because it was originally drawn to model the leakage of the exposed cable to the ground. However, R2 is really the combination of all loads to ground on L2. The most substantial part of R2 may actually be in the house, in the form of phantom loads of 120 V appliances connected between L2 and G.


Yes. The residual loads inside the house explain nicely the effect. So, it should suffice to disconnect every electrical device in the house and to measure the L1/L2 voltage again to prove that scenario. Curiously, the only difference with my scenario is how you interpret the R3 resistor (R2 in my schematic). ‭coquelicot‭ about 2 months ago

@coq: I misunderstood your R2. I thought you were trying to model the leakage due to the broken insulation. ‭Olin Lathrop‭ about 2 months ago

Thank you, @Olin. That explanation finally helped me understand the measurements! ‭Joel Lathrop‭ about 2 months ago


Well, here is was my logic tells me:

The corrosion has caused a high impedance in the L2 wire, but there should be also a "short" between the L1 and L2 wire. So that the voltage between L1 and L2 falls to 30V, and the voltage between G and L1 is 90V. In other words:

L1 - G = 120V = L1 - L2 + L2 - G = 30 + 90 = 120.

In the shematic below, I've noted +120V AC and -120V AC just to say that the voltage is in phase opposition.


One way to check that would be to scope the voltages together at L1 and L2 to observe that L1 and L2 are no more in phase opposition, but close to be in phase concordance. Also, it would be interesting to check the corrosion impedance of the L2 wire. In the schematic, I've given imaginary values, the point being that there is a ratio of 1:7 between the resistances:

30V = 240V / 8 = 240V / (1 + 7).


I think this is a good start, +1, but I disagree that the leakage is to L1, not ground. I think there is more to it than you show. ‭Olin Lathrop‭ about 2 months ago

A small bird told me that you have an answer :-) ‭coquelicot‭ about 2 months ago

The problem with leakage to ground is that apparently you'll always have a voltage between L1 and L2 not smaller than between L2 and G, that is 90V. So, I give up and wait for the answer of the real pro. ‭coquelicot‭ about 2 months ago

@coq: I'm an opamp kind of EE, not electrical power engineering. I do have an explanation that I think makes sense, but I was hoping one of the electrical power folks would jump in. I just got to a good break point in my day job work, so maybe I'll post my idea soon. ‭Olin Lathrop‭ about 2 months ago

Thank you for trying to help me understand the readings, @coq. ‭Joel Lathrop‭ about 2 months ago


When the outage occurred, I observed the following

  • The L1/G voltage difference was 120V (good).
  • The G/L2 voltage difference was 90V (bad).
  • The L1/L2 voltage difference was 30V (very bad!).

If the L2 wire was broken then, due to capacitive coupling between L1 and L2 (and the use of a high impedance voltmeter), the voltage on L2 will appear to be close in value to the L1 voltage. Hence you see a 30 volt difference between L1 and L2 and, L2 is circa 90 volts above ground.

Concusion: L2 is broken and naturally picks up a voltage due to capacitive coupling. I've seen this before.


Something here is missing to explain the L1/L2 voltage (say, a leakage from L2 to G). See the second schematic of Olin. The only difference is that you interpret his cap C1 in another way. But the logic is the same. Perhaps to complete this scenario, add another capacitive coupling to ground, so that L2 is in-between L1 and G. ‭coquelicot‭ about 2 months ago

Thank you for trying to help me understand the readings, @Andy. I had not thought of capacitive coupling potentially playing a factor. ‭Joel Lathrop‭ about 2 months ago

@coquelicot there is capacitive coupling between all the wires in the bundle that make up the cable. L2 is not ideally split halfway of course because that would make the L2 voltage become 60 volts. ‭Andy aka‭ about 2 months ago

@Andy. Yes, I'm sure this was evident for you, and this is probably hidden inside your terms "picks up". For cranks like me, it takes some time to understand because the pickup voltage is usually measured with a probe referenced to ground (not to the pickup source). So another cap coupling to ground is necessary for understanding how to close the circuit here. N.B As you pointed out, my term "in-between" above was inappropriate. I meant of course "somewhere between". ‭coquelicot‭ about 2 months ago

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