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With reference to this question I asked the other day: How can I measure a voltage between two conductive surfaces that are completely isolated?

There was a good answer there so I didn't want to continue in the comments but It didn't solve the main issue I had which is the 120V potential between the isolated case and the battery terminal. To quickly re-iterate, the battery pack is 350V and there are multiple layers of insulation between the pack and the case.

The multimeter used to measure the voltage was a Fluke 87V. This meter has a 10Meg input resistance. I've drawn the below diagram to illustrate this. I don't believe it's accurate because this is not a pure voltage source, more a measured voltage likely due to insulation leakage. Here, Rx is the unknown impedance between the terminal and the case. R1 is the meter input impedance. In this scenario, I measure 120V.

schematic

simulate this circuit – Schematic created using CircuitLab

Now, if I connect a 10k resistor across these points to see if there's any power behind this voltage, I now get the following two circuits when I measure voltage & current.

schematic

simulate this circuit

The voltage read is now 0.5V and when putting the meter in series with the 10k resistance, I read 50µA. My guess to what's happening is, the high input impedance of the multimeter is comparable to the insulation impedance between the case and the terminal, in that case, the meter's impedance has a large effect on the circuit, causing this voltage reading. When I connect the 10k Resistance, the voltage collapses by reducing the equivalent parallel resistance.

So my question is, what is actually happening here and is the 120V voltage I'm reading dangerous?

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    \$\begingroup\$ You already have a question on this exact problem at How can I measure a voltage between two conductive surfaces that are completely isolated? please do not post another, as that divides the awareness of the issue and what has already been sorted out by the community there. You sho uld not have so hastily accepted an answer which did not really address the issue; undo that. Instead of re-posting, edit your existing question to clarify the real issue. \$\endgroup\$ Oct 22 '20 at 0:02
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    \$\begingroup\$ @ChrisStratton But the answer did address the issue as I presented it. I contemplated continuing through comments with what i've posted here but didn't think that was good practice. Is it acceptable practice here to add an additional question to an already posted question even if my original question was answered? \$\endgroup\$
    – dos584
    Oct 22 '20 at 0:59
  • \$\begingroup\$ The "answer" there is little more than a statement of a basic fact; it's not really useful \$\endgroup\$ Oct 22 '20 at 1:01
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    \$\begingroup\$ @ChrisStratton Agreed. But that's just an unfortunate result of my bad question. He answered what I asked. I should have included the information from this post in the original post and asked a better question at the beginning. So should I have re-written my original question, even though the question was answered? \$\endgroup\$
    – dos584
    Oct 22 '20 at 1:25
  • \$\begingroup\$ What voltage reading do you get from the positive battery terminal to the case? \$\endgroup\$ Oct 22 '20 at 4:57
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So my question is, what is actually happening here and is the 120V voltage I'm reading dangerous?

Assuming a single leakage point inside the battery, we can calculate the voltage and leakage resistance at that point using simultaneous equations. The equivalent circuit is this:-

schematic

simulate this circuit – Schematic created using CircuitLab

For the voltage reading, Vbatt1 = (Rleak * 12 μA) + 120 V

For the current reading, Vbatt1 = (Rleak * 50 μA) + 0.5 V

Combining these equations we get Rleak = 3.145 MΩ and Vbatt1 = 157.7 V.

If the battery is made up of 96 cells each charged to ~3.95 V, these numbers could be caused by a high resistance contact to the case between cell numbers 40 and 41 (from the negative end).

At a maximum current of only 50 μA this isn't dangerous now, but it could be if the insulation breaks down further. As well as being a shock hazard it could unbalance the battery, or even become a dead short which might start a fire or cause the battery to explode!

Note that the same readings could be caused by multiple leakage points, and/or bulk leakage across surfaces or inside the insulation, particularly if a cell has been leaking electrolyte. If you can't find a single contact point then you should thoroughly inspect the battery internals.

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