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enter image description here

I was trying to apply Kirchhoff's rules to find out the reading that will be shown by voltmeter. But then, I realized that the current coming out of the voltmeter arm will be zero (assuming infinite resistance) and am stuck. What should be the right approach in this case?

My (failed)Approach: enter image description here

Simulation: I tried simulating it. Looks like the circuit is valid and Voltmeter would read 14V. This is the same result we get when we solve it through other methods. But KVL seems to not apply here or not giving this result. enter image description here

Please note: this is not a homework question. I was genuinely curious with this special case (note the current flowing from + to - in the 12V). I saw questions around KVL in both Physics and EE stack sites. If this is more suitable in EE, could someone help in moving it please? I don’t have enough reps to do that myself.

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    $\begingroup$ Are the two resistances those of separate resistors, i.e., not internal battery resistances? If so, the circuit should be drawn to show that. $\endgroup$
    – Ed V
    Aug 12, 2022 at 20:18
  • $\begingroup$ hi @EdV: Those are internal battery resistances. $\endgroup$
    – Janaaaa
    Aug 12, 2022 at 20:46
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    $\begingroup$ Thanks! So the answer is the same as if the resistors were external. $\endgroup$
    – Ed V
    Aug 12, 2022 at 21:57
  • $\begingroup$ Hi Janaaaa. I agree with John's answer, but if you're determined to use Kirchoff what you could do is replace the voltmeter by a resistance R then use KVL to find the voltage in terms of R. Then take the limit of R ⟶ ∞ to get the voltage shown by an ideal voltmeter. $\endgroup$ Aug 14, 2022 at 4:37

1 Answer 1

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An ideal voltmeter has no effect on the circuit, so

  1. Solve the circuit without the voltmeter connected.
  2. The voltmeter measures the difference between the nodes to which it connects.
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  • $\begingroup$ In that case, does this circuit violate KVL? There is just this one loop and 2 different power supplies. I could solve it using superposition rule, but couldn't still get how KVL applies here. $\endgroup$
    – Janaaaa
    Aug 12, 2022 at 21:17
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    $\begingroup$ @Janaaaa Consider the voltage drop due to the stated resistance (1Ω and 2Ω). $\endgroup$
    – John Doty
    Aug 12, 2022 at 21:28
  • $\begingroup$ I got 2A as the current circulating in the loop. So, there is a 2V drop at 1Ω and 4V drop at 2Ω. When summed with the source, I get 14V and 22V respectively. But since these are in parallel, I was expecting the voltage drop to be the same. I guess, I am still doing it wrong. $\endgroup$
    – Janaaaa
    Aug 13, 2022 at 6:28
  • $\begingroup$ You have a sign mistake. Sum of voltages around the loop should be zero. $\endgroup$
    – John Doty
    Aug 13, 2022 at 12:15
  • $\begingroup$ Sorry to bug again. I have updated the question with a picture of sign convention followed. I couldn’t see the mistake am making here. Could you let me know if sign convention should be any different when we deal with internal resistances please? $\endgroup$
    – Janaaaa
    Aug 13, 2022 at 16:57

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