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Supposed I have 3 resistor, R1=1 Ohm, R2=2 Ohm, R3=3 Ohm. I connect them with copper wire to form a ring R1-R2-R3-R1
I used a device to generate changing magnetic flux in that ring. So there must be induced current flow in the ring.
The problem is if I measure voltage between two ends of R1, what would that voltage be ? Since it is closed circuit then Ohm law indicate that I *R1 = I * (R2+R3)

EDIT: As suggested in comments, this question duplicate Can two voltmeters connected to the same terminals show different values? Circuit with induced EMF

qand
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    I*(R1 + R2 + R3) = dΦ/dt ,try this – user215805 Oct 28 '20 at 08:11
  • The equivalent circuit would consist of the three resistors and a voltage source for the induced voltage in series. Therefore your last statement would be incorrect. – Lars Hankeln Oct 28 '20 at 08:30
  • So by using I*(R1 + R2 + R3) = dΦ/dt, I can calculate current. But then how to calculate voltage between 2 ends of R1 ? – qand Oct 28 '20 at 08:38
  • @user215805 Doesn't that mean (R2+R3)*I is the voltage between 2 ends of R2 R3 ? So if I measured the voltage, it would be I*R1 then or I*(R2+R3) ? – qand Oct 28 '20 at 08:49
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    Yeah ,so its like we get different potential difference at same two points if we change our path and this is characteristic of non conservative field (i.e induce as an electric Field ) – user215805 Oct 28 '20 at 09:02
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    [Probably some help](https://electronics.stackexchange.com/questions/506590/can-two-voltmeters-connected-to-the-same-terminals-show-different-values-circui/507058#507058). Voltage is induced; not current and, how you take the measurement affects the reading. It's pretty much covered in the link. – Andy aka Oct 28 '20 at 09:12

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