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In the figure above,we have two cells $E_1$ and $E_2$ and $7$ resistances arranged as shown with $I_1$ current emitting from $E_1$ and $I_2$ emitting from $E_2$. Is it possible to find the equivalent resistance of such a circuit where there is more than one cell? If so,is my following understanding correct or flawed?

Since $I_1$ current is emitted from $E_1$ battery,this exact current should return to $E_1$. Hence $I_1$ current flows through $R_1,R_2,R_4$ and since the same current flows through them,therefore they are in series. With the same logic, $I_2$ current flows through $R_5,R_6,R_7$ and hence $R_5,R_6,R_7$ are in series. Only $I_3=I_1+I_2$ flows through $R_3$. Therefore,the equivalent resistance of the circuit is $(R_1+R_2+R_4) \parallel R_3 \parallel (R_5+R_6+R_7)$.

Kindly correct me if i am wrong by enlightening me with the proper concepts.

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  • $\begingroup$ Hint: use Kirchhoff's laws to set up 3 equations in 3 unknowns. Solve those equations to determine the current flow through each resistor. $\endgroup$ Oct 24, 2022 at 15:10
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    $\begingroup$ @DavidWhite I don't think the OP is asking about the currents in each resistor, but is asking if the approach gives the equivalent resistance of the circuit. $\endgroup$
    – Bob D
    Oct 24, 2022 at 15:14
  • $\begingroup$ @BobD, no, he's not. However, I have never heard of calculating the equivalent resistance for the type of circuit he has drawn. Is it possible to do so? If so, I would certainly learn something new if someone shows how to do such a calculation. $\endgroup$ Oct 24, 2022 at 18:10
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    $\begingroup$ @DavidWhite That's what he needs to be told, i.e., the "equivalent resistance" he proposes is meaningless. I will attempt to do so. $\endgroup$
    – Bob D
    Oct 24, 2022 at 18:43
  • $\begingroup$ @madness Try redrawing your circuit showing your equivalent resistance and the two cells $\endgroup$
    – Bob D
    Oct 24, 2022 at 18:45

1 Answer 1

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When you are attempting to determine the equivalent resistance of individual resistors you need to be able to replace the original resistors with the equivalent resistance in the circuit without affecting the potentials or potential differences in the circuit. So the question is, where would you locate your equivalent resistance in the circuit without affecting the potentials? The answer is you can't.

For example, refer to the figure below.

The equivalent resistance of the series combination of $R_1$ and $R_2$ is simply $R_{1}+R_{2}$. It can replace the two original resistors between nodes A and C without changing the potential at nodes A and C, or at any of the other nodes in the circuit as well.

But if you consider the equivalent resistance of $R_1$, $R_2$ and $R_4$ to be $R_{1}+R_{2}+R_4$ where would you put it in the circuit? Between nodes A and C? Between nodes G and F?. If you put it between A and C you will have a short (zero potential difference) between G and F. If you put it between G and F you would have a short (zero potential difference) between A and C. The decision will change the potentials and potential differences in the circuit. The same would apply to the right side loop.

For resistors to be in series, it is not sufficient for the current to be the same in all the resistors. All resistors must be connected end-to-end without other circuit components in between.

For resistors to be in parallel, each pair of terminals of the resistors must be connected to the same two nodes. None of the resistors can have a voltage source in series with it between the two nodes. Resistors $R_1$, $R_2$ and $R_4$ cannot be combined in parallel with $R_3$ because not only are the three resistors not in series, but cell $E_1$ is also between nodes C and F. Only if you replace cell $E_1$ with a short circuit can you combine $R_1$, $R_2$ and $R_4$ in series, and that series combination combined in parallel with $R_3$.

Hope this helps.

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  • $\begingroup$ Thank you very much for replying. I am pretty dumb so i could not get it still why $R_1,R_2,R_4$ could not be in series as you mentioned in your third para. If we didn't have the cell $E_2$,then surely $R_1,R_2,R_4$ would have been in series. But it seems that due to the presence of cell $E_2$,that is not possible anymore. Then what is the role of $E_2$ here in changing the combination of $R_1,R_2,R_4$ which would have been series if $E_2$ weren't there? $\endgroup$
    – madness
    Oct 25, 2022 at 6:55
  • $\begingroup$ I assume you mean E1. It prevents R1 and R4 from being connected end to end so that you can replace all three resistors with with a single equivalent resistor somewhere in the circuit. I’ll ask you one more time, where would you put the equivalent resistor? $\endgroup$
    – Bob D
    Oct 25, 2022 at 10:30
  • $\begingroup$ Thanks a lot,i get it now! $\endgroup$
    – madness
    Oct 25, 2022 at 13:32

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