While studying 'Parallel grouping of batteries with internal resistance' I came across its derivation and found that the equivalent emf to be Σ(E/r)/(Σ(1/r)) This is the final step of derivation

Why didn't we take the equivalent emf as just Σ(E/r) from the first step as we had the opportunity to do so. Why do we have to bring the Σ(1/r) to the numerator?


1 Answer 1


This is because if you check the original circuit used in the derivation, you will find that the external resistance R is in series with the internal resistances r (which are in parallel with each other). Hence if you calculate the net resistance you will find that it comes out to be the one in the denominator. To compare with I =E/R, you need to have to get to net resistance in the denominator so as to get the net Emf in the numerator. To adjust this, they have multiplied and divided by ∑ (1/r) to get the actual expression as the net resistance comes out to be R + 1/ ∑ (1/r). As r is in parallel with one another which gives it's net resistance parallel component as 1/ ∑(1/r). While R is in series with 1/ ∑ (1/r) which gives the total net resistance as R + 1/ ∑ (1/r). Hope my answer satisfies your needs.

  • $\begingroup$ So we have already guessed what the net resistance would be and we are adjusting the final result to match it. $\endgroup$ Nov 29, 2020 at 6:52
  • $\begingroup$ Yes you can say that. I don't think there is any other possible explanation. Please let me know if you find any. $\endgroup$ Nov 29, 2020 at 7:09
  • $\begingroup$ Sure. Thank you $\endgroup$ Nov 29, 2020 at 7:20
  • $\begingroup$ Please mark the answer as helpful if it helped you. Thanks. :-) $\endgroup$ Nov 29, 2020 at 12:49

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