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enter image description here

In the image :-

R is resistance in the circuit
r is the internal resistance of the battery
E is the EMF of the battery
V is the voltage across the resistor R
Arrows show the direction of current.

If I is the current flowing in the circuit then by Ohm's law I have
I = V÷R and I= (−E)÷r. Now, V/R = (-E/r)
therefore, V= (-E)×R÷r

The problem I'm getting is if r < R , then R/r > 1 which implies V>E.
How is it possible that the voltage across a resistor can be more than the emf of the battery? I must be mistaking somewhere but I'm unable to find it.

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closed as off-topic by Buzz, Jon Custer, John Rennie, ZeroTheHero, David Z Feb 9 at 8:43

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Your $\mathbf E$ is not the emf of the cell.
Assuming that $r$ is the internal resistance of the cell then $\mathbf E$ is the potential difference across the terminals of the cell.

Let $\mathcal E$ be the emf of the cell and $E$ the potential difference across the terminals of the cell.

enter image description here

Starting at the cell and going anticlockwise using Kirchhoff's voltage rule

$\mathcal E - IR - Ir = 0$

$\mathcal E - Ir = E$ is the potential difference across the cell terminals.

$ IR = V$ is the potential difference across the resistor R.

So substitution these two potential differences into the first equation one gets $E-V = 0 \Rightarrow E = V$ as expected.

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  • $\begingroup$ And one more doubt is : when we go anti-clockwise we would encounter positive terminal of R therefore IR should be positive and by the same argument Ir should be negative and so also EMF. $\endgroup$ – adesh mishra Feb 8 at 15:03
  • $\begingroup$ Voltage drop would be positive and voltage rise would be negative. Then the Ir and the EMF should be negative. Correct me if I’m wrong. $\endgroup$ – adesh mishra Feb 8 at 16:43
  • $\begingroup$ @adeshmishra I am sorry but I got the signs wrong across resistor $r$. I have now made the correction. $\endgroup$ – Farcher Feb 8 at 16:59
  • $\begingroup$ @adeshmishra With the correction made one can reason as follows: Start at the minus terminal of the cell and finish up at the plus terminal there is an increase of potential $\mathcal E$. Then for the resistor $R$ the direction of the current indicates a drop in potential from right to left so it is $-IR$ and the same for the other resistor $-Ir$. Another way of doing it is to say that you enter the minus terminal of the cell so it is $-\mathcal E$, enter at the plus sign for the two resistors so it is $+IR$ and $+Ir$. In both cases the voltages add up to zero and you get equivalent equations $\endgroup$ – Farcher Feb 8 at 17:05
  • $\begingroup$ Your answer has really helped me so much. Thank you. $\endgroup$ – adesh mishra Feb 8 at 17:19

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