Consider a cell having terminal voltage $V'$ and EMF, $V$ having internal resistance, $r$ and current, $I$ flows through the circuit in which an external resistance, $R$ is connected.

Potential drop across internal resistance= Ir

$$V= V' + Ir = IR + Ir$$


Now if the cell is in an open circuit then $I=0$ and hence,

But we know that the Emf of the cell is equal to the Terminal Voltage of the cell in an open circuit, i.e.,

Why such a contradiction in my concept?

I know how to derive $V=V'$. I don't need that. I want to know why using this concept makes Emf, $V= 0$ rather becoming $V=V'$.

  • $\begingroup$ $V=I(R+r)$ applies to a closed circuit, where $I$ cannot be zero. $\endgroup$ Commented Sep 17, 2019 at 15:39

3 Answers 3


Note that: The electromotive force (emf) is the potential difference of a source when no current is flowing. EMF has nothing to do with whether the terminals are connected to form a circuit or not. EMF is an inherent characteristic of a cell due to certain chemical reactions necessary for the driving of electrons to give rise to EMF. EMF is the voltage generated by a cell or by the magnetic force according to Faraday's Law.

So here

$$V=EMF$$ $$V' = EMF- Ir$$

  • $\begingroup$ This answer is relevant to why the EMF is equal to the potential difference in a opened circuit. But, it doesn't pointed that why @Mrithun Jay's deviated equation is contradicted with the normal situation. $\endgroup$ Commented Jun 18, 2018 at 2:54
  • $\begingroup$ @Osal, i think it's not the equation which is deviated but the conclusion i took from the equation V = V' is wrong. EMF is the reason why we have terminal voltage across the source. As correctly pointed out by Yashas, V' = EMF - Ir. So, when the source is not connected to any circuit, V' = V and not the other way around, i.e., V = V'. Ofcourse, mathematically they are same but if consider the source then, V' = EMF or V but if consider a circuit element like resistor, R then V' the terminal voltage which happen to be appeared across R becomes 0 due to no current across it. $\endgroup$ Commented Sep 27, 2019 at 15:48

You noted that I is 0 for open circuit but didn't consider the resistance which is really high. So going further. V=I(r+R) For R>>r V=IR The current is 0 when reistance is infinite. However they are limits so lets consider I slightly greater than 0 and V slightly lesser than infinity. Simply, I here is given by I=V/R which brings us back to V=(V/R)*R i.e V=V not zero


in an open circuit, there is no flow of electricity. ergo, no voltage. But capacity is the same


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