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. $\endgroup$ – David White Sep 17 '19 at 15:39

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. $\endgroup$ – Osal Thuduwage Jun 18 '18 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. $\endgroup$ – Perspicacious Sep 27 '19 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.