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I understand that the set up of a battery consists of very good, but not ideal conductor, and therefore, some internal resistance exists. Also, I get that emf is the PD that would exist if the internal resistance is zero. What I don't get, however is why EMF and PD are equal in case of an open circuit (circuit with infinite resistance). I'm sure I'm missing something here. Help is appreciated.

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It's not really correct. Better to say "PD is equal to EMF", because EMF is constant most of the time. Intuitively, the charge can't go anywhere, so you have a constant separation of charge. Quantitatively - $V=\mathcal{E}-I r$. In the case of open circuit $I$ is zero, thus $V=\mathcal{E}$. –  cth May 19 at 15:02
    
I don't understand the first part of cthulu's "constant separation of charge" argument, but the last sentence really summarizes the solution. –  BMS May 20 at 3:03
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@BMS - What I mean is that there is a separation of charge between the terminals of the battery which causes the PD even when the circuit is open. –  cth May 20 at 8:05

2 Answers 2

The electromotive force of a battery is equal to the potential difference between its terminals in an "open circuit", when there is no current being drawn. The potential difference between the terminals generally drops when the current is being drawn.

let me know if you need more explanation.

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As you stated, we can think of a real battery as an ideal one with an internal resistance $R_i$. This battery is then connected to an external circuit with resistance $R$. Those 2 resistors form a voltage divider. If the EMF has a value of $V$ then the voltage measured across the external resistance is $V*R/(R+R_i)$.

This voltage is equal to the EMF of the battery when $R_i = 0$ (because $R/R=1$), and also when $R=\infty$ (because in that case $R_i$ can be ignored in comparison to $R$).

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