# Why is EMF equal to PD when circuit is “open”?

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.

• 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 '14 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 '14 at 3:03
• @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 '14 at 8:05

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 $R$).