Why is the EMF=terminal voltage in an open circuit?

Question

Now a trivial answer would be that the current is 0.

V= E-Ir (where v is terminal voltage, e is emf, I is current, and r is internal resistance) Since i=0, V=E

But one thing I believe we are neglecting here is that ohms law does not work when either of three quantities is 0 (in this case the current). Simply because I can write V=IR (where R is external resistance), this would mean v=0 which is not the case.

Also there's a way I've figured out to solve this but in that case we assume the external resistance to be infinity in open circuit

V=IR --> I=V/R (where R is very close to infinity)

E=I(R+r)--> (V/R)(R+r)

Since R is very much greater than r we can ignore the value of r, this would imply

E= (V/R)(R) --> E=V

Is this a correct way to tackle this problem or do I have a flaw in my understanding?

Answer 1

I'm not quite sure what you're getting at. Suppose we represent a physical cell by an ideal voltage source of emf $\varepsilon$ in series with a resistor $r$.

If we don't connect the terminals of the physical cell to anything, i.e. we have an open circuit, then as you say there is no current through the cell. By Ohm's law there is no voltage acrosss the resistor $r$ since $v_{r} = ir = 0$, and the total voltage across the phyical cell is then just $\varepsilon$.

If now we connect the physical cell to a load such that there is now a non-zero current, there is a voltage drop of $v_{r} = ir$ across the internal resistance. So traversing the physical cell, the potential difference between the terminals is the sum of the changes of potential, namely $v_{terminal} = -ir + \varepsilon$.

Ohm's law still applies if $v, i$ or $R$ are zero, the only time we need to be careful is if we get a $\frac{0}{0}$ form.

For instance, an ideal wire in a circuit has current flowing through it, but zero resistance. By $v=iR$ the voltage across it is zero. But then if we rearrange Ohm's law into the form $i=\frac{v}{R}$, $i$ is undefined - physically, this because the current could take any value! There is no electric field so the charge carriers simply persist in their state of motion.