Update: Thanks Spehro Pefhany for finding the issue!

I've got a formula deduced which should - in theory - suggest that a battery with no external load connected should have an infinite external voltage, however in practice this obviously isn't the case!
Steps are listed below to get to the conclusion, any ideas where something's amiss?

Consider the following series resistors:

The ratio between \$R_1\$ and \$R_2\$ is the same as \$V_1\$ and \$V_2\$, with \$V_1\$ and \$V_2\$ being the voltage across resistors \$R_1\$ and \$R_2\$, as shown below:

$$ \begin{aligned} &R_1 : R_2\\\\ =\ &\frac{V_1}{I} : \frac{V_2}{I}\\\\ =\ &V_1 : V_2 \end{aligned} $$ Therefore, in terms of the two series resistors in the diagram, this equation is valid: $$ \frac{V_1}{V_2} = \frac{R_1}{R_2} $$ Now, consider the following schematic of a battery with internal resistance \$R_{\text{int}}\$ (excuse the multiple cells), and external resistor \$R_{\text{ext}}\$:

Resistors \$R_{\text{int}}\$ and \$R_{\text{ext}}\$ and their subsequent voltages can be substituted from the \$\frac{V_1}{V_2} = \frac{R_1}{R_2}\$ formula, into: $$ \begin{aligned} &\frac{V_\text{ext}}{V_\text{int}} = \frac{R_\text{ext}}{R_\text{int}}\\\\ \therefore\ &V_\text{ext} = \frac{R_\text{ext}}{R_\text{int}} \times V_\text{int} \end{aligned} $$ We've ended up with a formula to calculate voltage across \$R_{\text{ext}}\$, which is the same as the external voltage of the battery. However, what if \$R_{\text{ext}}\$ were to burn out or be disconnected, but voltage across it (the battery) is still measurable?
Essentially \$R_{\text{ext}}\$ now has a resistance of \$\infty\$, as no current is flowing through, meaning the battery has no external load attached.

Therefore, working out the limit of \$V_{\text{ext}}\$ as \$R_{\text{ext}}\$ approaches \$\infty\$ : $$ \begin{aligned} V_{\text{ext}} &= \lim_{R_{\text{ext}}\to\infty} \left(\frac{R_\text{ext}}{R_\text{int}} \times V_\text{int}\right)\\\\ &= \frac{\infty}{R_\text{int}} \times V_\text{int}\\\\ &= \infty \text{V} \end{aligned} $$ This is an unusual result, and obviously doesn't work in real life. The math seems alright, but something has to be amiss. Any suggestions?

  • \$\begingroup\$ Are you taking the limit or simply setting R_ext equal to infinity? That's pretty different... \$\endgroup\$
    – Samuel
    Commented Aug 20, 2015 at 18:54
  • 1
    \$\begingroup\$ As you defined, the Vext is not V through the battery, but the V through the Rext. \$\endgroup\$ Commented Aug 20, 2015 at 18:56

1 Answer 1


Your formula is wrong.

The correct voltage across the external resistor is

$$ \begin{aligned} &\frac{V_\text{ext}}{V_\text{int}} = \frac{R_\text{ext}}{R_\text{int}+R_\text{ext}}\\\\ \end{aligned} $$

So the voltage Vint approaches Vext when Rext approaches infinity.

  • \$\begingroup\$ That simple.. I've started to look for division by zero somewhere. \$\endgroup\$
    – Eugene Sh.
    Commented Aug 20, 2015 at 19:00
  • \$\begingroup\$ Thanks Sphero! Can you just explain how the correct formula was deducted? \$\endgroup\$
    – baharini
    Commented Aug 20, 2015 at 19:02
  • \$\begingroup\$ You called Vext as the voltage through the Rext. It isn't the V through the battery. At the first moment (loaded battery) the V through the battery is Vint + Vext (as you called). At the second moment there is no more Vext, so the voltage through battery will be Vint. \$\endgroup\$ Commented Aug 20, 2015 at 19:03
  • \$\begingroup\$ Makes sense. Explanation much appreciated, missed that ;) \$\endgroup\$
    – baharini
    Commented Aug 20, 2015 at 19:08

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