# Understanding the effect of internal resistance on voltage

I am having trouble understanding the effect of internal resistance on voltage. I have a circuit which is a real voltage source modelled as an ideal voltage source with EMF of Vs in series with an internal resistance of resistance Rs, and this real voltage source is attached to a load (Rload) of negligible resistance.

I have encountered the following contradiction: using I=V/R gives I=Vs/(Rs+Rload) and since Rload is negligible, this is equal to I=Vs/Rs. Using V=IR and substituting I=Vs/Rs to find voltage across the internal resistor gives, V=(Vs/Rs)*Rs, which gives V=Vs, ie that the voltage across the internal resistor is equal to the emf provided by the ideal voltage source. I am also familiar with the formula V=Vs-IRs, where V is the voltage across Rload. Substituting I for Vs/Rs, gives V=Vs-(Vs/Rs)Rs which is V=Vs-Vs, V=0. This means that the internal resistance has "used up" all over the volatge before it was even able to leave the real voltage source.

This seems wrong to me. Is it really true that this is the case whenever Rload is negligible? And is it really true that when Rload is negligible it is also that case that EMF of the ideal voltage source (Vs) is equal to the voltage across the internal resistor? If so, is this this only true when Rload is negligible?

Your first paragraph is clear and correct. The answer to each question in your second paragraph is "Yes". The point is that $$R_\text {load}=0$$ means that you have short-circuited the cell. You have connected a wire of zero resistance across the cell terminals. There can be no voltage drop across a zero resistance for any finite current through it. All the voltage drop is across the internal resistance of the cell.
If you like you can look on $$R_\text {load}$$ and $$R_\text s$$ as forming a potential divider, with input voltage $$V_\text s$$. You'll see that $$R_\text s$$ gets the whole of $$V_\text s$$ across it, and $$R_\text {load}$$ gets none.