This is College Physics Answers with Shaun Dychko. This potentiometer is adjusted such that with an unknown EMF, <i>Ex</i>, we have a resistance <i>Rx</i> that we do know because we read off of our potentiometer here. And it’s multiplied by some current that we don’t know and don’t care about, because then we replace this unknown EMF with a standard EMF which we do know. And that’s going to equal the same current because we haven’t change this EMF up here. The same current multiplied by new resistance <i>Rs</i> which we get by adjusting this potentiometer. And we can divide these two on equations and get the unknown EMF divided by the standard EMF equals <i>I</i> times <i>Rx</i> over <i>I</i> times <i>Rs</i>. And the <i>I</i> is cancel and so this is the ratio of the resistances. And so we can multiply both sides by the <i>Es</i>, the standard EMF and we get the unknown EMF equals the standard EMF multiplied by the resistance when you have the unknown EMF divided by resistance with the standard EMF. And so this is three Volts times ten Ohms divided 15 Ohms. And this gives 2.00 Volts, must be the unknown EMF. Now, what if the standard EMF was instead 3.100 Volts, what would the resistance <i>Rx</i> be given a standard EMF of 3.1 Volts assuming that <i>Rs</i> bounces at the same resistance that I did before? And so we figured out what the <i>Ex</i> is the unknown EMF. Or at least it was unknown before. Now it’s two Volts, so we can substitute that in. We’re rearranging this formula here in blue and solving for <i>Rx</i> by multiplying both sides by <i>Rs</i>. So we multiply by <i>Rs</i>, multiply by <i>Rs</i> and we get <i>Rx</i> equals <i>Rs</i> times <i>EMFx</i> divided by EMF standard. So it’s 15 Ohms times two Volts divided by 3.1 Volts giving 9.68 Ohms.