Question

Calculate the $\textrm{emf}_\textrm{x}$ of a dry cell for which a
potentiometer is balanced when $R_\textrm{x} = 1.200\textrm{ }\Omega$ , while an alkaline standard cell with an emf of 1.600 V requires $R_\textrm{s} = 1.247\textrm{ }\Omega$ to balance the potentiometer.

Final Answer

$1.540\textrm{ V}$

### Solution video

# OpenStax College Physics for AP® Courses, Chapter 21, Problem 58 (Problems & Exercises)

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Video Transcript

This is College Physics Answers with Shaun Dychko. This is a circuit diagram for potentiometer I have drawn it slightly differently than the way it's drawn here but it's the same thing; we have this resistance

*R x*which is varied such that the current through this galvanometer or the current through this branch containing the unknown*emf**ε x*is zero. And when that happens, we know that the potential is only that of this unknown*emf*; there's no drop across any internal resistance in it or anything because there's no current through it at all and so it's a direct measurement of the*emf*and since this and this are in parallel, the potentials are the same and so if only we knew what this current*I*was then we can multiply that by this known*R x*and then figure out what the potential is across this branch and then that therefore is the same as the potential across the branch with the*ε x*. So we figure out what that current is by replacing this*emf*with a known*emf*a standard, letter*s*I think stands for 'standard' some known*emf*and then adjust the resistance such that the current is zero through this standard*emf*and then this*R s*for standard is known and then we can figure out this current*I*from that because the current across this branch will be the same as the current across this known*emf*, which we know is*ε s*and that equals the current times*R s*and then we can solve for*I*by dividing both sides by*R s*and then this current is going to be substituted in here to figure out what*emf x*is and we do that here. So the unknown*emf*is the standard*emf*divided by the standard resistance multiplied by the resistance when the unknown*emf*is in that circuit and this works out to 1.600 volts divided by 1.247 ohms multiplied by 1.200 ohms and that is 1.540 volts is the unknown*emf*.