A potentiometer balances at 1.200 ohm of slide wire for an unknown cell, and at 1.247 ohm for a standard cell of 1.600 V. What is the unknown emf?

1. At balance no current passes through the galvanometer, so none passes through the cell being tested. 2. With no current, the cell delivers its full emf with no internal drop subtracted. 3. That emf equals the IR drop along the balancing length of wire, so emf equals I times R for each balance. 4. The same steady current I runs along the wire in both balances, so dividing one relation by the other cancels it. 5. This leaves the unknown emf as 1.600 V times the ratio 1.200 ohm divided by 1.247 ohm. 6. The ratio is 0.9623, so the unknown emf is (1.600 V)(0.9623) = 1.54 V. 7. 1.66 V comes from inverting the ratio, the trap when the standard and unknown balances are swapped. 8. 1.30 V and 1.44 V follow from neither arrangement of the two balancing resistances. _Source: OpenStax College Physics (CC BY 4.0), Ch 21 "Circuits and DC Instruments", section 21.5 Null Measurements_