For determination of current we use a coil of standard resistance. To measure a cell's emf a potentiometer is used since in a potentiometer measurement no current is flowing. Find the emf of a cell which balances against a length of 180 cm of the potentiometer wire. Potentiometer. Were the solution steps not detailed enough? A cell is characterised by its emf ε. ← Prev Question Next Question → The emf of a cell may be defined as the terminal voltage of the cell when not under load, that is, delivering no current. If reading of voltmeter is V then error is V – V’ which can be positive, negative or zero. POTENTIOMETER OBJECT: To measure the Emf of an unknown cell using a potentiometer. Difference Between Potentiometer & Voltmeter The potentiometer and the voltmeter both are the voltage measuring device.The significant difference between the two is that the potentiometer measures the emf of the circuit whereas voltmeter measures the end terminal voltage of the circuit. The error $I-I^{\prime}$can be found and calibration curve is obtained by plotting ammeter reading  on x axis and error on y axis. All high potential terminals of primary and secondary circuit should be connected at same point of Potentiometer. The current flowing through the primary circuit is 2A and resistance per unit length is. When we use a voltmeter across a cell, a small current flow through the voltmeter and we are getting only the terminal potential difference of the cell The negative terminals of these two cells are connected to two terminals 1 and 2 of a two way key. APPARATUS: Slide wire potentiometer, battery or power supply, rheostat, double pole-double throw (DPDT) switch, standard cell, unknown cell, galvanometer, key switch and a multimeter. © 2007-2021 Transweb Global Inc. All rights reserved. Download India's Leading JEE | NEET | Class 9,10 Exam preparation app, Uses of Potentiometer | Comparison of emf of two cells using Potentiometer || Class 12, JEE & NEET, If unknown potential difference V is balanced for length $\ell$ than $V=x \ell=\left(\frac{E_{0}}{\ell_{0}}\right) l$, If the length of potentiometer wire is changed from L to L’ then the new balancing length is $\ell^{\prime}=\left(\frac{L^{\prime}}{L}\right) \ell$ If length is increased L’ > L so $l^{\prime}>l$ and if length is decreased L’ < L so \$l^{\prime}