Why potentiometer be preferred over a voltmeter for measurement of emf of a cell.?

See below

We know that the terminal voltage of a cell is the potential difference between its electrodes.

A voltmeter should not be used to measure the emf of a cell as a voltmeter draws some current from the cell during measurement. When voltmeter is connected across the cell it amounts to closed circuit measurements and there is inner potential drop due to internal resistance of the cell. Therefore, the measurement of cell's emf will not be accurate

To measure a cell's emf, use of a potentiometer is preferred since in potentiometer-measurement no current flows through the cell. It is an open circuit measurements. The emf of cell will be measured accurately.

Other advantages of using a potentiometer are

• Sensitivity of potentiometer is high as compared to that of voltmeter.
• The resistance of potentiometer becomes infinite at the time of measurement. Whereas in case of voltmeter the resistance is very high but still a measurable number.

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Look at the figure below

A potentiometer is a circuit that is used to measure EMF of a cell. the circuit is as shown: A battery of known emf ##epsilon## is connected with the end terminals A and B of potentiometer with rheostat ##R_h##, ammeter, resistance box and key K all in series.

The ends of a resistance ##R_1## are connected to terminal A and jockey J through galvanometer. The resistance ##R_1## is connected to cell ##E_1## and key ##K_1## as shown.

Step 1. Close key K. Adjust the resistance ##R## from resistance box so that the potential difference across the potentiometer wire is greater than the EMF of the cell, which is to be measured.

Suppose ##R_P## is the resistance of the potentiometer wire of length ##=L## (100cm for a lab model Meter potentiometer and Length of the potentiometer wire varies from 5 to 10m). Current through the potentiometer ##I=epsilon/(R+R_p)## Potential drop across potentiometer wire##=epsilon/(R+R_p)xxR_p## .....(1)

Step 2. Close key ##K_1##. Current flows through resistance ##R_1## due to which potential difference is developed across this resistance.

Step 3. Adjust the position of jockey on potentiometer wire such that there is no deflection seen on the galvanometer. Indicates a balanced circuit and that the potentiometer is another null- measurement device.

Let this position of jockey be is J.

Let length AJ of potentiometer wire be ##=l##. Let ##K## be the potential gradient of potentiometer wire. From (1)

##K=epsilon/(R+R_p)xxR_p/L## .........(2)

With Galvanometer showing no deflection, EMF of the cell is equal to the potential difference ##V## across ##R_1##. Now ##V=Kxxl## ......(3). Inserting value of ##K## from (2)

##V=epsilonxxR_p/(R+R_p)xxl/L##