Name: Conversion of a Galvanometer Into a Voltmeter
Uploaded: 2019-07-19
Duration: 7 min 18 s
Description: This video helps us understand the working of a galvanometer. It also teaches us how to convert it into a voltmeter of the desired range both theoretically a...

Question

Describe how a potentiometer is used to compare the emfs of two cells by combination method.

Accepted Answer

A battery of stable emf $E$ is used to set up a potential gradient, along the potentiometer wire.

The potential difference across the length, $= V$

Length of wire $= L$

The positive terminal of the cell $1$ is connected to the higher potential terminal $A$ of the potentiometer; the negative terminal is connected to the galvanometer $G$ through the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The cell $2$ is connected across the remaining two opposite terminals of the reversing key. The other terminal of the galvanometer is connected to a pencil jockey. The emf $E_{1}$ should be greater than the emf $E_{2}$ this can be adjusted by trial and error.

The EMFs of two cells can be compared using the sun and difference method.

When two cells are connected such that the positive terminal of the first cell is connected to the negative terminal of the second cell, the emf of the two cells are added up and the effective emf of the combination of two cells is $E_{1} + E_{2}$ . This method of connecting two cells is called the sum method.

When two cells are connected such that their negative terminals are together or their positive terminals are connected together, then their emf oppose each other and the effective emf of the combination of two cells is $E_{1} - E_{2}$ (Considering $E_{1} > E_{2}$ This method of connecting two cells is called the difference method.

Circuit for sum and difference method is connected as shown in below figure When keys $K_{1}$ and $K_{3}$ are closed the cells $E_{1}$ and $E_{2}$ are in the sum mode. The null point is obtained using the jockey.

Let $I_{1}$ be the length of the wire between the null point and point $A$ . This corresponds to the emf $(E_{1} + E_{2})$ .
$∴ E_{1} + E_{2} = K I_{1} - - - - (1)$

Where K is the potential gradient along the potentiometer wire.

Now the keys $K_{1}$ and $K_{3}$ are kept open and the keys $K_{2}$ and $K_{4}$ are closed. In this case, the two cells are in the difference mode.

Again the null point is obtained. Let $I_{2}$ be the length of the wire between the null point and point $A$ .

This corresponds to emf $(E_{1} - E_{2})$ .

$∴ E_{1} - E_{2} = K I_{2} - - - - - (2)$

Dividing equation $(1)$ by equation $(2)$ ,

$\frac{E_{1} + E_{2}}{E_{1} - E_{2}} = \frac{I_{1}}{I_{2}}$

By componendo and dividendo method :

$\frac{(E_{1} + E_{2}) + (E_{1} - E_{2})}{(E_{1} + E_{2}) + (E_{1} - E_{2})} = \frac{I_{1} + I_{2}}{I_{1} - I_{2}} \frac{E_{1}}{E_{2}} = \frac{I_{1} + I_{2}}{I_{1} - I_{2}}$

Thus, the emf of two cells can be compared using the sum and difference method.

Describe how a potentiometer is used to compare the emfs of two cells by combination method.

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