How can e.m.f. of two cells be compared using potentiometer ?

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Q. How can e.m.f. of two cells be compared using potentiometer ?

Puducherry TnboardTamil Nadu HSC (DGE) Board 2017Subjective· 5mImportance★★★★★

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Concept understanding — Comparison of EMF

Comparison of EMF — The Intuition

Imagine you have two batteries, but you don't know their voltages. You want to find out which is stronger, and by how much. The obvious way is to connect a voltmeter across each one. But a voltmeter draws a small current, and that tiny current causes a voltage drop inside the battery itself (due to its internal resistance). So the reading you get is not the true EMF — it's the terminal voltage, which is slightly less.

If you want the true EMF — the voltage the battery would produce if no current flowed through it — you need a method that doesn't draw any current at all. That's exactly what a potentiometer does.

The Core Idea

A potentiometer is a long uniform wire with a known potential difference across its ends. By sliding a contact along the wire, you can "tap off" any fraction of that total potential difference. When you connect a cell to this tapped voltage, you adjust the sliding contact until the galvanometer shows zero deflection — meaning no current flows through the cell. At that exact point, the potential difference across the tapped length of wire exactly balances the cell's EMF.

Because no current flows through the cell at balance, the internal resistance of the cell has no effect. You are measuring the true EMF.

Comparing Two Cells

Now suppose you have two cells, with EMFs $\varepsilon_1$ and $\varepsilon_2$ . You do the balancing experiment for each one, using the same potentiometer wire with the same total voltage across it.

For the first cell, the balancing length is $l_1$ . That means the potential drop per unit length of the wire (call it $k$ ) times $l_1$ equals $\varepsilon_1$ :

\varepsilon_1 = k \, l_1

For the second cell, the balancing length is $l_2$ :

\varepsilon_2 = k \, l_2

Since $k$ is the same in both cases (same wire, same total voltage), you can divide the two equations:

\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}

\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}

That's the entire principle. You don't need to know $k$ , you don't need to know the total voltage across the wire — you only need the two balancing lengths.

Why This Works — The Precision

The key condition is that the potentiometer wire must have a uniform cross-section and uniform resistivity, so that the potential drop per unit length is constant. If the wire is not uniform, the ratio of lengths won't equal the ratio of EMFs.

Also, the total voltage across the wire must remain constant during both measurements. If the driving cell (the one supplying the potentiometer) fluctuates, $k$ changes and the comparison becomes unreliable. …

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Since the potential drop along a potentiometer wire is directly proportional to length, balancing each cell in turn (with a galvanometer showing zero deflection) against a length of the wire gives the ratio of the two emfs as the ratio of their balancing lengths.

Principle of the potentiometer

A potentiometer is a long, uniform resistance wire (usually 4–10 m, laid out on a metre-scale board) through which a steady current $I$ is maintained by a driver circuit — a battery (with emf greater than either of the two cells to be compared), a rheostat, and a key, connected across the full length of the wire. Because the wire is uniform, this steady current produces a uniform potential gradient (potential drop per unit length) along it,

$k=\frac{V}{L}\;(\text{volts per metre})$

where $V$ is the potential difference across the whole wire of length $L$ .

Comparing the emfs of two cells

Let the two cells to be compared have emfs $E_1$ and $E_2$ . Using a two-way key, connect the positive terminal of each cell (in turn) to the starting (zero) end of the potentiometer wire, with its negative terminal joined through a galvanometer to a sliding contact (jockey) that can touch the wire at any point.

For the first cell, the jockey is slid along the wire until the galvanometer shows no deflection; let this balancing length (from the starting end) be $l_1$ . At this null point, the potential drop across the length $l_1$ of the wire exactly equals the emf of the cell (since no current is then drawn from the cell, so there is no potential drop within the cell itself):

$E_1=k\,l_1$

Without changing the driver-circuit current, the key is switched to connect the second cell, and the new balancing length $l_2$ is found similarly:

$E_2=k\,l_2$

…

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Q.How can e.m.f. of two cells be compared using potentiometer ?

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