Potentiometer — how to compare EMFs of two cells

Question

Explain how a potentiometer is used to compare the EMFs of two cells. In an experiment, the null point for cell 1 is obtained at 65 cm and for cell 2 at 52 cm. Find the ratio of their EMFs.

(NEET 2022, similar pattern)

Solution — Step by Step

A potentiometer is a device that measures potential difference without drawing any current from the source being measured. It uses a uniform resistance wire (length $L$ ) connected to a driver cell that maintains a steady current.

The potential drop per unit length along the wire is constant:

k = \frac{V}{L} \text{ (volts per cm)}

where $V$ is the total potential drop across the wire.

The cell whose EMF we want to measure is connected across a section of the wire via a galvanometer. We slide the jockey along the wire until the galvanometer shows zero deflection — this is the null point.

At the null point, the EMF of the cell exactly equals the potential drop across that length of wire:

\varepsilon = k \times l

where $l$ is the balancing length.

For cell 1: $\varepsilon_1 = k \times l_1$

For cell 2: $\varepsilon_2 = k \times l_2$

Since $k$ is the same (we don’t change the driver cell or current between measurements):

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

$l_1 = 65$ cm, $l_2 = 52$ cm

\frac{\varepsilon_1}{\varepsilon_2} = \frac{65}{52} = \frac{5}{4}

\boxed{\varepsilon_1 : \varepsilon_2 = 5 : 4}

Why This Works

The potentiometer works because at the null point, no current flows through the cell being tested. This means we’re measuring the true EMF (not terminal voltage, which would be less due to internal resistance). This is the potentiometer’s biggest advantage over a voltmeter — a voltmeter draws current and gives a reading slightly less than the actual EMF.

The uniform wire ensures that resistance (and hence potential drop) is directly proportional to length. So comparing lengths is the same as comparing potential drops.

Alternative Method — Direct calculation if k is known

If the driver cell is 2 V across a 100 cm wire:

$k = 2/100 = 0.02$ V/cm

$\varepsilon_1 = 0.02 \times 65 = 1.30$ V

$\varepsilon_2 = 0.02 \times 52 = 1.04$ V

Ratio: $1.30/1.04 = 5/4$ (same answer).

For NEET, remember: potentiometer questions almost always reduce to ratios of balancing lengths. You rarely need to calculate $k$ explicitly. If the question asks “compare EMFs,” just write $\varepsilon_1/\varepsilon_2 = l_1/l_2$ and substitute. This saves precious exam time.

Common Mistake

A very common error: students forget the condition that the driver cell EMF must be greater than the EMF of the cell being measured. If the driver cell’s potential drop across the full wire is less than the unknown EMF, you’ll never find a null point — the jockey will slide to the end without the galvanometer reaching zero. NEET questions sometimes test this: “Why is no null point obtained?” Answer: the driver cell EMF is too low.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct calculation if k is known

Common Mistake

Try These Next