Null method: no current drawn. Compare EMFs of two cells.

Potentiometer — How It Measures EMF Accurately

Question

A potentiometer wire of length 1 m is connected to a driver cell of EMF 2V. A cell of EMF 1.02V is balanced at a length of 51 cm on the potentiometer wire. Calculate the EMF of another cell that balances at 40 cm.

Also explain: why does a potentiometer give a more accurate reading of EMF than a voltmeter?

Solution — Step by Step

At the balance point, the potential difference across the potentiometer wire exactly equals the EMF of the cell being tested. No current flows through the galvanometer — this is the null condition. This is the whole trick of the potentiometer.

The potential drop per unit length (potential gradient) is uniform along the wire. So EMF is directly proportional to the balancing length:

\frac{E_1}{E_2} = \frac{l_1}{l_2}

This works because the wire has uniform resistance per unit length, so equal lengths give equal potential drops.

We know $E_1 = 1.02\text{ V}$ balances at $l_1 = 51\text{ cm}$ , and the second cell balances at $l_2 = 40\text{ cm}$ .

E_2 = E_1 \times \frac{l_2}{l_1} = 1.02 \times \frac{40}{51}

E_2 = 1.02 \times \frac{40}{51} = \frac{40.8}{51} = \mathbf{0.8\text{ V}}

A voltmeter always draws a small current from the cell. When current flows, there’s a voltage drop across the internal resistance $r$ of the cell, so the voltmeter reads $E - Ir$ , not $E$ itself.

At the null point of a potentiometer, the galvanometer reads zero — meaning zero current is drawn from the cell being tested. So the terminal voltage equals the actual EMF. This is why the potentiometer is called a null method instrument.

Why This Works

The potentiometer works on the principle that potential drop across a uniform wire is proportional to its length. When we connect the driver cell across the full wire, we get a steady potential gradient $k = V/L$ (volts per metre) along the wire.

When the test cell is connected in opposition and we slide the jockey to find the null point, we’re essentially finding the length whose potential drop exactly matches the test cell’s EMF. No net current flows, so the internal resistance of the test cell plays no role whatsoever.

The first cell (1.02V in this problem) is called the standard cell — it’s a reference with precisely known EMF. This is often a Weston cadmium cell in labs. In board problems, whenever you see one cell with a given EMF and balancing length, it’s being used to calibrate the potential gradient.

Alternative Method — Using Potential Gradient

Instead of the ratio method, we can explicitly calculate the potential gradient $k$ first.

From the standard cell: $E_1 = k \times l_1$

k = \frac{E_1}{l_1} = \frac{1.02}{51} = 0.02 \text{ V/cm}

Now for the second cell:

E_2 = k \times l_2 = 0.02 \times 40 = \mathbf{0.8 \text{ V}}

Both methods give the same answer — but in problems that ask you to find the potential gradient separately, this approach is cleaner. JEE Main 2024 Shift 2 had a question where you had to find $k$ explicitly before comparing cells.

Common Mistake

Students mix up which cell is which. The cell connected across the entire wire (with the rheostat) is the driver cell — it creates the potential gradient. The cells being measured are the ones connected to the jockey circuit. Many students accidentally put the standard cell’s EMF where the driver cell’s voltage should go, and then wonder why the numbers don’t work. Draw the circuit before writing any equation.

A second frequent error: using the driver cell’s EMF (2V here) in the ratio formula. The ratio $E_1/E_2 = l_1/l_2$ only applies to the two cells being compared — both measured against the same wire under the same potential gradient. The driver cell just sets up that gradient; it doesn’t enter the comparison directly.

Potential gradient: $k = \frac{V}{L}$ (V/m or V/cm)

EMF comparison: $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$

Internal resistance: $r = R\left(\dfrac{l_1 - l_2}{l_2}\right)$

where $l_1$ = balancing length (open circuit), $l_2$ = balancing length (with external resistance $R$ )

The internal resistance formula is CBSE 2025 pattern — expect it alongside the EMF comparison in the same question. Practise both parts together.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Potential Gradient

Common Mistake

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