Potentiometer

Setting up the potential gradient. A potentiometer is a long, uniform resistance wire (often several metres, wound in sections on a board) connected in series with a driver cell of EMF $\varepsilon_d$ and a rheostat — this is the primary circuit. It drives a steady current $I$ through the wire, and because the wire's resistance per unit length is uniform, the potential falls at a constant rate along its length. This constant rate is the potential gradient, $k$:

where: $V$ is the total potential difference across the wire, $L$ is its total length, and $k$ is in volts per metre. Because the drop is uniform, the p.d. between the wire's starting end A and any point a distance $l$ along it is simply $V_l = kl$ — directly proportional to $l$.

Balancing an unknown EMF. To measure an unknown EMF or p.d., that source is connected — through a galvanometer, forming the secondary circuit — between end A and a sliding jockey on the wire, with its positive terminal on the same side as the driver cell's positive terminal. The jockey is slid until the galvanometer shows zero deflection: the balance point, at length $l$. Here the wire's potential drop from A to the jockey exactly cancels the unknown EMF, so no current flows in the secondary circuit:

where: $\varepsilon$ is the EMF (or p.d.) being measured and $l$ is the balance length. Because zero current is drawn from the unknown source at the exact balance point, this is the true EMF, not a terminal voltage reduced by an internal-resistance drop — the reason a potentiometer beats a voltmeter for this job.

Comparing two EMFs. Keeping the primary circuit (and hence $k$) unchanged, a two-way key swaps between two cells of EMF $\varepsilon_1$ and $\varepsilon_2$, giving separate balance lengths $l_1$ and $l_2$:

The ratio of EMFs equals the ratio of balance lengths, so $k$ itself never needs to be known explicitly.

The goal of this derivation is to turn two measured balance lengths — one with the cell open-circuited, one with a known resistance box drawing current through it — into a formula for the cell's internal resistance, using only Ohm's law and the balance condition $\varepsilon = kl$ already established above; no new physics is introduced, only algebra connecting the open-circuit and loaded readings.

Step 1. Balance the cell (EMF $\varepsilon$, internal resistance $r$) on open circuit — nothing else is connected to it, so no current flows through it and its terminal voltage equals its full EMF. The balance length is $l_1$, giving $\varepsilon = kl_1$.

Step 2. Now connect a known resistance box, set to $R$, directly across the same cell. Current $I$ flows through the loop formed by the cell and $R$, and the cell's terminal voltage drops to $V = \varepsilon - Ir$, which is smaller than $\varepsilon$ because of the internal drop $Ir$. Re-balance the potentiometer against this reduced terminal voltage; the new (shorter) balance length is $l_2$, giving $V = kl_2$.

Step 3. Take the ratio of the two balance conditions, which eliminates $k$ entirely: $\dfrac{\varepsilon}{V} = \dfrac{kl_1}{kl_2} = \dfrac{l_1}{l_2}$.

Step 4. Express the same ratio using circuit equations. Since $V = IR$ and $\varepsilon = I(R+r)$, dividing gives $\dfrac{\varepsilon}{V} = \dfrac{R+r}{R}$.

Step 5. Equate the two expressions for $\varepsilon/V$ from Steps 3 and 4, then solve for $r$:

where: $r$ is the internal resistance, $R$ is the known resistance-box value, $l_1$ is the open-circuit balance length and $l_2$ is the balance length with $R$ connected. Because $l_2 < l_1$ whenever $r > 0$, the formula always yields a positive resistance for a real cell.

A student first balances a cell (EMF $\varepsilon$, internal resistance $r$) on open circuit at $l_1 = 75\ \text{cm}$ against a potential gradient set by the primary circuit. She then connects a resistance box set to $R = 5\ \Omega$ across the cell and finds the new balance length is $l_2 = 60\ \text{cm}$. Find the internal resistance of the cell, and also find the potential gradient $k$ if the driver-circuit current gives the cell an EMF reading of $\varepsilon = 0.5\ \text{V}$ at balance.

Step 1. Apply the internal-resistance formula from the derivation: $r = R\,\dfrac{l_1-l_2}{l_2} = 5 \times \dfrac{75-60}{60} = 5 \times \dfrac{15}{60}$.

Step 2. Simplify: $5 \times 0.25 = 1.25\ \Omega$.

Step 3. For the potential gradient, use the open-circuit balance condition $\varepsilon = kl_1$ with $\varepsilon = 0.5\ \text{V}$ and $l_1 = 0.75\ \text{m}$ (converting $75\ \text{cm}$ to metres): $k = \dfrac{\varepsilon}{l_1} = \dfrac{0.5}{0.75} = 0.4\ \text{V/m}$. This matches the gradient used to balance a $1.25\ \text{m}$ length against a $0.5\ \text{V}$ cell in the standard EMF-balance case ($\varepsilon = kl = 0.4 \times 1.25 = 0.5\ \text{V}$), showing the same wire and driver setup can serve both applications.

The cell's internal resistance is $\mathbf{1.25\ \Omega}$, with a potential gradient of $\mathbf{0.4\ \text{V/m}}$. Notice the balance length shrank from $75\ \text{cm}$ to $60\ \text{cm}$ once current was drawn through $R$ — exactly what is expected since the terminal voltage $V = \varepsilon - Ir$ must be smaller than the open-circuit EMF.

• Confusing the potentiometer with a voltmeter. A voltmeter always draws some current and slightly loads the circuit; a potentiometer at the exact balance point draws zero current from the source being measured — this null-method property is its entire advantage.
• Letting the potential gradient $k$ change between readings. If the driver-circuit rheostat or battery is disturbed while taking successive balance lengths, $k$ is no longer constant and the ratio methods (EMF comparison, internal resistance) both fail.
• Getting cell polarity wrong. The positive terminal of the cell being balanced must connect to the same end of the wire as the positive terminal of the driver cell, or no null point exists anywhere on the wire.
• Applying $\varepsilon = kl$ when current is still flowing in the secondary circuit. This relation is valid only exactly at the true null (zero-deflection) balance point, not at an approximate reading.

Ask the tutor: Try a potentiometer numerical with your own balance-length data and ask the tutor to check whether your ratio setup and polarity are correct before you compute the final answer.