The Potentiometer
A voltmeter that draws no current.
Lesson at a glanceShow the key points
- Uniform wire, steady current: potential drop ∝ length.
- At balance no current is drawn: V = V₀l/L.
- Calibrate with a standard cell: V = (l/l₀)ℰ.
- ℰ₁/ℰ₂ = l₁/l₂; internal resistance r = R(l − l′)/l′.
An ideal voltmeter would draw no current at all. A potentiometer achieves exactly that by balancing the unknown voltage against a known one until nothing flows.
The stretched-wire potentiometer
A long uniform wire AB (5–10 m, laid in 1 m lengths joined by thick copper strips) is fixed to a board beside a metre scale. A driver circuit, a strong cell with a key and a rheostat, sends a steady current through it, so the potential falls evenly from A to B. A galvanometer is connected to a jockey that can touch the wire at any point P.
To measure the potential difference between points a and b (a higher), join a to A and b, through the galvanometer, to the jockey. Slide the jockey until the galvanometer shows no deflection. At that point P has the same potential as b, so Vab = VAP. Because the wire is uniform, the drop is proportional to length: with V₀ across the whole length L,
Calibration
To know V₀, balance a standard cell of accurately known emf ℰ: if it balances at l₀, then ℰ = V₀l₀/L, so V₀ = Lℰ/l₀ and any later reading is V = (l/l₀)ℰ. (No current flows through the standard cell at balance, so its terminal voltage is its emf.)
On a 10 m wire, a standard cell of 1.018 V balances at 509 cm. An unknown cell balances at 742 cm. Find V₀ and the unknown emf.
Uses
Comparing two emfs
Balance the first cell at l₁, then the second at l₂ (a two-way key swaps them). Since ℰ₁ = V₀l₁/L and ℰ₂ = V₀l₂/L,
Two cells balance at 6.00 m and 4.50 m on the same potentiometer. Find the ratio of their emfs.
Internal resistance of a cell
Connect a known resistance R across the cell through a key K. With K open, no current flows from the cell, so it balances at l with ℰ = V₀l/L. Close K: current ℰ/(R + r) flows and the terminal voltage drops to ℰR/(R + r), balancing at a shorter l′. Dividing the two:
A cell balances at 7.50 m on open circuit and at 6.25 m when shunted by 10 Ω. Find its internal resistance.
- 1Uniform wire, steady current: V ∝ length; at balance no current is drawn.
- 2V = V₀l/L; calibrate with a standard cell.
- 3ℰ₁/ℰ₂ = l₁/l₂; internal resistance r = R(l − l′)/l′.
Check yourself
Keep exploring
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