A voltmeter measures potential difference by drawing a small current — but even a "small" current affects the circuit being measured and means you're reading the terminal voltage, not the true EMF. The Potentiometer solves this problem elegantly: at the balance point, it draws zero current from the cell being tested. This makes it the most accurate instrument for measuring EMF, comparing EMFs, and determining internal resistance. It is a null method — the instrument reading is zero at the measurement point — and this is its fundamental advantage over the voltmeter. In NEET and JEE, potentiometer problems involving comparison of EMFs and determination of internal resistance are standard, direct-mark questions.
1. Principle of the Potentiometer
A long uniform resistance wire AB is connected to a driver cell (battery) of EMF E_d. A steady current I flows through the wire, setting up a uniform potential gradient (voltage per unit length) k:
The potential at any point P at distance l from A: V_P = kl
When the jockey (sliding contact) is at the balance point — a point where the potential of the jockey equals the EMF of the test cell — no current flows through the galvanometer.
2. Application 1 — Comparison of EMFs
Two cells of EMF E₁ and E₂ are connected one at a time to the galvanometer circuit. Balance lengths l₁ and l₂ are found:
Example: E₁ balances at l₁ = 60 cm, E₂ balances at l₂ = 40 cm:
E₁/E₂ = 60/40 = 3:2 = 1.5
Important: The driver cell's EMF must be greater than both E₁ and E₂ for balance to be achievable on the wire.
3. Application 2 — Determination of Internal Resistance
The cell of unknown internal resistance r is connected to a shunt resistance R via a key. Two balance lengths are found:
- l₁: key open (I = 0 through cell) → measures EMF E directly
- l₂: key closed (current flows through R) → measures terminal voltage V = E − Ir
Example: R = 10 Ω, l₁ = 75 cm (open), l₂ = 60 cm (closed):
r = 10 × (75 − 60)/60 = 10 × 15/60 = 2.5 Ω
4. Advantages of Potentiometer over Voltmeter
| Potentiometer | Voltmeter |
|---|
| Null method — zero current drawn at balance | Always draws current (finite resistance) |
| Measures true EMF (no current → V = E) | Measures terminal voltage V = E − Ir (slightly less than EMF) |
| Can compare two EMFs directly | Cannot directly compare two EMFs |
| Higher accuracy (sensitivity depends on wire length) | Limited accuracy (depends on resistance value) |
5. Sensitivity of a Potentiometer
Sensitivity is the ability to detect small differences in EMF. A potentiometer is more sensitive if its potential gradient k is smaller (smaller change in V per cm of wire).
Ways to increase sensitivity:
- Use a longer wire (same voltage across more length → smaller k)
- Reduce the driver cell voltage (lower V across wire → smaller k)
- Use a high-resistance series resistor in the driver circuit (reduces current → reduces k)
Practice Questions
Q1 (NEET/Board): In a potentiometer experiment, the balance point for a standard cell of EMF 1.02 V is found at 68 cm. If an unknown cell balances at 52 cm, find its EMF.
E₁/E₂ = l₁/l₂ → E₂/1.02 = 52/68
E₂ = 1.02 × 52/68 = 0.78 V
Q2 (Board): In a potentiometer circuit, with the key open, balance length = 75 cm. With a shunt resistance of 5 Ω, balance length = 60 cm. Find the internal resistance of the cell.
r = R(l₁ − l₂)/l₂ = 5 × (75 − 60)/60 = 5 × 15/60 = 1.25 Ω
Q3 (MCQ): A potentiometer measures EMF more accurately than a voltmeter because:
A) It has a long wire
B) It uses a null method and draws no current from the cell at balance
C) It uses a more sensitive galvanometer
D) It has lower resistance
Answer: B) It uses a null method and draws no current from the cell at balance. At the balance point, the potential of the jockey equals the EMF of the test cell — no current flows. With no current flowing, there is no voltage drop across the internal resistance, so the reading equals the true EMF.
Q4 (JEE Main): A potentiometer wire of length 100 cm has a potential difference of 10 V across it. A cell of EMF 2 V balances at length l. Find l. If the wire length is increased to 400 cm (same voltage), what is the new balance length?
k₁ = 10/100 = 0.1 V/cm. E = k₁ × l → 2 = 0.1 × l → l = 20 cm
New k₂ = 10/400 = 0.025 V/cm. New l = E/k₂ = 2/0.025 = 80 cm
Sensitivity improves (longer wire → smaller gradient → more precise measurement).
