PotentiometerJEE Main

Measure EMF without drawing current — interactive Physics simulation for IIT-JEE.

Concept

A potentiometer compares EMFs by sliding a jockey along a uniform current-carrying wire until the galvanometer reads zero. At balance the unknown cell drives no current, so you measure its true EMF — not the terminal voltage a voltmeter would sag it to. The wire acts as a smooth voltage divider with gradient $k$ volts per cm.

Key formula

\varepsilon = k\,l_{balance}, \qquad k = \frac{V_{wire}}{L} = \frac{\varepsilon_{driver}R_{wire}}{(R_{wire}+R_{series})L}, \qquad \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}

Derivation

Uniform wire ⇒ potential drops linearly along it. The jockey taps off $V(l) = kl$ .

When $kl = \varepsilon_{cell}$ , no current flows through the galvanometer branch (null). Because zero current crosses the cell, its internal resistance drops nothing — the reading is the open-circuit EMF. Internal resistance itself: $r = R(l_1/l_2 - 1)$ using balance lengths with and without a shunt R.

Scenarios to explore

Potentiometer — Measure true EMF with zero current draw — ε = k·l.

Real-world applications

Comparing EMFs of two cells (ratio of balance lengths).
Measuring internal resistance of a cell.
Calibrating voltmeters & ammeters (they draw current; this doesn't).

JEE exam tips

Sensitivity ↑ when gradient k ↓ (longer balance length) — add series resistance to the driver.
ε₁/ε₂ = l₁/l₂ needs the SAME gradient — don't touch the rheostat between readings.
Internal resistance: r = R(l₁ − l₂)/l₂ with shunt R across the cell.

Common mistakes

Balance length beyond the wire — driver voltage too low; increase driver EMF or reduce series R.
Confusing potentiometer (null, no current) with voltmeter (draws current).
Forgetting the driver EMF must EXCEED the unknown EMF.

Exam traps to avoid

If the unknown cell's polarity is reversed, no balance point exists anywhere on the wire.
A potentiometer is 'ideal voltmeter of infinite resistance' ONLY at the balance point.