Q.On a potentiometer, a cell of unknown internal resistance $r$ gives a balance point at $300\ \text{cm}$ when its circuit is open. When a resistance of $5\ \Omega$ is connected across the cell, the balance point shifts to $250\ \text{cm}$. Find the internal resistance of the cell.

The potentiometer measures emf by a null-deflection (zero-current) method, so it draws no current from the cell and gives the true emf, not just terminal voltage.

A potentiometer is used to measure the emf of a cell. Its working principle is based on comparing the emf of the unknown cell against a known potential gradient along a uniform wire, using a balance-point (null-deflection) method: at the balancing length, no current is drawn from the cell being measured, so inte …

The 10 mV source balances against the potential across 40 cm of wire. With driving current I = 2/(R+10), the drop across 40 cm is I·10·(40/100) = 4I = 0·01 V, giving R = 790 Ω. Option (c).

Step 1 — driving-loop current (emf 2 V, negligible internal resistance, wire 10 Ω in series with R):

I = 2/(R + 10).

Step 2 — potential drop across the full 100 cm wire = I × 10 V; across 40 cm it is a fraction 40/100:

V₄₀ = I × 10 × (40/100) = 4I.

Step 3 — at balance this equals the unknown emf, 10 mV = 0·01 V: …

A potentiometer mainly measures potential difference (and EMF).

A potentiometer works on the principle that the potential drop across a length of uniform wire carrying a steady current is proportional to that length. By finding the balance (null) point it compares an unknown potential difference or EMF against a known one, drawing no current from the source at balance. Thus …

A potentiometer measures potential difference without drawing any current at the balance point, making it more accurate than a voltmeter, which always draws some current.

A voltmeter, however high its resistance, is still a finite-resistance device connected across the points being measured, so it always draws a small current and hence slightly changes the very potential difference it is trying to measure — introducing an error, particularly noticeable when measuring the emf of a cell with internal resistance. A potentiometer, by contrast, works on a null-deflection (balance) principle: at the balance point, no current is drawn from the source be …

A potentiometer measures emf accurately because it works on a null method drawing no current from the cell.

A voltmeter always draws a small current, so it reads terminal voltage (emf minus the internal-resistance drop), not the true emf. A potentiometer is balanced until no current flows through the cell being tested; at balance the cell delivers zero current, its internal resistance causes …

Potential drop per unit length along the potentiometer wire.

The potential gradient of a potentiometer wire is defined as the fall in potential per unit length of the wire:

$K = \frac{V}{L}$ …

The device described is the potentiometer.

A potentiometer works on the principle that the potential drop along a uniform wire is proportional to its length ($V \propto l$) when a steady current flows. Because it draws no current from the cell at the balance point, it acts as an ideal voltmeter and can:

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A potentiometer measures emf accurately because at the balance point no current is drawn from the cell being tested, ensuring the terminal voltage equals the true emf. The answer is (B).

Potential gradient K = V/L falls as L rises (V fixed), and the balance length l = emf/K, so l grows.

For a potentiometer, the potential gradient is $K = \dfrac{V}{L}$, where $V$ is the (constant) potential difference maintained across the wire and $L$ its length. The balancing length for an emf $\varepsilon$ is $l = \varepsilon/K$. If $L$ increases while $V$ stays fixed, $K$ decrease …

A potentiometer measures e.m.f. exactly because, at balance, it draws no current from the cell — a voltmeter always draws some current and so under-reads.

Why a voltmeter can't measure true e.m.f. exactly: A voltmeter, however high its resistance, always draws a small but non-zero current $i$ from the cell being measured. This current causes a potential drop $ir$ across the cell's internal resistance $r$, so the voltmeter actually reads the terminal potential difference $V = \varepsilon - ir$, which is slightly less than the true e.m.f. $\varepsilon$.

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A potentiometer measures potential difference/emf without drawing current at balance, so it acts as an ideal voltmeter.

An ordinary voltmeter has finite resistance and always draws a small current from the circuit it is measuring, so it slightly reduces the reading (it cannot measure true emf).

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