JEE Main Chemistry High Weightage Chapters and Topics 2025

Potentiometer - Principle And Applications - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Syllabus

Quick Facts

Determine the internal resistance of a cell is considered one the most difficult concept.
7 Questions around this concept.

Solve by difficulty

MEDIUM

The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5 $\Omega$ and is 1m long. The resistance which must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is

$1985 \Omega$

$1990 \Omega$

$1995 \Omega$

$2000 \Omega$

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Concepts Covered - 0

Potentiometer

Potentiometer

The potentiometer is a device which does not draw current from the given circuit and still measures the potential difference.
It is a device used to measure e.m.f of a given cell and to compare e.m.f's of cells
It is also used to measure the internal resistance of a given cell

Potentiometer consists of wire of length 5 to 10 meters arranged on a wooden block as parallel strips of wires with 1-meter length each and end of wires are joined by thick coppers. The wire has a uniform cross-section and is made up of the same material. A driver circuit that contains a rheostat, key, and a voltage source with internal resistance r. The driver circuit sends a constant current (I) through the wire.

Potential across the wire AB having length L is given as V=IR, Where R is the resistance of the wire AB

Since the driver circuit sends a constant current (I) through the wire So $V \ \ \alpha \ \ R$

Using $R=\frac{\rho L}{A}$ we can say that $R \ \ \alpha \ \ L$ since area and resistivity are constant.

Therefore we get V is proportional to length. I.e $V \propto L$

The secondary circuit contains cell/resistors whose potential is to be measured. Whose one end is connected to a galvanometer and another end of the galvanometer is connected to a jockey which is moved along the wire to obtain a point where there is no current through the galvanometer. So that potential of the secondary circuit is proportional to the length at which there is no current through the galvanometer. This is how the potential of a circuit is measured using the potentiometer.

Comparison of emf of cell

For the above figure

$l_1$ is the balancing length obtained when cell with emf $E_1$ is included in the secondary circuit. That is key is at position 1. $l_2$ is the balancing length obtained when cell with emf $E_2$ is included in the secondary circuit. That is key is at position 2.

So since $E \ \ \alpha \ l$ we get

$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$

With the help of the above ratio, we can compare the emf of these cells.

Determine the internal resistance of a cell

Determine the internal resistance of a cell

Note-The cell in the secondary circuit has emf E and internal resistance r

Here $l_1$ is the balancing length obtained when key K' is open that is we include only the cell in the secondary circuit. So corresponding potentials of wire of balancing length $l_1$ is E. And we know that $E \ \ \alpha \ \ l_1..... (1)$

Similarly $l_2$ is the balancing length obtained when key K' is closed that is both cell and R' is connected in the secondary circuit.So corresponding potentials of wire of balancing length $l_2$ is V.

And we know that $V \ \ \alpha \ \ l_2 .... (2)$

or we can say that $IR' \ \ \alpha \ \ l_2\\ \Rightarrow \frac{E}{r+R'}*R ' \ \ \alpha \ \ l_2 .... (3)$

So taking ratio of equation (1) to equation (2)

we get $\frac{E}{V}=\frac{l_{1}}{l_{2}}$

$\frac{E}{\frac{ER'}{r+R'}}=\frac{l_{1}}{l_{2}} \\ \\ \\ \Rightarrow \frac{r+R'}{R'}=\frac{l_{1}}{l_{2}}\\$

Then the internal resistance is given by

$r=(\frac{l_{1}-l_{2}}{l_{2}})R'$

or $r=(\frac{E}{V}-1)R'$

Comparison of resistances

Comparison of resistances

The balance point is at a length l₁ cm from A when jockey J is plugged in between Y and X, while the balance point is at a length l₂ cm from A when jockey J is plugged in between Y and Z.

Then we get a ratio of resistances as

$\frac{R_{2}}{R_{1}}=\frac{l_{2}-l_{1}}{l_{1}}$