Potentiometer has multiple uses. It is based on the principle that when constant current flows through a wire of uniform area of cross-section then potential drop along its length is directly proportional to corresponding length.
i.e. \( V \propto l \) ; \( V = K l \)
Various uses of potentiometer on :
- To measure potential differences between any two points in an electrical network
- To compare emf of a cell
- To measure emf of a cell
- To measure internal resistance of a cell etc:
Answer the following question related to potentiometer.
In the following set up of potentiometer, the value of potential gradient is :
AP is balancing length for the primary cell of emf E, Resistors of AB length of the wire \( P_{AB} = 8\,\Omega \), and length AB (L) = 10 m.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.16 volt/m
The potential gradient is the potential drop per unit length of the potentiometer wire, \( k = I\,R_{AB}/L \), where \( I \) is the current from the driver circuit.
From the figure, the driver cell of 2 V, the resistance \( R = 2\,\Omega \) and the wire of resistance \( 8\,\Omega \) are in series. The driver cell is taken as ideal. The secondary circuit (E, r, X and G) draws no current from the wire at balance, so it does not affect \( k \).
\( \displaystyle I = \frac{2\,\mathrm{V}}{2\,\Omega + 8\,\Omega} = 0.2\,\mathrm{A} \)
\( \displaystyle k = \frac{I\,R_{AB}}{L} = \frac{0.2\,\mathrm{A} \times 8\,\Omega}{10\,\mathrm{m}} = 0.16\,\mathrm{V\,m^{-1}} \)
So the potential gradient is 0.16 volt/m.
Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.
NCERT (Pre-rationalised edition, 2022-23):
Potentiometer – “… Since the wire is uniform, the potential difference between A and any point at a distance l from A is
\( \displaystyle \varepsilon(l) = \phi\, l \qquad (3.89) \)
where φ is the potential drop per unit length.”