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.
The length of potentiometer wire AB is 10 m and its resistance is \( 8\,\Omega \). If balancing length on the potentiometer is 225 cm, the value of emf of the primary cell is :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.36 volt
At balance the emf of the primary cell equals the potential drop across the balancing length: \( E = k\,l \), where \( k \) is the potential gradient of the wire.
From the figure, the driver cell \( V = 2\,\mathrm{V} \) and \( R = 2\,\Omega \) are in series with the wire of resistance \( 8\,\Omega \). The driver cell is taken as ideal.
\( \displaystyle I = \frac{V}{R + R_{AB}} = \frac{2\,\mathrm{V}}{2\,\Omega + 8\,\Omega} = 0.2\,\mathrm{A} \)
\( \displaystyle k = \frac{I\,R_{AB}}{L} = \frac{0.2\,\mathrm{A} \times 8\,\Omega}{10\,\mathrm{m}} = 0.16\,\mathrm{V\,m^{-1}} \)
\( l = 225\,\mathrm{cm} = 2.25\,\mathrm{m} \)
\( \displaystyle E = k\,l = 0.16\,\mathrm{V\,m^{-1}} \times 2.25\,\mathrm{m} = 0.36\,\mathrm{V} \)
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.”