A Potentiometer wire of length \( L \) and a resistance \( r \) are connected in series with a battery of emf \( E_0 \) and a resistance \( r_1 \). An unknown emf E is balanced at a length \( l \) of the potentiometer wire. The emf E will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{E_0 r}{(r+r_1)} \cdot \frac{l}{L} \)
In a potentiometer the unknown emf is balanced by the potential drop across a length \( l \) of the wire, so \( E = k\,l \), where \( k \) is the potential gradient (potential drop per unit length) along the wire.
The wire (resistance \( r \)) and the resistance \( r_1 \) are in series with the driving battery \( E_0 \), taken as ideal.
\( \displaystyle I = \frac{E_0}{r + r_1} \)
\( \displaystyle V_{\mathrm{wire}} = I r = \frac{E_0\,r}{r + r_1} \)
\( \displaystyle k = \frac{V_{\mathrm{wire}}}{L} = \frac{E_0\,r}{(r + r_1)\,L} \)
\( \displaystyle E = k\,l = \frac{E_0\,r}{(r + r_1)} \cdot \frac{l}{L} \)
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 – “… A current I flows through the wire which can be varied by a variable resistance (rheostat, R) in the circuit. 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 \( \phi \) is the potential drop per unit length.”