In a potentiometer, a cell is balanced at 240 cm. On shunting the cell with \( 2\,\Omega \) resistance, the balancing comes at 120 cm. What is the internal resistance of the cell ?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2\,\Omega \)
At balance a potentiometer draws no current from the cell, so the balancing length is proportional to the voltage across the cell. Without the shunt this is the emf \( E \); with the shunt \( R \) a current flows and the balance gives the terminal voltage \( V \).
\( E = k\,l_1 \)
\( V = k\,l_2 \)
\( \displaystyle V = \frac{E\,R}{R + r} \;\Rightarrow\; \frac{l_1}{l_2} = \frac{E}{V} = \frac{R + r}{R} \)
\( \displaystyle r = R\left(\frac{l_1}{l_2} - 1\right) \)
\( \displaystyle r = 2\left(\frac{240}{120} - 1\right) = 2 \times 1 = 2\,\Omega \)
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 – “… We can also use a potentiometer to measure internal resistance of a cell [Fig. 3.28 (b)]. … From Eq. [3.94(a)] and [3.94(b)] we have
\( \displaystyle \frac{R + r}{R} = \frac{l_1}{l_2}, \qquad r = R\left(\frac{l_1}{l_2} - 1\right) \qquad (3.95) \)
Using Eq. (3.95) we can find the internal resistance of a given cell.”