Question:
In the given potentiometer circuit arrangement, the balancing length AC is measured to be \(250\ \mathrm{cm}\). When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes \(400\ \mathrm{cm}\). The ratio of the emf of two cells, \(\dfrac{\varepsilon_1}{\varepsilon_2}\) is :
In the given potentiometer circuit arrangement, the balancing length AC is measured to be \(250\ \mathrm{cm}\). When the galvanometer connection is shifted from point (1) to point (2) in the given diagram, the balancing length becomes \(400\ \mathrm{cm}\). The ratio of the emf of two cells, \(\dfrac{\varepsilon_1}{\varepsilon_2}\) is :
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With the galvanometer at (1) the potentiometer balances only epsilon_1; at (2) it balances the series combination epsilon_1 + epsilon_2. Balancing length is proportional to emf.
Updated On: Nov 4, 2026
- \(\frac{5}{3}\)
- \(\frac{8}{5}\)
- \(\frac{4}{3}\)
- \(\frac{3}{2}\)
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The Correct Option is A
Solution and Explanation
Potential gradient is constant, so emf \(\propto\) balancing length.
Galvanometer at (1): only cell \(\varepsilon_1\) is balanced: \(\varepsilon_1 = k(250)\).
Galvanometer at (2): the two cells in series are balanced: \(\varepsilon_1 + \varepsilon_2 = k(400)\).
Therefore \(\varepsilon_2 = k(400-250) = k(150)\).
\(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{250}{150} = \dfrac{5}{3}\).
Answer: (1) \(\frac{5}{3}\)
Galvanometer at (1): only cell \(\varepsilon_1\) is balanced: \(\varepsilon_1 = k(250)\).
Galvanometer at (2): the two cells in series are balanced: \(\varepsilon_1 + \varepsilon_2 = k(400)\).
Therefore \(\varepsilon_2 = k(400-250) = k(150)\).
\(\dfrac{\varepsilon_1}{\varepsilon_2} = \dfrac{250}{150} = \dfrac{5}{3}\).
Answer: (1) \(\frac{5}{3}\)
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