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Internal resistance of a cell using a potentiometer with two shunt resistances

This type of question, asked 4 times: 15 May 2023 S1, 19 Apr 2025 S2, 20 Apr 2025 S1, 12 May 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    15 May 2023 S1

    48
    A cell balances against a length of 150 cm on a potentiometer wire when it is shunted by a resistance of 5 Ω5\,\Omega. But when it is shunted by a resistance of 10 Ω10\,\Omega, then balancing length increases by 25 cm. The balancing length when the cell is in an open circuit is

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  2. Question 2

    19 Apr 2025 S2

    35
    To determine the internal resistance of a cell with potentiometer, when the cell is shunted by a resistance of 5Ω5\Omega, the balancing length is 250 cm. When the cell is shunted by 20Ω20\Omega, the balancing length of potentiometer wire is 400 cm. The internal resistance of the cell is

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  3. Question 3

    20 Apr 2025 S1

    43
    A null point is obtained at 200 cm on potentiometer wire when cell in secondary circuit is shunted by 5Ω5\Omega. When a resistance of 15Ω15\Omega is used for shunting, null point moves to 300 cm . The internal resistance of the cell is

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  4. Question 4

    12 May 2026 S2

    130
    To determine the internal resistance of a cell by using a potentiometer, the null point is obtained at 11m when cell is shunted by 3 Ω3\,\Omega resistance and at 1.51.5m length when cell is shunted by 6 Ω6\,\Omega resistance. The internal resistance of the cell is

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