In the potentiometer, when the cell in the secondary circuit is shunted with 4 Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell i $s ___$ Ω.
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Question

In the potentiometer, when the cell in the secondary circuit is shunted with 4 Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell i $s ___$ Ω.

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G1 Physics · Accepted Answer

Let E be the emf of the cell, r its internal resistance and K the potential gradient of the potentiometer wire. When the cell is shunted by 4 Ω, its terminal potential difference is V₁ = E×4/(r+4). At balance, V₁ = 120K. Therefore, 4E/(r+4) = 120K. (1) When the shunt is 12 Ω, V₂ = E×12/(r+12). At balance, V₂ = 180K. Thus, 12E/(r+12) = 180K. (2) Divide equation (2) by equation (1): [12/(r+12)]/[4/(r+4)] = 180/120 = 3/2. Therefore, 3(r+4)/(r+12) = 3/2. Hence, 2(r+4) = r+12. Thus, r = 4 Ω. Correct…

In the potentiometer, when the cell in the secondary circuit is shunted with 4 Ω resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with 12 Ω resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell i s ___ Ω.