In a potentiometer, when the cell in the secondary circuit is shunted with a resistance, the balance is obtained at a length of 120 cm of the wire. When the same cell is shunted with a resistance, the balance is shifted to a length of 180 cm. The intern

Question

In the potentiometer, when the cell in the secondary circuit is shunted with $4 Ω$ resistance, the balance is obtained at the length of 120 cm of the wire. When the same cell is shunted with a $12 Ω$ resistance, the balance is shifted to a length of 180 cm. The internal resistance of the cell is ______ $Ω$ . [2026]

Smart Achievers · Accepted Answer

(3)

$Let E is emf and r is internal resistance of cell.$

$\frac{E \cdot 4}{r + 4} = 120 K$

$\frac{E \cdot 12}{r + 12} = 180 K$

$\Rightarrow \frac{1}{3} \cdot \frac{r + 12}{r + 4} = \frac{2}{3}$

$r + 12 = 2 (r + 4)$

$\Rightarrow r = 4$

Solution

1 12

2 3

3 4

4 6

Ans.
(3)

$Let E is emf and r is internal resistance of cell.$

$\frac{E \cdot 4}{r + 4} = 120 K$

$\frac{E \cdot 12}{r + 12} = 180 K$

$\Rightarrow \frac{1}{3} \cdot \frac{r + 12}{r + 4} = \frac{2}{3}$

$r + 12 = 2 (r + 4)$

$\Rightarrow r = 4$

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Solution

1 12

2 3

3 4

4 6

Ans. (3) Let E is emf and r is internal resistance of cell. E·4r+4=120K E·12r+12=180K ⇒13·r+12r+4=23 r+12=2(r+4) ⇒r=4

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Ans.
(3)

$Let E is emf and r is internal resistance of cell.$

$\frac{E \cdot 4}{r + 4} = 120 K$

$\frac{E \cdot 12}{r + 12} = 180 K$

$\Rightarrow \frac{1}{3} \cdot \frac{r + 12}{r + 4} = \frac{2}{3}$

$r + 12 = 2 (r + 4)$

$\Rightarrow r = 4$