JEE Main 2026 Jan 28 Shift 1, Physics Q3: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

JEE Main2026Jan 28, Shift 1Physics

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 in shunted with 12 $Ω$ resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ____ $Ω$ .

A
3
B
4
C
12
D
6

Solution

Let $E$ be the EMF and $r$ the internal resistance. When the cell is shunted with external $R$ , the terminal voltage across $R$ is $V = \frac{E R}{R + r}$ , and this is balanced at length $ℓ$ , so $V = K ℓ$ with $K$ being the potential gradient.

For $R = 4 Ω$ , $ℓ = 120$ cm: $\frac{4 E}{r + 4} = 120 K$ .

For $R = 12 Ω$ , $ℓ = 180$ cm: $\frac{12 E}{r + 12} = 180 K$ .

Divide the first equation by the second to eliminate $E$ and $K$ :

$\frac{4 ( r + 12 )}{12 ( r + 4 )} = \frac{120}{180} = \frac{2}{3}$

$\frac{r + 12}{3 ( r + 4 )} = \frac{2}{3} \Rightarrow r + 12 = 2 (r + 4) \Rightarrow r = 4 Ω$ .

Option (2).

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JEE Main 2026 Jan 28 Shift 1, Physics Q3: Wheatstone Bridge, Ammeter, Voltmeter And Potentiometer

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