JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Current Electricity

In the potentiometer, when the cell in the secondary circuit is shunted with

4Ω

resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell in shunted with

12Ω

resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ______

Ω

Answer: (B)

Solution

Let

E

be the emf and

r

be the internal resistance of the cell. When shunted with resistance

R

, the terminal voltage is

V = \frac{E R}{r + R}

, which is proportional to the balance length.

For the two cases:

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

and

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

Dividing these equations:

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

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

r + 12 = 2 (r + 4)

r + 12 = 2 r + 8

r = 4Ω

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option