In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of $2 \Omega$ the balancing length becomes 120 cm. The internal resistance of the cell is

JEE Main · 2005 · 08 May 2005 · Offline · Current Electricity

In a potentiometer experiment the balancing with a cell is at length 240 cm. On shunting the cell with a resistance of

2Ω

the balancing length becomes 120 cm. The internal resistance of the cell is

Answer: (D)

The balancing length is proportional to the potential tapped, so the open-cell emf corresponds to

240

cm and the terminal voltage across the

2 Ω

shunt to

120

cm, giving

\frac{E}{V} = \frac{240}{120} = 2

. For a shunted cell

\frac{E}{V} = 1 + \frac{r}{R}

, so

2 = 1 + \frac{r}{2}

, which gives the internal resistance

r = 2 Ω

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option