Question

 Draw a circuit diagram to determine internal resistance of a cell in the laboratory?

Answer 1

A cell of emf E (internal resistance r) is connected across a resistance box (R) through key K₂.

When K₂ is opened, the balance length is obtained at length AN₁ = l₁.E= Φ l₁       ....(1)

When K₂ is closed, the balance length is obtained at AN₂ = l₂.

 Let V be the terminal potential difference of the cell.

 Then V = Φ l₂           .....(2)

From equations (1) and (2), we get:

\[\frac{E}{V} = \frac{l_1}{l_2}\]      ....(3)

E= I(r +R) and V = IR

\[\frac{E}{V} = \frac{r + R}{R}\]     ......(4)

From (3) and (4), we get:

`(R+r)/R = I_1/I_2`

Therefore, we have `E/V= I_1/I_2 `

∴ `r = R (E/V - 1)`

∴ `r = R (I_1/I_2 - 1)`