Question

a cell of emf 2 V is connected to an external resistance of 5 ohms when pd recorded at its ends is 1.8V.calculate internal resistance of cell

Answer 1

Let ‘r’ be the internal resistance of the cell. The internal and external resistance are in series. So, equivalent resistance is, R = r + 5

Thus, the current through the circuit is, I = 2/(r + 5)

And, the voltage drop across external resistance is = I × 5 = 10/(r + 5)

Which is given to be 1.8 V

So,

10/(r + 5) = 1.8

=> r = 0.56 Ω