A cell of emf 1 . 8 volts gives a current of 17 A when directly connected to an ammeter of resistance 0 . 06 Ω . Internal resistance of the cell is

Step-by-step solution

Let the Internal resistance of the cell is r ,   the ammeter connected in series showing a reading of 17   A is the current flowing in the circuit having resistance, 0 . 06   Ω Therefore, E = I r + I R ⇒ r = E - I R I = 1 . 8 - 17 × 0 . 06 17 = 0 . 0458   Ω ⇒ r = 0 . 0458   Ω = 0 . 046   Ω