A cell of emf `1.1 V` and internal resistance `0.5 Omega` is connected to a wire of resistance `0.5 Omega`. Another cell of the same emf is connected in series bur the current in the wire remain the same .Find the internal resistance of second cell

Answer

`1 Omega`

Explanation

`I = (epsilon)/(R + r) = (1.1)/(0.5 + 0.5) = 1.1A`
`I = (1.1+1.1)/(0.5 + 0.5+ r) = (2.2)/(1 + r)`
`:. 1.1 =(2.2)/(1 + r) or 1 + r = 2 or r = 1 Omega`