Question

A cell of EMF 1.1V and internal resistance 0.5 ohm is connected to a wire of resistance 0.5 ohm another cell of the same emf id connected in series but the current remains the same. find the internal resistance of second cell.​

Answer 1

the internal resistance of the emf is 1

Explanation:

firstly find the current in the first circuit using V=iR

V=1.1,R=0.5+0.5=1,then i=1.1.

now in the second circuit another battery of same voltage and unknown resistance is connected in series let the unknown resistance is x .

now the net emf in the circuit is 1.1+1.1=2.2

the ney resistance is x+0.5+0.5=x+1

then solve this 2.2=(1.1)(1+x) then x=1.