Q.
A galvanometer connected with an unknown resistor and two identical cells in series each of emf 2V shows a current of 1A. If the cells are connected in parallel, it shows 0.8A. Then the internal resistance of the cell is
In series, the current i=R+nrnE 1=R+2r2×2 R+2r=4...(i)
Again, in parallel the current i=R+nrE 0.8=R+2r2 0.8=2R+r2×2
From Eq. (i) 0.8=2(4−2r)+r4 0.8=8−3r4 8−3r=5 r=1Ω