A few storage cells in series are to be charged from a 200 V dc supply. The emf of each cell is 2.5 V, Internal resistance `0.1 Omega` and the charging current is 8A. In this arrangement how many cells can be charged and what extra resistance is required to be conncted in the circuit?

Explanation

Suppose, the maximum no. of cells=x and extra resistance =R.
Here, E=2.5V,r=`0.1 Omega` I=8A
`therefore` By the condition,` Ex+Ir:x+IR=200 `
or, `(2.5 +8 times 0.1)x+8R=200`
or,`3.3x+8R=200`
Now if we divide 200 by 3.3 the quotient is 60 and remainder is 2.
So, the replied number of cells=60 and 2V is the potential difference across the extra resistance R.
i.e., 8R=2 or, `R=2/8=0.25 Omega`