A set of `4` cell each of emf `2V` and internal resistance `1.5 Omega` are connected across an external load of `10 Omega` with 2 rows, two cells in eacg row. Calculate the current in each row and potential difference across `10 Omega`

Answer

`0.245A, 0.245 A , 4.9 V`

Explanation

Here, `epsilon = 2V, r = 1.5 Omega, m = 2n = 3,R= 10`
`I= (mn epsilon)/(mR + nr) = (2 xx 3xx 2)/(2 xx 10 + 3 xx 15) = (12)/(24.5) = 0.49 A`
As two branches of cells are identical , the current through each branch is
`I_(1) = I_(2) = I//2 = 0.49//2 = 0.245 A`

Potential difference across `R = IR = 0.49 xx 10 = 4.9 V`