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

`0.175 A, 0.175A, 3.5V`

Explanation

Here `epsilon = 2V, r= 1.5 Omega`
`n = 2 , m = 2`
`I = (mn epsilon )/(mR + nr) = (2 xx 2xx 2)/(2 xx 10 + 2 xx 15) = (8)/(23) = 0.35 A`

As two branches are identical current in each branch is `I_(1) = I_(2) = I//2 = 0.35//2 = 0.175 A`
potential difference across `R`
`IR = 0.35 xx 10 = 3.5 V`