Explain 'Mixed Connection' of cells and derive an expression for its equivalent emf and current.

Explanation

n. cells of emf `epsilon_(1) , epsilon_(2) ....epsilon_(n) and r_(1) , r_(2) ....r_(n)` internal resistance are connected in a series. .m. rows of above series are joined in parallel to form a mixed connection as shown in figure.

`rArr` Now total emf of each series is,
`epsilon = sum_(i=l)^(n) epsilon_(i)`
`rArr` If total (equivalent) internal resistance of mixed connection is r then
`(1)/(r) = (1)/(r.) + (1)/(r.) + (1)/(r.) ` + ..... m times = `(m)/(r.) = (m)/(sum r_(i)) `
`therefore r = (sum r_(i))/(m)`
`rArr` All rows are parallel to each other, So total emf `epsilon = sum epsilon_(1)`
`therefore` Current I =` (" Total emf" )/("Total resistance " )`
`I = (sum epsilon_(i))/(R + (sum r_(i))/(m) ) `
where R is the series resistance given in the circuit.