Obtain an expression for equivalent resistance of two resistors connected in parallel.

Explanation

Consider two cells of emf `epsi _(1) and epsi_(2)` and internal resistances `r _(1) and r _(2)` connected in parallel as shoen in fig(a). Let ` I _(1) and I _(2)` be the curents through the cells. At `B_(1)` currents `I _(1) and I _(2)` flow in wheres the current I flows out.
`therefore I = I_(1) + I_(2)" "...(1)`
Let `V_(1) and V _(2)` be the potentials at `B _(1) and B _(2)` respectively.
Pd between first cell, `V - V _(1) = V _(2) = epsi _(1) - I _(1) r _(1)`
`implies I _(1) = (epsi _(1) -V)/(r _(1)) " "..(2)`
Pd between second cell, `V = V_(1) -V_(2) =epsi _(2) - I _(1) r _(2)`
`implies I _(2) = (epsi _(2) -V)/(r _(2))" "...(3)`
`therefore` Eq1uation (1) becomes
`I = ( epsi _(1)- V )/( r _(1)) + (epsi _(2 ) - V)/( r _(2))`
`1= (epsi _(1))/( r _(1)) - (V)/( r _(1))+ (epsi _(2))/( r _(2))- ( V )/( r _(2))`
`I = ((epsi _(1))/( r _(1)) + ( epsi _(2))/( r _(2))) - ((V)/(r _(1)) + ( V)/( r _(2)))`
` I = ((epsi _(1))/( r _(1)) + (epsi _(2 ))/( r _(2 ))) - V ((1)/( r _(1)) + (1)/( r _(2)))`
`I = ((epsi_(1) r _(2) + epsi _(2) r _(1 ))/( r _(1 ) r _(2)))- V (( r _(1) + r _(2))/( r _(1) r _(2)))`
`V ((r _(1) + r _(2))/( r _(1 ) r _(2))) = (( epsi_(1) r _(2) + epsi_(2) r _(1))/( r _(1) r _(2)))-1`
`V = ((e _(1)r _(1) + epsi _(2 ) r _(1))/( r _(1) r _(2)))((r _(1) r _(2))/( r _(1) + r _(2))) - I ((r_(1) r _(2))/( r _(1) + r _(2)))`
`V = (( epsi _(1) epsi _(2) + epsi _(2)r _(1))/( r _(1) r _(2))) -I ((r _(1) r _(2))/( r _(1)+ r _(2)))`
Let the parallel combination of two cells be replaced by a single cell between `B _(1) and B _(2).` Let `epsi _(eq)` be its emf and `r _(eq)` be its internal resistance. Then
`V = epsi _(eq) - I r _(eq) " "...(5)`
From equation (4) and (5), we have
`epsi _(eq)= ( epsi _(1) r _(2) + epsi _(2) r _(1))/( r _(1) + r _(2)) " "...(6)`
`r _(eq) = (r _(1) r _(2))/( r _(1) + r _(2))" "...(7)`
Equation (6) and (7) can also be written as
`(epsi _(eq))/( r _(eq))=(epsi _(1))/(r _(1)) + ( epsi _(2))/( r _(2))" "...(8)`
`(1)/( r _(eq)) = (1)/( r _(1)) + (1)/( r _(2))" "...(9)`