What is parallel connections of cells? Derive the equivalent equation of parallel connections of two cells.

Explanation

`rArr` If positive terminal of given cells are connected at one point and negative terminal of cell connected with second point then such connection is called parallel connection of cells.

`rArr` As shown in figure, two cells with emf `epsilon_(1) and epsilon_(2)`
and respectively internal resistance `r_(1) and r_(2)` are connected in parallel between `B_(1) and B_(2)` .
`rArr` Current in cell with emf `epsilon_(1) " is " I_(1) ` and with emf `epsilon_(2)" is " I_(2)` .
` rArr` Total current at `B_(1)` be,
`I = I_(1) + I_(2) " "` .... (1)
`rArr` Let potential at `B_(1) and B_(2)` be `V(B_(1)) and V(B_(2))` respectively.
`rArr` Potential difference between two cell `V(B_(1)) - V(B_(2))`
will be equal ,
`V(B_(1)) - V(B_(2)) - epsilon_(1) - I_(1) r_(1)`
`V(B_(1)) - V(B_(2)) - epsilon_(2) - I_(2) r_(2)`
`V(B_(1))- V(B_(2)) = V `
` V = epsilon_(1) - I_(1) r_(1)`
`I_(1) = (epsilon_(1) - V)/(r_(1)) " "` ....(2)
similarly,
`I_(2) = (epsilon_(2) - V)/(r_(2)) " "`....(3)
`I = I_(1) + I_(2)`
`= (epsilon_(1) - V) /(r_(1)) + (epsilon_(2) - V)/(r_(2))`
`= (epsilon_(1))/(r_(1)) - (V)/(r_(1)) + (epsilon_(2))/(r_(2)) - (V)/(r_(2))`
`therefore I = (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) - V((1)/(r_(1)) + (1)/(2)) `
`V ((r_(1) + r_(2))/(r_(1) r_(2))) = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) ) - 1 `
`therefore V ((r_(1) + r_(2))/(r_(1) r_(2))) = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) ) - 1 `
`therefore V = ((epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) xx (r_(1) r_(2))/(r_(1) + r_(2)) ) - I ((r_(1) r_(2))/(r_(1) + r_(2)) )`
`therefore V = (epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) - I ( (r_(1) r_(2))/(r_(1) + r_(2)) ) " " `....(4)
Let cell with emf `E_(eq)` and internal resistance req is connected between `B_(1) and B_(2) ` then,
`V= epsilon_(eq) - Ir_(eq) " " ` ... (5)
Comparing equation (4) and (5),
`epsilon_(eq) = (epsilon_(1) r_(2) + epsilon_(2) r_(1))/(r_(1) r_(2)) " " ` ..... (6)
`r_(eq) = (r_(1) r_(2))/(r_(1) + r_(2)) " " ` (7)
By using equation (6) and (7),
`(epsilon_(eq))/(r_(eq)) = (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) `
`(I)/(r_(eq)) = (1)/(r_(1)) + (1)/(r_(2))`
here we have considered that positive terminal of both cell are joined together.
If negative terminal of second cell is connected to positive terminal of first cell then in equation `epsilon_(2) ` should be replaced by `- epsilon_(2)` .
`rArr` If .n. cell are connected in parallel then, equivalent internal resistance,
`(1)/(r_(eq)) = (1)/(r_(1)) + (1)/(r_(2)) + .... + (1)/(r_(n))`
and `(epsilon_(eq))/(r_(eq) ) = (epsilon_(1))/(r_(1)) + (epsilon_(2))/(r_(2)) + ..... + (epsilon_(n))/(r_(n))`
`rArr` Special case : When two cell having equal emf and equal resistance are connected in parallel equivalent emf will be equal to individual cell (one cell).
`epsilon_(eq) = (epsilon_(1) r_(1) + epsilon_(2) r_(2))/(r_(1) + r_(2))`
Let `epsilon_(1)= epsilon_(2) = epsilon and r_(1) = r_(2) = r `
`therefore epsilon_(eq) = (epsilon r + epsilon r)/(r + r)`
= `(2 epsilon r )/(2 r)`
`therefore epsilon_(eq) = epsilon`