A student connects a cell of emf `epsilon_(2)` and internal resistance `r_(2)`, with a cell of emf `epsilon_(1)` such that their combinationhas a net internal resistance less then `r_(1)`. This combination in the connected across a resistance `R`. Draw a circuit of the 'set up' and obtain an external for the current flowing through the resistance `R`

Answer

`(epsilon_(1)r_(2) + epsilon_(2)r_(1))/((r_(1) xx r_(2)) + R (r_(2) + r_(1))`

Explanation

As the effective internal resistance of two cells is less than the internal resistance `r_(1)` of one cells so the cell are connected in parallel between the two point `R_(1) and R_(2)`. The circuit is be the potential difference between point `R_(1) and R_(2)`. The potential difference between the termianl of first cell is

`V = V_(B_1) - V_(B_2) = epsilon_(1) - I_(1) r_(1) orI_(1) = (epsilon_(1) - V)/(r_(1))`
Potential difference between the terminal of second cell is
`V = V_(B_1) - V_(B_2) = epsilon_(2) - I_(2) r_(2) or I_(2)= (epsilon_(2) - V)/(r_(2))`
Hence current in external resistance
`I = I_(1) + I_(2) = (epsilon_(1) - V)/(r_(1)) + (epsilon_(2) - V)/(r_(2))`
`= ((epsilon_(1))/(r_(1)) +(epsilon_(2))/(r_(2))) - V ((1)/(r_(1)) + (1)/(r_(2)))`
or `I =((epsilon_(1))/(r_(1)) +(epsilon_(2))/(r_(2))) - IR ((1)/(r_(1)) + (1)/(r_(2))) [ :' V = IR]`
or `I[1 + R((1)/(r_(1)) + (1)/(r_(2)))] = ((epsilon_(1))/(r_(1))+(epsilon_(2))/(r_(2)))`
or `I [ (r_(1)r_(2) + R (r_(2) r_(1)))/(r_(1)r_(2))] = (epsilon_(1)r_(2) + epsilon_(2)r_(1))/(r_(1)r_(2))`
or `I = (epsilon_(1)r_(2) + epsilon_(2)r_(1))/((r_(1) xx r_(2)) + R (r_(2) + r_(1))`