Obtain an expression for the equivalent emf and internal resistance of two cells connected in parallel.

Explanation

Let `epsi_0`, and `epis_2` be the emfs of the two cells connected in parallel.
Let `r_1 , r_2` be their internal resistances respectively.
`I_1` and `I_2` are the currents leaving the positive electrodes of the cells.
At B, currents I, and I, flow in whereas the current I flows out. Therefore `I = I_1 +I_2`
Let `V_1 and V_2` be the potentials at `B_1 and B_2` respectively.
P.d across the first cell is `V_1 -V_2 = V = epsi_1 - I_1 -r_1`
P. d across the second cell is also` V_1 -V_2 = V =epsi_0 , -I_1 r_1`

Hence `I_i =( epis_1 - V )/(r_1 ) and I_2 = ( epsi_0 -V)/( r_2)`
`I=I_1 +I_2 =(epsi_1-V)/(r_1) +(epsi_0 -V)/(r_2)`
`I=(epis_1)/(r_1) - (V )/(r_1 ) - (V) /( r_1) + (epsi_2)/( r_2) -(V )/(r_2)`
` I=((epsi_0)/(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) -I`
` V(r_1 +r_2) = epsi_1 r_1 + epsi_2 r_1 - I r_1 r_2`
` V= (epsi_1 r_2 +epsi_2 r_1)/(r_1 +r_2)-I ((r_1r_2)/(r_1 +r_2))`
If the parallel combination of cells is replaced by a single cell between B, and B, of emf `epis_p`, and internal resistance `r_p` then,` V = epsi_0 -1 r_p`
comparing the above equations
` epis_p = (epsi_1r_2+epsi_2 r_2)/(r_1 +r_2) and r_p =(r_1 r_2 )/( r_1 +r_2) OR (1 )/(r_P) = (1)/(r_1) +(1)/(r_2)`