In the circuit shown in Fig the sources have `emf's xi_(1) = 1.0 V` and `xi_(2) = 2.5 V` and teh resistances have the values `R_(1) = 10 Omega` and R_(2) = 20 Omega. The internal resistances of the sources are neglibile . Find a potential differences `varphi_(A) - varphi_(B)` between the plates `A` and `B` of the capacitance `C`.

Explanation

As the capacitor is fully charged, no current flows though it, So current
`i = (xi_(2) - xi_(1))/(R_(1) + R_(2)) (as xi_(2) gt xi_(1)`)
And hence, `varphi_(A) - varphi_(B) = xi_(1) - xi_(2) + i R_(2)`
`= xi_(1) - xi_(2) + ( xi_(2) - xi_(1))/(R_(1) + R_(2)) R_(2)`
`= (( xi_(1) - xi_(2)) R_(1))/(R_(1) + R_(2)) = 0.5 V`