Find a potential difference `varphi_(A) - varphi_(B)` between the plates of a capacitor `C` in the circuit shown in Fig, if the sources have `emf's E_(1) = 40 V, E_(2) = 1.0 V` and the resistances are equal to `R_(1) = 10 Omega, R_(2) = 20 Omega`, and `R_(3) = 30 Omega`. The internal resistances of the sources are negligble.

Explanation

Indicate the currents in all the branches using charge conservation as shwon in the figure. Applying loop rule, `(-Delta varphi = 0)` in the loops `12341` and `15781`, we get
`-xi_(1) + i_(3) R_(1) - (i_(1) - i_(3)) R_(2) = 0`....(1)
and `(i_(1) - i_(3)) R_(2) - xi_(2) + i_(1) R_(3) = 0` .....(2)
Solving Eqs. (1) and (2), we get
`i_(3) = (xi_(1) (R_(2) + R_(3)) + xi_(2) R_(2))/(R_(1) R_(2) + R_(2) + R_(2) R_(3) + R_(3) R_(1))`
Hence, the sought p.d.
`varphi_(A) - varphi_(B) - xi_(2) - i_(3) R_(1)`
`= (xi_(2) R_(3) (R_(1) + R_(2)) - xi_(1) R_(1) (R_(2) + R_(3)))/(R_(1) R_(2) + R_(2) R_(3) + R_(3) R_(1)) = -1 V`