Find a potentail difference `varphi_(1) - varphi_(2)` between points 1 and 2 of the circuit shown in Fig if `R_(1) 1= 10 Omega, R_(2) = 20 Omega, E_(1) = 5.0 V`, and `E_(2) = 2.0 V`. The internal resisances of the current sources are negligible.

Explanation

Assume the curent flow, as shown. Then potentials are as shown. Thus,
`varphi_(1) - varphi_(2) - IR_(1) + xi_(1) - IR_(2) - xi_(2)`
or, `I = (xi_(1) - xi_(2))/(R_(1) + R_(2))`
`varphi_(2) = varphi_(1) - IR_(1) + xi_(1)`
So, `varphi_(1) - varphi_(2) = -xi_(1) + (xi_(1) - xi_(2))/(R_(1) + R_(2)) R_(1)`
`= -(xi_(1) R_(2) + xi_(2) R_(1))//(R_(1) + R_(2)) = -4 V`