Find the charge and energy stored on the capacitor C in the following circuit.

Explanation

As usual steady current will flow through C . So there is no potential drop across `R_(3)` and it.ll be removed form the circuit r edraw below. But don.t removes `epsi_(2)` because it always has a potential difference of its own.
As usual ` I = (epsi_(1))/(R_(1) + R_(2))`
Now caclculate potential drop across the capacitor as follow :
`V_(a) + epsi_(2) - IR_(1) = V_(b) or , V_(a) - V_(b) = IR_(1) - epsi_(2) = (epsiR_(1))/(R_(1) + R_(2)) - epsi_(2)`
So , charge on the capacitor C is `C(V_(a) - V_(b)) = C[(epsi_(1)R_(1))/(R_(1) + R_(2)) - epsi_(2)]`
And the energy stored ` = (1)/(2)C(V_(a) - V_(b))^(2) = (1)/(2)C[(epsi_(1)R_(1))/(R_(1) + R_(2))- epsi_(2)]`