http://i.imgur.com/ZCcaZpq.png I have the solution manual, so this is more just a question of why, rather than what. I understand that with the switch open: vA = 3v vB = vC = 0v i0 = 0.25 mA When the switch closes it creates a short that makes the two 6kΩ resistors irrelevant. What I don't understand is why anything should change then. The solution manual tells me that a current would flow from the capacitor from B to C, but why? It seems like nodes B and C should be at 0v both before and after the switch is closed.