1. The problem statement, all variables and given/known data In the circuit shown in the figure each capacitor initially has a charge of magnitude 3.00nC on its plates. A) After the switch S is closed, what will be the current in the circuit at the instant that the capacitors have lost 80.0% of their initial stored energy? Image attached 2. Relevant equations C=Q/V V-IR 3. The attempt at a solution Knowing that V = Q/C I tried to calculate the voltage V = (3x10-9C)(1/15x10-12F + 1/20x10-12F + 1/10x10-12F) V = 1x1024 Then using V=IR I = V/R I got 4x1022 It says this is wrong and I'm not sure what else to try. Help would be appreciated.