Find the Current After the Capacitors Have Lost 80% of Stored Energy

[Northbysouth]

Homework Statement

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

Homework Equations

C=Q/V
V-IR

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.

 

[gneill]

Your calculated voltage looks awfully high. Better check your arithmetic.

I don't see where you've accounted for the 80% drop in energy.