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

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SUMMARY

The discussion focuses on calculating the current in a circuit after capacitors have lost 80% of their stored energy. Initially, each capacitor has a charge of 3.00 nC. The voltage was incorrectly calculated as 1 x 1024 V, leading to an erroneous current of 4 x 1022 A. The correct approach requires accounting for the 80% energy loss, which was not addressed in the initial calculations.

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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.
 

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

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

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