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## Homework Statement

A 2.0-microfarad capacitor is charged to 250 V. It is then connected to an uncharged 1.0 microfarad capacitor through a 2.2 ohm resistor, by closing switch S in the figure .

Find the total energy dissipated in the resistor as the circuit comes to equilibrium. Hint: Think about charge conservation.

Express your answer using two significant figures.

The answer to the question is 21 millijoule.

## Homework Equations

C=Q/V

Capacitators in series carry the same charge.

P=I*V

I=V/R

P=I^2*R

U=1/2 * Q * V^2 (energy stored in a capacitor)

## The Attempt at a Solution

I don't really know what to do. I guess charge moves from the charged capacitor to the uncharged one, until the both charges on the capacitors are equal. Moving charge is a current, and this current should give an energy dissipation in the resistor.

I tried some calculations on paper, but there's not much use writing them here, since I seem to need help with the basic approach of this problem.

Thanks in advance,

Merlione.