- #1

Falcons

- 23

- 1

## Homework Statement

I am asked to find the charge on each cap in this circuit when S

_{1}is closed:

I am given that C

_{1}=1μF, C

_{2}=2μF, C

_{3}=3μF, and C

_{4}=4μF. V=12V

## Homework Equations

Caps in series have an equivalent capacitance of C

_{eq}=C

_{1}C

_{2}/(C

_{1}+C

_{2}), and caps in parallel have an equivalent capacitance of C

_{eq}=C

_{1}+C

_{2}. The capacitance equals the voltage across the cap times the charge "held" in the cap. Parallel capacitors have the same voltage, whereas caps in series have the same charge.

## The Attempt at a Solution

I found the equivalent capacitance of the circuit to be 2.083μF, and using that and C=QV, found that the total charge on the circuit is 25μC. I worked backwards from there, knowing that caps in parallel have the same V, and found the charge on each equivalent cap for the series to be 9μC for C

_{13}and C

_{24}. This is where I get lost. The book has the answer to be 9μC on both C

_{1}and C

_{3}and 16μC for C

_{2}and C

_{4}. Assuming that I found the correct charge on the circuit, 25μC, this violates the law of charge conservation, doesn't it? Any help is appreciated, thanks.

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