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## Homework Statement

A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 4.6 μF, C2 = 1.7 μF, C3 = 5.7 μF, and C4 = 2.7 μF. The battery voltage is V = 12 V. 3)

What is Q5, the charge on capacitor C5?

## Homework Equations

Parallel:

Cequ=C1+C2

Vequ=V1=V2

Qequi=Q1+Q2

Series:

Cequ=1/c1+1/c2

vequ=v1+v2

Qequi=Q1=Q2

C=Q/V

## The Attempt at a Solution

I found the total capacitance for C234=4.009. Then I figured that C234 was in a series with C15.....Where C15 the parallel capacitors C1 and C5. Is this logic correct? I added them appropriately and got 2.79 for C12345. Knowing these two groups had equal charges I multiplied C234(Qtotal)=V234 and C15(Qtotal)=V15. Q total 12v*C12345. From there I figured solved for Q5.. Q5=(C5)*(V15) because C1 and C5 have the same voltage. I got Q5 to be 16.75microC which is wrong.

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