How Do You Calculate the Charge on Capacitor C5 in a Mixed Circuit?

[mrshappy0]

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.

 

[Clever-Name]

Could you post a picture of the circuit? It's difficult to interpret the question without the diagram.

 

[mrshappy0]

I forgot to do that. Oops.

 

[SammyS]

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.

C₁ is not in parallel with C₅ .

 

[mrshappy0]

How is it not parallel with C5? I understand parallel to be this and that's what I see...

So the total capacitance would be (1/C1234+1/C5)^-1 ?

 

[SammyS]

How is it not parallel with C5? I understand parallel to be this and that's what I see...

So the total capacitance would be (1/C1234+1/C5)^-1 ?

If those capacitors were resistors, would you say they're in series or say they're in parallel ? (in either figure)

 

[Clever-Name]

Think of parallel and series in this way.

To be in series, there is only one way for the current to get from start to finish, and that's through the circuit elements you're interested in. (A)

However, when two things are in parallel the current can either go through one, or the other, and still end up completing the circuit (B)

See my drawing

Apply this to your circuit, what conclusion do you have to make when it comes to the current passing through C1 and C5

 

[mrshappy0]

Oh I see... so a parallel series is when the current is split and can go through one of multiple paths. That helps a lot. I wish I had tried that possibility before resorting to the forum for help. Thanks.

 

[mrshappy0]

nice visualization. thanks!