Combination of Capacitors

  • Thread starter mrshappy0
  • Start date
  • #1
99
0

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.
 
Last edited:

Answers and Replies

  • #2
380
1
Could you post a picture of the circuit? It's difficult to interpret the question without the diagram.
 
  • #3
99
0
I forgot to do that. Oops.
Circuit.jpg
 
  • #4
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012

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 .
 
  • #5
99
0
How is it not parallel with C5? I understand parallel to be this and thats what I see....
Capacitors.jpg


So the total capacitance would be (1/C1234+1/C5)^-1 ?
 
  • #6
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,331
1,012
How is it not parallel with C5? I understand parallel to be this and thats what I see....
Capacitors.jpg


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)
 
  • #7
380
1
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
 

Attachments

  • #8
99
0
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.
 
  • #9
99
0
nice visualization. thanks!
 

Related Threads on Combination of Capacitors

  • Last Post
Replies
0
Views
5K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
9
Views
975
  • Last Post
Replies
1
Views
691
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
4
Views
7K
Replies
7
Views
4K
  • Last Post
Replies
4
Views
3K
Replies
10
Views
4K
Replies
8
Views
23K
Top