Total Capacitance. Parallel or Series?

AI Thread Summary
The discussion focuses on determining the total capacitance of a network of capacitors arranged in both series and parallel configurations. Initially, the user is unsure whether to treat the capacitors as being in series or parallel. It is clarified that the capacitors in each pair are in series, and after calculating the equivalent capacitance for these pairs, the user can then treat the resulting capacitors as being in parallel. The final solution involves simplifying the circuit step-by-step, ultimately leading to the correct total capacitance calculation. The conversation concludes with encouragement for the user's upcoming midterm exam.
ZedCar
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Homework Statement



In the attached question about total capacitance of the network, because of the layout I wasn't sure if this should be totalled as capacitors in parallel or capacitors in series.

I was thinking perhaps parallel, but am not sure.

Thank you! :smile:


Homework Equations





The Attempt at a Solution


 

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:) good way to start is by noticing that you have 4 pairs of capacitors (upper-left,upper-right,lower-left, and lower-right) .. each pair there is connecting in series .. find the equivalent capacitor for each of the four pairs, then replace them with their equivalents .. then what will you notice? can you work fron there (hopefully yes) .. :)


[NOTE] :
to figure out whether the two capacitors are in series or parallel :

1- if they are sharing one end (node) , then they are in series ..
2- it they are sharing two ends (nodes), then they are in parallel..
 
Thanks BigStar,

so I could do an "in series" calculation on each pair so I have a figure for four capacitors.

Then with these four figures I could calculate using an "in parallel" calculation...
 
you are missing something here!
correct me if i got u wrong .. are you suggesting that after you get the equivalent capacitors (four of them, one at the up-right, one at up-left, one at low-right, and low-left) you calculate them as if they all were in parallel?
 
That's what I had thought, but obviously I'm not taking something into account...
 
:( yes that is not correct .. you want to try again?

lets see what we know and what we have ..

first, you know that capacitors in series and parallel look something like the following figure:

http://img175.imageshack.us/img175/66/95306873.jpg

as I mentioned in a previous reply:

from the figure, let's start with the capacitors in " series ":

as you can notice, one end from a capacitor is connecting to one end of the other capacitor .. and that's always the case whenever you deal with capacitors in series ..

what about capacitors in " parallel "? :

from the figure you can see that one end from a capacitor is connecting to an end of the other capacitor (and that's not just this) also the other ends from the two capacitors are connecting.. and that's how is goes with capacitors in parallel ..


now we get to your problem .. it looks like the following figure:

http://img710.imageshack.us/img710/8066/36727868.jpg

..

we agreed that, the each of the four pairs is connecting in series .. then your problem will reduce to the following figure :


http://img185.imageshack.us/img185/6946/94146670.jpg


to this point you are doing a good job .. so now can you tell me what should you do next?

(** hint **: what if you want to take the two capacitors at the right half together , and the capacitors at the left hand together?)
 
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Thanks, yes I understand it now.

Redraw the circuit with the two 10uF capacitors in series as a single 5uF capacitor. On the left are two 5uF capacitors in parallel, in series with two 5uF capacitors in parallel on the right.

Redraw the circuit again replacing the two 5uFs in parallel with a single 10uF capacitor.

There are then two 10uFs in series which is 5uF.
 
:) that is exactly the perfect answer ..
 
Thanks very much BigStar...!
 
  • #10
;) welcome ZedCar .. >> (wish me good luck for my midterm tomorrow!)
 
  • #11
thebigstar25 said:
;) welcome ZedCar .. >> (wish me good luck for my midterm tomorrow!)

Very best of luck...!
 
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