# Help on Total Capacitance

#### Electro_explorer

Joined Jan 18, 2008
3
Hi to all fellow explorers of Electronics !

I am new to the subject of Electronics, however I am quite allured by the potential I see in them.
Because of my being new to the study of Electronics I have come across a problem I can not solve myself.
In the name of knowledge exchange I call for your altruistic instinct and pursue your help.

The problem has to do with finding the total capacitance of a complex structure that consists of some capacitors of the same capacitance.
Here you can see the schematic I have made with a java applet from Phet.
For I understand the complexity of the problem I request not a complete solution but more of some advice so that I can devise a solution myself.

#### SgtWookie

Joined Jul 17, 2007
22,210
The value of capacitors in series are calculated like resistors in parallel.
For just two caps:
Rich (BB code):
       C1 x C2
Ctot = -------
C1 + C2
For more than two: Ctot = 1 / ( 1 / C1 ) + ( 1 / C2 ) + .... ( 1 / Cn)
The value of capacitors in parallel are calculated like resistors in series.
ie: Ctot = C1 + C2 + ... Cn
Once you get that, things will become much easier.

#### thingmaker3

Joined May 16, 2005
5,084
Looks like a wheatstone bridge, but with caps...

#### Electro_explorer

Joined Jan 18, 2008
3
In fact I think there is strong resemblance to a Wheatstone bridge.
I would now like to ask, can the total capacitance be found by using just the typical rules of parallel and serial connection?
For example, let' s say the horizontal capacitor is not functional because there is no potential at its connectors, each two capacitors in a leg are connected in serial and the one couple of serial capacitors is parallel to the other couple to the opposite leg.
What do you think?

#### thingmaker3

Joined May 16, 2005
5,084
You betcha! 1/(1/(20,000uF+20,000uF)+1/(20,000uF+20,000uF)). Don't even need a calculator if you have decent pattern recognition skill.

#### Electro_explorer

Joined Jan 18, 2008
3
You betcha! 1/(1/(20,000uF+20,000uF)+1/(20,000uF+20,000uF)). Don't even need a calculator if you have decent pattern recognition skill.
Great! Thank you thingmaker3!

I 'd just like to point out in that your solution you consider the first capacitors on each leg parallel to each other, the second capacitors on each leg parallel to each other. Then you consider these two groups are serial to each other.

Isn' t this solution a bit different than mine? ( two serial capacitors on each leg and these two groups parallel to each other )

The fact is that the result is the same for both solutions ( 20 mF ).

I believe that it' s exactly the same thing. Just because I am new to the subject I' d like a confirmation.