Equivalent capacitance A hard one?

In summary, the homework statement says that calculating equivalent capacitance between two capacitors can be difficult. The attempt at a solution suggests finding those that can be treated as just parallel or just in series, calculating their equivalent capacitance, and then treating them as one capacitor. There are several different routes that could be taken, but the most simplified would be to add one capacitor and assume a voltage is applied across terminals a and b.f
  • #1
Equivalent capacitance... A hard one!?

Homework Statement


Calculate Equivalent capacitance between A and B:
[PLAIN]http://scsupport.org/moodle/file.php/13/Chapter26/imgch26q14.gif [Broken]
EDIT: If you can't see the picture, see 3rd post for file attachment, sorry!


Homework Equations


C1 = 3.55 µF
C2 = 1.90 µF


The Attempt at a Solution


I know how to solve for equivalent capacitance with capacitors in series and parallel, but not both at the same time! Unless I'm seeing the problem wrong this is a complicated problem that I have never seen before! My mind is blown...

Any help at how to approach this would be greatly appreciated!
 
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  • #2


Hard to help without seeing a picture - but usual approach is to simplify the task by finding those that can be treated as just parallel or just in series, calculating their equivalent capacitance - and then treat them as one capacitor.
 
  • #3


Shoot, is the picture not showing up for you guys? It must've been cached in my browser and not visible to you guys since it requires a login. Diagram attached to this reply.

C1 = 3.55 µF
C2 = 1.90 µF
 

Attachments

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  • #4


If you know how to do it as resistors, then treat each as a resistor with value 1/(k.C) if you prefer.

Once you have a value for that 'resistance', 'X', reconvert it to C ( = 1/k.X)
 
  • #5


I suggest redrawing the circuit more symmetrically for yourself to reviel the role played by the 8 uF capacitor that is neither in series nor parallel with any other capacitor.
 
  • #6


You could use a Delta-Y transformation on one of the Δ configurations and then proceed as usual to identify parallel and serial opportunities.

Or you could slap a voltage source V between a and b, creating a third loop, and solve the mesh equations for the current being driven by that source. The impedance of the circuit is then V/I. I'd start by assigning symbols Z1, Z2, Z3 to the capacitor impedances and finding a symbolic result... you may find that the result has an obvious form that you can exploit.
 
  • #7


Applying KVL to the circuit, could something be said about Vc in relation to V1 and V2 (Vc is the voltage across the 8µF capacitor)? Looking at several different routes, I think I found a result that would massively simplify this problem.
 
  • #8


Applying KVL to the circuit, could something be said about Vc in relation to V1 and V2 (Vc is the voltage across the 8µF capacitor)? Looking at several different routes, I think I found a result that would massively simplify this problem.

Could be! :wink:
 
  • #9


I'm not exactly sure how to apply KVL to the circuit because there's no voltage source.
 
  • #10


I'm not exactly sure how to apply KVL to the circuit because there's no voltage source.

For purposes of analysis it's perfectly acceptable to add one as long as it doesn't change the results. You can assume a voltage Vab is applied across terminals a and b. It won't change the impedance.
 

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