Calculating the total capacitance of a network of capacitors

[martin25p2]

Homework Statement

I would like to know how to calculate the total capacity of the capacity network displayed in the uploaded picture.

I know how to calculate the total capacity if the capacity are in parallel or in series, but I am confused by the capacity Nr 6. I know how to calculate the total capacity for the case if they are all the same, please give me an advice how to solve for the case of different capacities.

Homework Equations

C = Q/U

The Attempt at a Solution

The part between 2-3-6 and 4-6-5 are not connected to the source, so their total charge must be zero.

 

[magoo]

I think you mean capacitor network and you're trying to find the total capacitance between the two leads (one going to #2 and the other to #3).
I would start by reviewing what is the resulting capacitance if:
- two capacitors are in series
- two capacitors are in parallel

Just focus on parts of the circuit to start.

 

[martin25p2]

Thank you for your response, but it is not that simple. Please take a closer look at the capacitor number 6: if it would not be for it, the network would be easily solveable as you wrote, but that is not the case, or I fail to see how one can redraw the network in a simpler way. I do know how to calculate the total capacity if the capacitors are in parallel or in series, but that is not that easy here.

I think you mean capacitor network and you're trying to find the total capacitance between the two leads (one going to #2 and the other to #3).
I would start by reviewing what is the resulting capacitance if:
- two capacitors are in series
- two capacitors are in parallel

Just focus on parts of the circuit to start.

 

[magoo]

There is a wye to delta conversion. 4-5-6 form a wye as drawn. You can convert them to a delta. Do you know how to do that?

 

[berkeman]

Thank you for your response, but it is not that simple. Please take a closer look at the capacitor number 6: if it would not be for it, the network would be easily solveable as you wrote, but that is not the case, or I fail to see how one can redraw the network in a simpler way.

If the values of the capacitors have certain symmetries, you may still be able to collapse it. But in the general case, just write the KCL equations to solve the problem.

Ground the right node and write the KCL equations for each of the other nodes. Put in a test voltage of 1Vac at a constant frequency at the left node, and calculate the total current. That will end up leading you to the total capacitance.

EDIT -- or use the hint from @magoo ...

 

[martin25p2]

Thank you for the hint with wye to delta conversion. I have just read about it how to do it for resistances, I assume it will be similar done with the impedance of capacitors...will this be still correct for DC? What are the KLC equations ?

 

[berkeman]

What are the KLC equations ?

The KCL equations are "Kirchoff's Current Law" equations. They are often written as the sum of all currents out of each node must total to zero. Have you learned about them yet? How about phasor impedances? Or are you in a DC circuits class right now?

https://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

 

[martin25p2]

Thanks, I am familiar with Kirchhoff's Circuit Law, just not with the abbreviation. I solved that exercise for the case of identical capacities, but I was wondering how to do it for different ones. The wye to delta transformation is the way to go. (I did not know of that before). For DC the impedance becomes infinity, but I think one should still be able to do the transformations. Thank you very much for your help. Greetings from Germany :^)

 

[eprofu]

The problem is solved here: https://www.physicsforums.com/threads/capacitor-networks.714601/