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Greeting friends I am having some trouble to calculate the capacitance equivalence in this circuit due to the short circuit? Calculate the capacitance equivalence in the circuit across terminals A and C: -

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My solution so far: -

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    \$\begingroup\$ We expect you to show your work and where you are stuck. This is not a free homework service. Please edit your question or it will be closed due to lack of effort. Please remove the duplicate image while you are editing. \$\endgroup\$ – Transistor May 27 at 16:04
  • \$\begingroup\$ Are both diagrams the same? Ditto the above comments. As usual with problems like this, look hard at the diagram and then simplify it by showing parallel components as one component - I can see several straight away. \$\endgroup\$ – Andy aka May 27 at 16:05
  • \$\begingroup\$ Sorry guys, i did not post my attempt. The problem is there is a short circuit at C, so i guess no current will flow. But i am asked to find the equivalence capacitance at C! \$\endgroup\$ – user3919758 May 27 at 16:16
  • \$\begingroup\$ Simplify it by removing anything that is unimportant to the question and looking for parallel and series capacitors that can become 1 capacitor. Look harder at it. Use your eyes. If necessary use crayons to mark the nodes that might be common to two capacitors. \$\endgroup\$ – Andy aka May 27 at 16:22
  • \$\begingroup\$ You lost the original picture. \$\endgroup\$ – Andy aka May 27 at 16:23
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The very easiest thing you can do is to unclutter the original picture and start the process of simplifying like this: -

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Can you see what I've done above? Can you take it another step? Can you see that the 3 nF is shorted out? Can you see that the three 10 nF capacitors can be rationalized into one capacitor?

OK more help is needed: -

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Getting a precise answer requires doing things carefully and neatly and not putting crappy scribbles on a piece of paper hoping for the best. Do it my way and don't show me another scribbled diagram until you are much more skilled about thinking about these things. I'm being hard on your because you are not listening.

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  • \$\begingroup\$ Thank you sir! Your way is much easier! Appreciate your help. Based on previous comment reached, i understand there is no short circuit presented while calculating capacitance. Am i right? \$\endgroup\$ – user3919758 May 27 at 17:39
  • \$\begingroup\$ I'm not sure what previous comment you mean. Stick with the same format as the original diagram rather than jump into redrawing it yourself because that is a step that can introduce errors. I don't think your redraw has made errors but you need to simplify as much as possible (I've only done 50% so far) before finalizing an answer. \$\endgroup\$ – Andy aka May 27 at 17:42
  • \$\begingroup\$ Well you'll have to take that up with Chu - I'm not going to get involved in every side issue unless it's glaring. Regards the two 4nF capacitors - can you see what has happened to them in my answer? \$\endgroup\$ – Andy aka May 27 at 17:44
  • \$\begingroup\$ Please refer to the above photo and please comment \$\endgroup\$ – user3919758 May 27 at 18:02
  • \$\begingroup\$ You have it wrong. Now listen... modify the original diagram like I have done and do it the way I have done to avoid more mistakes. I'll post one more slide of what to do next and I expect you to see why I have done it my way and why I'm recommending doing it my way and not scribbles on a piece of paper. Accuracy is needed. \$\endgroup\$ – Andy aka May 27 at 21:06

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