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Equivalent capacitance

  • Thread starter jishnu
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  • #1
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Thread moved from the technical forums, so no Homework Template is shown
The picture attached with this post shows a combination of capacitors, we are asked to find out the equivalent capacitance between the points A and B.
Provided that all the capacitors are identical and has an individual capacitance C.
Can anyone help me provide basic ideas of solving these kind of problems( I mean a general approach). [emoji4]
 

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  • #2
Simon Bridge
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A general approach would be to use the kirkoff circuit laws.
If you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.

For instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.

Notice that node 4 is the same as node 0.

So those first two capacitors are in parallel.

Redraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.

The point of these exercizes is to get you to spot these sorts of patterns.
 
  • #3
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A general approach would be to use the kirkoff circuit laws.
If you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.

For instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.

Notice that node 4 is the same as node 0.

So those first two capacitors are in parallel.

Redraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.

The point of these exercizes is to get you to spot these sorts of patterns.
In that case, I am getting the answer as (2/3) *C
Is that correct..?
 
  • #4
cnh1995
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  • #5
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No.
Then could you please help me solve this...
 
  • #6
sophiecentaur
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Then could you please help me solve this...
Show us your redrawn diagram as @Simon Bridge recommended - that could be your problem.
Alternatively: We all know how to calculate the equivalent to two series or parallel components but there is also a useful formula for transforming a star of three resistors (or Reactances) to a delta of three equivalent resistors (Reactances) and vice versa. This link explains it. You can use that to change your circuit into two deltas which makes the problem easier (visually) and shows up some symmetry which allows you to eliminate some components for the final result. I think that link should help you enough to solve the prob and you will then have another tool in your tool box.
 
  • #7
Simon Bridge
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In that case, I am getting the answer as (2/3) *C
Is that correct..?
I don't know: I did not do the problem. Others say "no". Maybe they are correct maybe not.

I cannot help you if you do not explain your reasoning.
 
  • #8
74
14
A general approach would be to use the kirkoff circuit laws.
If you don't know about those, you had best look for ways to redraw the diagram so it looks simpler.

For instance ... see that box of capacitors in the middle... number the nodes around them 1 through 4, clockwise, from the top left. Label the node closest to A as 0, and the one closest to B as 5.

Notice that node 4 is the same as node 0.

So those first two capacitors are in parallel.

Redraw it like that ... you'll end up with sort-of nested parallel capacitors. You know a rule for parallel capacitors.

The point of these exercizes is to get you to spot these sorts of patterns.
I didn't understand how will the nested arrangement of capacitors looks like... Can u please explain more on that or provide a suitable diagram, the final results that I have posted seems to be wrong to me because I couldn't make any logic out of that, please anyone help me out with this..!
 
  • #9
cnh1995
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I didn't understand how will the nested arrangement of capacitors looks like... Can u please explain more on that or provide a suitable diagram, the final results that I have posted seems to be wrong to me because I couldn't make any logic out of that, please anyone help me out with this..!
Put a voltage source between a and b and analyse the diagram. Can you find any series/parallel connection?
 
  • #10
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Put a voltage source between a and b and analyse the diagram. Can you find any series/parallel connection?
Not exactly. I couldn't figure out one from the diagram the problematic capacitors for me are X and Y(figure 1), but I have doubt whether is it possible to redraw the circuit as to the one I have done in figure 2 then I can find 3 parallel and 1 series circuit of capacitors from that. Am I right in redrawing that circuit? Please explain..
1528701064507.jpg
 

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  • #11
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Not exactly. I couldn't figure out one from the diagram the problematic capacitors for me are X and Y(figure 1), but I have doubt whether is it possible to redraw the circuit as to the one I have done in figure 2 then I can find 3 parallel and 1 series circuit of capacitors from that. Am I right in redrawing that circuit? Please explain.. View attachment 226812
Sorry, this is the figure 2 that I have mentioned in my previous post
1528701309204.jpg
 

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  • #12
CWatters
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Fig 2 is not correct. Q and R are not in parallel in the original. You are trying to make too many simplifications in one go. I would start by just combining Y and S (or P and X) and redraw that.
 
  • #13
Simon Bridge
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No.

Note... from diagram 1, nodes 2 and 5 are the same.
So Y and S are parallel... you can collapse them to one capacitor between nodes 5 and 3, node 2 vanishes.

Redraw.

Q connects between 1 and 5 while R connects between 4 and 3... so Q and R are NOT parallel.

Take more care.
 
  • #14
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No.

Note... from diagram 1, nodes 2 and 5 are the same.
So Y and S are parallel... you can collapse them to one capacitor between nodes 5 and 3, node 2 vanishes.

Redraw.

Q connects between 1 and 5 while R connects between 4 and 3... so Q and R are NOT parallel.

Take more care.
Sir, I am having one doubt, if the node 2 collapses then what happens to the connection between node 2 and node 3..?
I want to know whether the newly attached diagram (diagram 3) can be considered correct after collapsing. Can I proceed the same way for the capacitors P and X ?
1528715015823.jpg
 

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  • #15
Simon Bridge
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Last diagram looks good.
The connection between 2 and 3 becomes a connection between 5 and 3.

The key is that wires are part of the node they are attached to. So two nodes joined by a wire are, in fact, the same node.

This lets you simplify the diagram... you are allowed to draw curved lines for wires/connections.

Now see if there are other capacitor pairs you can collapse this way.
 
  • #16
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Last diagram looks good.
The connection between 2 and 3 becomes a connection between 5 and 3.

The key is that wires are part of the node they are attached to. So two nodes joined by a wire are, in fact, the same node.

This lets you simplify the diagram... you are allowed to draw curved lines for wires/connections.

Now see if there are other capacitor pairs you can collapse this way.
Thank you sir. [emoji4]
I think in that way we can collapse node 4 and combine P and X with parallel connected capacitors formula to achieve even simpler circuit diagram.
1528786736280.jpg
 

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  • #17
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Show us your redrawn diagram as @Simon Bridge recommended - that could be your problem.
Alternatively: We all know how to calculate the equivalent to two series or parallel components but there is also a useful formula for transforming a star of three resistors (or Reactances) to a delta of three equivalent resistors (Reactances) and vice versa. This link explains it. You can use that to change your circuit into two deltas which makes the problem easier (visually) and shows up some symmetry which allows you to eliminate some components for the final result. I think that link should help you enough to solve the prob and you will then have another tool in your tool box.
Using this conversion I have redrawn my diagram can you please tell me whether its correct or not?
I also want to know can we use resistor combination formulas and wheatstone's bridge concept to deal with reactance(I mean in simplifying and solving them).....!
1528788005266.jpg
 

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Last edited:
  • #18
sophiecentaur
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Using this conversion I have redrawn my diagram can you please tell me whether its correct or not?
I also want to know can we use resistor combination formulas and wheatstone's bridge concept to deal with reactance(I mean in simplifying and solving them).....! View attachment 226863
That looks fine. Now notice that the two parallel capacitors are placed symmetrically across the two paths 0,2,5 and 0,4,5. The PD across them is zero (when all values are equal) because there is no current flowing across, so you now have two pairs of series C in parallel. No sums actually required now!
PS Resistor formulae also apply to Reactances and Impedances (with care of course - big C = small reactance)
 
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  • #19
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I have tried to solve this question in the both ways suggested
1. Collapsing the nodes
2. Using nodal analysis and conversion
But, still I got some problem. Those two ways are giving me different answers to the same problem. Here I am attaching my answers with this post, can anyone please help me to figure out where I went wrong...!
Here I changed the capacitance as
P = Q = R = S = X = Y = "C" (each has same capacitance)
1528857784414.jpg
 

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  • #20
Simon Bridge
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Recheck your working -- especially where you are rearranging the capacitors in method 2. You seem to have a blindness there.
I am getting the same answers by the two methods.

The point of this sort of exercise is to get you to understand how circuits work ... not to find the answer by correctly applying the rules and formulas. The actual answer is irrelevant to the task in front of you.
 
  • #21
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Recheck your working -- especially where you are rearranging the capacitors in method 2. You seem to have a blindness there.
I am getting the same answers by the two methods.

The point of this sort of exercise is to get you to understand how circuits work ... not to find the answer by correctly applying the rules and formulas. The actual answer is irrelevant to the task in front of you.
I am a bit confused sir, in case of the second method I used the mesh analysis(considering the reactance of the capacitor) to convert the available two stars in to delta then as suggested I removed the capacitors connecting 0-2-5 and 0-4-5 (ref post 18) since the PD between them is zero,
There I am stuck with the problem that whether I have to continue solving with reactance(using resistor equations) or by using the new capacitance( using capacitor combination formulas) please help...
 
  • #22
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Please anybody..... Help me out with this problem....[emoji846]
 
  • #23
epenguin
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Please anybody..... Help me out with this problem....[emoji846]
You appear to be OK in #16. You were probably on the right lines in #3 but then made a mistake. If you had written out your working there you might either have seen the error yourself, or someone could have pointed it out to you.
Go back to #16 - that is quite a simple circuit. It should look even simpler if you put the actual values in (In terms of C).
 
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  • #24
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You appear to be OK in #16. You were probably on the right lines in #3 but then made a mistake. If you had written out your working there you might either have seen the error yourself, or someone could have pointed it out to you.
Go back to #16 - that is quite a simple circuit. It should look even simpler if you put the actual values in (In terms of C).
I have solved the circuit of #16 in #19 by substituting the actual C values, but my confusion is there in the 2nd method(nodal analysis) of #19. Can you please help me figure out where I went wrong from there onwards, because I am getting different answers in those two methods for the same problem.
 
  • #25
epenguin
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Sorry I missed that you had given the answer in the left-hand page of #19. I think that is correct, notice it is something that you could have worked out in your head (actually I did). For the other method it is useful to know the right answer beforehand, but I will leave it to those sponsoring this approach. :oldsmile:
 

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