Equivalent capacitance homework problem

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The discussion revolves around finding the equivalent capacitance between points A and B in a complex circuit involving capacitors. Participants clarify that the circuit is a bridge configuration, which complicates the application of simple series and parallel capacitance formulas. It is emphasized that proper identification of nodes is crucial, as misidentifying them can lead to incorrect conclusions about capacitor arrangements. The conversation also highlights the need for more advanced circuit analysis methods, such as Kirchhoff's laws, when straightforward reduction is not possible. Ultimately, understanding the bridge circuit's symmetry is key to solving the problem effectively.
  • #61
gracy said:
But then I get confused which one will be denominator and which one will be numerator.
It doesn't matter which should be numerator or denominator. Just pick any ratio on one side, use the same on the other side.
 
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  • #62
Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.
 
  • #63
gracy said:
But then I get confused which one will be denominator and which one will be numerator.
You want to see if the two voltage dividers in the bridge circuit will produce the same voltage division. So a simple procedure is to make ratios of the components that make up each voltage divider. Compare the ratios. You can't go wrong if you select corresponding pairs (use symmetry and common sense).
 
  • #64
gneill said:
You can't go wrong if you select corresponding pairs (
You mean if I have selected C1 as numerator for one potential divider I shall select the one which is adjacent to it but at the other side of bridge as numerator of second potential divider,Right?
 
  • #65
gracy said:
Let say if I would have taken
C1/C2=C4/C3
After putting the values
I could have thought the bridge is not balanced.
You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.
 
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  • #66
cnh1995 said:
You can call the two resistors on one side as 'upper' and 'lower'. So, upper/loewer on one side=upper/lower on the other side.
Its just a memory trick..To understand the bridge balance conceptually, you should analyse it using voltage divider.
 
  • #67
logspot.com%2F-jehPxcGkMYs%2FTfgZWoCbTII%2FAAAAAAAAAQY%2F4eEQrl6bCcQ%2Fs1600%2Fbridge%2Bresistor.gif


upper/lower on one side=upper/lower on the other side.And if bridge is of this sort
proxy.php?image=http%3A%2F%2Fimg.tfd.com%2Fggse%2F4c%2Fgsed_0001_0017_0_img4261.png

upper left/upper right on one side=lower left/lower right on the other side
 
  • #68
gracy said:
logspot.com%2F-jehPxcGkMYs%2FTfgZWoCbTII%2FAAAAAAAAAQY%2F4eEQrl6bCcQ%2Fs1600%2Fbridge%2Bresistor.gif


upper/lower on one side=upper/lower on the other side.And if bridge is of this sort
proxy.php?image=http%3A%2F%2Fimg.tfd.com%2Fggse%2F4c%2Fgsed_0001_0017_0_img4261.png

upper left/upper right on one side=lower left/lower right on the other side
Right. Once you're fluent, you won't need this trick.
 

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