Equivalent capacitance odd circuit

  • Thread starter RandRange
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  • #1
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I'm having a time figuring out how to solve this one, hope someone can help me out

Homework Statement


"Calculate the equivalent capacitance between X and Y on the circuit in the figure. C1=C3=C4=C5=8uF, C2=20uF"


Homework Equations





The Attempt at a Solution


I can't figure out a reduced circuit straight away. I tried to see what will be current at the ends of each capacitor and each of them as different ones so I can't consider any of them to be in parallel.
 

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Answers and Replies

  • #2
gneill
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Sometimes redrawing the circuit can help you to visualize a solution. Try moving terminal X to the top (centered over C2) and terminal Y to the bottom (centered below C2). Nodes 1 and 2 remain where they are.
 
  • #3
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I simplified the circuit for you here. Hope this helps!
 

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  • #4
rude man
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Considering the symmetry of the network, what can you say about the ratio of potentials at nodes 1 and 2 if you "ground" Y (define V_Y = 0) and apply sinusoidal voltage V to X ?

Use that simplification to compute I = jwCV where I is the current from the source V and C is your equivalent capacitance.
 
  • #5
gneill
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I simplified the circuit for you here. Hope this helps!

Do not do the Original Poster's homework for them. This is a Forum rule. Your first redrawing is okay, but you should let the OP draw conclusions regarding how the capacitors might be combined.
 
  • #6
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Thank you very much for your answers! I feel so stupid for not being able to redraw the circuit myself :P
 
  • #7
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OK, so this is a "Wheatstone Bridge". The more you know...
Now since C1/C5=C4/C3 the voltage at points 1 and 2 (both sides of C2) is equivalent, thus there is not current flowing through C2 and I can just ignore it. Then I end up with C1 and C5 on one side (1/Ceq_a=1/C1+1/C5) and C4 and C3 on the other (1/Ceq_b=1/C4+1/C3) and the total Ceq=Ceq_a+Ceq_b.
 
  • #8
gneill
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OK, so this is a "Wheatstone Bridge". The more you know...
Now since C1/C5=C4/C3 the voltage at points 1 and 2 (both sides of C2) is equivalent, thus there is not current flowing through C2 and I can just ignore it. Then I end up with C1 and C5 on one side (1/Ceq_a=1/C1+1/C5) and C4 and C3 on the other (1/Ceq_b=1/C4+1/C3) and the total Ceq=Ceq_a+Ceq_b.

There ya go! :approve:

Actually you have a choice of methods. Since those two potentials are equal you can either remove the capacitor C2 or replace it with a short. No current will flow in either case :wink:
 
  • #9
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OK, so this is a "Wheatstone Bridge". The more you know...
Now since C1/C5=C4/C3 the voltage at points 1 and 2 (both sides of C2) is equivalent, thus there is not current flowing through C2 and I can just ignore it. Then I end up with C1 and C5 on one side (1/Ceq_a=1/C1+1/C5) and C4 and C3 on the other (1/Ceq_b=1/C4+1/C3) and the total Ceq=Ceq_a+Ceq_b.

It is capacitors, There is no current at all in any place in this circuit

the weatstone is applicable just to Resistants
 
  • #10
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It is capacitors, There is no current at all in any place in this circuit

the weatstone is applicable just to Resistants
Of course there is no currents, but the weatstone brigde works here!!!
Not duo to zero current, but because of zero voltage of C2.
 
  • #11
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Of course there is no currents, but the weatstone brigde works here!!!
Not duo to zero current, but because of zero voltage of C2.
Here you can watch a correct solving this problem, using weatstone bridge.
 
  • #12
Merlin3189
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The original "Wheatstone" bridge was resistive, but the concept of bridge circuits has expanded to cover circuits with reactive arms. The idea of a balance point where no current flows in the bridge is common. And as gneil says, it may therefore be replaced by open or short for that condition.
There is no "proper" method. You just use what you know about the circuit to help in whatever way you choose.

It may not be quite correct to call it a Wheatstone bridge, but I don't think anyone would be misled or fail to understand if you did. The principle is the same.
 
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