How to find the equivalent capacitance in a complex circuit?

In summary, when analyzing a network with symmetrical capacitors, the voltage across the joining capacitor will be equal to the voltage across any other capacitor in the network. This means that the joining capacitor will not factor into the analysis and the network can be simplified to two series capacitances placed in parallel to another two series capacitances. Additionally, a more rigorous proof using Kirchoff's second law and integrating equations has been provided.
  • #1
neik
15
0
http://img.photobucket.com/albums/v64/neik7/asg.jpg

this is neither a parallel nor a series circuit
i duno how to find the equivalent capacitance from a to b
can anyone give me some hint ? :cry:
thanks in advance
 
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  • #2
Here's a hint : consider the symmetry about the central joining capacitor. What would be the voltage across it when the network is connected to a voltage source ?

If you can't see it, think about the situation when all the capacitances are replaced by equal resistors, and the network is connected to a voltage source. Is there any current across the joining resistor ? Is there any voltage across it ?

From your conclusion about the voltage across the joining capacitance, what conclusion can you draw ? Can you now reduce the network to a much simpler one ?
 
  • #3
Curious3141 said:
If you can't see it, think about the situation when all the capacitances are replaced by equal resistors, and the network is connected to a voltage source. Is there any current across the joining resistor ? Is there any voltage across it ?

i guess what you'r trying to say is there is no current across the central capacitance ? but i still don't understand why :yuck:
 
  • #4
neik said:
i guess what you'r trying to say is there is no current across the central capacitance ? but i still don't understand why :yuck:

Let's not talk about current (though you are correct, there is none). Let's restrict ourselves to talking about voltage. See if you agree with me here :

a) A capacitor can only get charged when there is a potential difference applied across its terminals

b) Only when a capacitor is capable of getting charged in the above fashion will it factor into a network being analysed.

c) In the given network, if you draw a horizontal line through the joining capacitance, the top and bottom halves are exactly identical and indistinguishable.

d) Since the top and bottom are indistinguishable, it makes no sense to assume that they're going to behave differently electrically.

e) Similarly, if you draw a vertical line through the joining capacitance, the left and right halves of the network are again exactly identical. What happens on the left happens on the right.

f) Adding up the logical inferences, you can conclude that the voltage on either plate of the joining capacitance is exactly the same.

g) Using a), the joining capacitance cannot be charged, and by b), it does not factor into the analysis

h) The network reduces to 2 series capacitances placed in parallel to another 2 series capacitances.

Agree ?
 
  • #5
oki
thanks a lot :rofl:
 
  • #6
Here's a more rigorous proof :

The charge Q carried by a capacitance C with voltage V across its plates is given by [itex]Q = CV[/itex]. The current across such a capacitor with a varying voltage is the first differential wrt time, viz. [tex]I = C\dot{V}[/tex]

Let a voltage V be placed across the network (V on the left, ground on the right), and let the currents and voltages as labelled be the result. Then,

[tex]I_1 = C(\dot{V} - \dot{V_1})[/tex] --eqn 1
[tex]I_2 = C(\dot{V} - \dot{V_2})[/tex] --eqn 2
[tex]I_3 = C(\dot{V_1})[/tex] --eqn 3
[tex]I_4 = C(\dot{V_2})[/tex] --eqn 4

Further, by Kirchoff's second law,
[tex]I_1 - I_3 = C(\dot{V_1} - \dot{V_2})[/tex] --eqn 5
[tex]I_1 + I_2 = I_3 + I_4[/tex] --eqn 6

Making the substitutions into eqn 6 and simplifying,

[tex]\dot{V} = \dot{V_1} + \dot{V_2}[/tex] --eqn 7

Taking eqn 1 - eqn 3,

[tex]I_1 - I_3 = C(\dot{V} - 2\dot{V_1})[/tex] --eqn 8

Comparing eqn 8 to eqn 5 and simplifying,

[tex]\dot{V} + \dot{V_2} = 3\dot{V_1}[/tex] --eqn 9

Using the result from eqn 7 in eqn 9 and simplifying,

we get [tex]\dot{V_1} = \dot{V_2} = \frac{1}{2}\dot{V}[/tex]

From this we can discern that [tex]I_1 = I_2 = I_3 = I_4[/tex] and the current across the joining capacitor is zero.

Just for completeness, we've proved that [tex]\dot{V_1} = \dot{V_2}[/tex]. One more step remains :

[tex]\dot{V_1} = \dot{V_2}[/tex]

Integrate both sides wrt t, taking bounds from 0 to T :

[tex]V_1(T) - V_1(0) = V_2(T) - V_2(0)[/tex]

Since the initial voltages can be assumed to be equal,

[tex]V_1(T) = V_2(T)[/tex]

and our analysis is complete.
 

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1. What is equivalent capacitance?

Equivalent capacitance refers to the combined capacitance of two or more capacitors that are connected in a circuit. It represents the total amount of charge that can be stored in the circuit.

2. How is equivalent capacitance calculated?

The equivalent capacitance of capacitors connected in series can be calculated by adding the reciprocals of the individual capacitances and taking the reciprocal of the sum. For capacitors connected in parallel, the equivalent capacitance is simply the sum of the individual capacitances.

3. Why is equivalent capacitance important?

Equivalent capacitance is important because it allows us to simplify complex circuits and analyze them more easily. It also helps us determine the total energy stored in a circuit and the time constant of the circuit.

4. How does equivalent capacitance affect the overall capacitance of a circuit?

Equivalent capacitance increases when capacitors are connected in parallel and decreases when they are connected in series. This means that by changing the configuration of capacitors in a circuit, we can alter the overall capacitance of the circuit.

5. What are some real-life applications of equivalent capacitance?

Equivalent capacitance is used in many electronic devices, such as radios, computers, and smartphones. It is also used in power grids to store energy and in electric cars to regulate the flow of electricity. Equivalent capacitance is also important in designing and analyzing circuits in telecommunications, aerospace, and other industries.

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