# [help] equivalent capacitance

1. Feb 21, 2005

### neik

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 ?

Last edited: Feb 21, 2005
2. Feb 21, 2005

### Curious3141

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. Feb 21, 2005

### neik

i guess what you'r trying to say is there is no current across the central capacitance ? but i still dont understand why :yuck:

4. Feb 21, 2005

### Curious3141

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. Feb 21, 2005

### neik

oki
thanks alot :rofl:

6. Feb 21, 2005

### Curious3141

Here's a more rigorous proof :

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

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,

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

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

Making the substitutions into eqn 6 and simplifying,

$$\dot{V} = \dot{V_1} + \dot{V_2}$$ --eqn 7

Taking eqn 1 - eqn 3,

$$I_1 - I_3 = C(\dot{V} - 2\dot{V_1})$$ --eqn 8

Comparing eqn 8 to eqn 5 and simplifying,

$$\dot{V} + \dot{V_2} = 3\dot{V_1}$$ --eqn 9

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

we get $$\dot{V_1} = \dot{V_2} = \frac{1}{2}\dot{V}$$

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

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

$$\dot{V_1} = \dot{V_2}$$

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

$$V_1(T) - V_1(0) = V_2(T) - V_2(0)$$

Since the initial voltages can be assumed to be equal,

$$V_1(T) = V_2(T)$$

and our analysis is complete.

#### Attached Files:

• ###### asg.jpg
File size:
6.1 KB
Views:
87
Last edited: Feb 21, 2005