Breakdown voltage in capacitance

[Axioms]

Each capacitor in the combination shown has a breakdown voltage of 15V. What is the breakdown voltage of the combination?

 

 

...........20uF................................................20uF

----------/-------- ........10uF............------------/------------

A..........................--------/--------- .................................----------B

---------/---------.........................-----------/-------------

.........20uF...........................................20uF

 

Sorry this is the best I could do for the circuit diagram. It is capacitors between two points A and B. Full stops are there to keep the values in place they have no other meaning. The dashes represent the wire of the circuit.

 

I calculated Ceq to be 20/3uF

 

1/Ceq = 1/(20+20) + 1/(10)+ 1/(20 + 20)

so Ceq = 20/3uF

 

There is no more information in the question and I am not sure how to calculate the breakdown voltage of the combination. I think I recall something that delt with the distance and voltage being to close and high respectively. I am sure it is a simple solution but I have no idea how to calculate it. Any help would be appreciated.

 

Here is a better representation of the circuit

Edited May 12, 2012 by Axioms

[Joatmon]

Just a couple of points.

This looks like homework.

It's past my bedtime.

I'll see if I can start you off. If the capacitors were resistors I expect you could solve this problem by making the network equivalent circuit out of three resistors. Using the voltage divider principle you could determine the ratios of the voltages across each part. Bearing in mind that no resistor must have more than 15 volts across it you could determine the input minimum voltage that produces a maximum 15V across any resistor.

 

You can do much the same with the circuit made of capacitors BUT WITH ONE IMPORTANT DIFFERENCE. That's enough information for now - have a go and show some working - I'm off to bed (12.30 UK time)

[Axioms]

I did do voltage division and I got the right answer but I'm not sure if the logic is right.

 

This is what I did to try and solve the breakdown voltage.

 

I took the 10uF capacitor and applied 15V . I did voltage division with the equivalent circuit to find out the maximum voltage that the circuit can handle.

 

 

Calculation:

 

(10uF)/(20/3uF)*15 = 22.5V

 

Thus you can apply a maximum voltage of 22.5V to the circuit before the 10uF capacitor will break.

[Joatmon]

That looks ok.

I would have looked at it a different way which, to me ,seems easier.

I would consider an equivalent circuit made up of 3 capacitors in series 40, 10, 40 (all in microfarads).

I would then say the smallest capacitor (10yF) would have the largest voltage across it (15V).

I would then say the 40 yF capacitors would each have 15/4 volts across them.

All being in series the total voltage across the circuit will be 15/4 +15 +15/4 = 22.5V.

Edited May 13, 2012 by Joatmon