Voltages

Discussion in 'Electronics Homework Help' started by IlanSherer, Jan 4, 2018.

1. IlanSherer

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Jan 4, 2018
Hello  I need to find a voltage of every capacitor, I succedded to find voltages of C1, C2 and C3.
Now i'm stuck in voltage of C4, it means, I'm trying to find it but I can't, because of diagonal.

Thanks a lot 2. Harald KappModeratorModerator

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Nov 17, 2011
You can sequentially find equivalent circuits by using the equations for capacitors in series and in parallel:

C89 = C9 || C8 (|| meaning is parallel to)
C789 = C7 + C89 (+ meaning in series with)
C6789 = C6 || C789
etc.

The subscript don't have to be as long as I used them - this I did only to indicate which capacitors are involved in the equivalent circuit. You can use shorter subscripts as long as you know how the equivalent capacitors refer to teh original circuit.

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3. IlanSherer

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Jan 4, 2018
Thanks for the response I know about that, but I don't know how I can apply this in finding voltages of C4 and above (C5, C6 and more), I'm really sorry for bothering, it's my first exercise which has diagonals (not really my first time, i did solve exercises with diagonlas in the past, but i only had to find balancing capacitors/voltages).

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5. IlanSherer

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Jan 4, 2018
Thanks for the response Well, to be honest (and I'm really sorry), it's complicated for me (our lecturer has not reached this level yet, i'm sure it will happen soon in a lecture one day), I don't know why he gave us this exercise but it seems that i have no choice, so where I can learn it from the basics (only about diagonals of course)? It will help me greatly if you have a website to recommend me, I'm also trying to get a good book right now.

6. Harald KappModeratorModerator

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Nov 17, 2011
This is how it's done, example for 3 capacitors, cf. my post #2: As @dorke said: don't worry about the diagonals, it's only a graphical representation. Identify the parallel and series circuits and replace them by single components with equivalent values.
With the above example:
• Determine the charge on C423 (hint: C = Q/V)
• From this value determine the charge distribution between C1 and C23
• From the charge on C23 determine the charge on C2 and C3
• From the charges on C1, C2 and C3 determine the respective voltages from Q = C*V

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7. dorke

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Jun 20, 2015
Well,
this is merely a different and somewhat compact way of drawing.
These questions are a kind of trick question for the novice ,
since the novice is used to see drawings in "squares forms ",
they would sometimes use arches instead of straight lines to create confusion.
Or even build circuits to look 3 dimensional !

All that matters is to locate the devices endpoints and redraw in any way that would simplify the circuit for you(mostly square-like).

Here is an attempt to "square a triangle",phase by phase for C4.
You can do it for C9 as a drill. IlanSherer and Harald Kapp like this.
8. Ratch

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Mar 10, 2013
You have 5 loops and 4 nodes in that circuit, therefore use node analysis to figure out the voltage at each node. The voltage difference between each node is the voltage across each capacitor connecting any two nodes. You will have 4 differential equations and 4 unknowns. You can also try to solve it by charge analysis as others have suggested, but that method is confusing and prone to error. If you make an attempt to solve the problem and redraw your schematic so that I can see the subscripts, I will show you my solution.

Ratch

Last edited: Jan 5, 2018
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9. (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
There is no need for a little resistance or a lot of time...

All you need to understand is what DC voltage appears at the point between two arbitrary capacitors (initially at 0V) when a DC voltage exists at the ends.

They form a voltage divider producing the same voltage as a pair of resistors with a resistances inversely proportional to the capacitances.

So, 1 uf and 2 uf could be replaced with resistors 1 ohm and 0.5 ohms.

Then work out the voltage on the nodes. It will be the same as in the capacitance problem.

In this particular case it probably doesn't require you to solve any equations (least of all differential equations) if you don't want to because it can be treated stepwise as resistors in series and parallel.

10. Harald KappModeratorModerator

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Nov 17, 2011
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11. IlanSherer

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Jan 4, 2018
Thanks for the responses Finally I figured out how it works, and thanks to all of you!
Yes, I really let the diagonals change my point of view, I'll have to solve a lot of exercises like the current exercise.

Again - thanks a lot!
Have an excellent week   