IlanSherer
- Jan 4, 2018
- 4
- Joined
- Jan 4, 2018
- Messages
- 4
Thanks for the responseYou 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.
Thanks for the responseDon't let the "diagonals" confuse you.
Draw them in the "straight" way for more clarity,like so, (fill in the reset and calculate).
View attachment 38314
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.Hello
Can you help me please about the next exercise:
View attachment 38307
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
This is how it's done, example for 3 capacitors, cf. my post #2:
View attachment 38350
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
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.
View attachment 38353
To @Ratch, @dorke, @duke37 , @(*steve*) : I moved your posts to a new thread: https://www.electronicspoint.com/threads/voltages-continuation-of-the-discussion.286932/
as I think this discussion, interesting as it is, will not help to enlighten the op of this thread. Please continue the discussion in the new thread. This here thread should be used to support the op only.
Thanks for understanding,
Harald
Thanks for the responsesYou 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