Voltage of a Capacitor using Kirchhoff's Loop Rule

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unknownv2
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(Sorry, meant to post this in Introductory Physics Homework, if someone could please move it.)


1. Homework Statement


Use Kirchhoff's Loop Rule to solve for the voltages of the five capacitors in the network shown below.

Circuit schematic of a 24.0V battery connected across a network of five capacitors: C1=3.00μF and C2=1.00μF on the left branch, C3=2.00μF and C4=8.00μF on the right branch, and C5=4.00μF as a bridge between the two branches.

Homework Equations



For Capacitance, C = q/v
Kirchoff's juction law states that the total current into a junction must equal the total current leaving the function and that the sum of the potential differences around any loop or closed path is zero.

Kirchhoff loop rule equation: sum from k=1 to n of V_k equals 0, written as plain typeset mathematics.


The Attempt at a Solution


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My work can be found here: (Warning large image)

https://i.imgur.com/RMDZrAP.jpg

The solution requires a system of equations to solve and I have:

1) ## -V_1 - V_5 - V_2 + V_A = 0##
2) ## -V_3 - V_5 - V_4 + V_A = 0##
3) ## -Q_5 + Q_4 + Q_2= 0##
4) ## -Q_1 - Q_3 + Q_5= 0##1 and 2 translate to:
1) ##\frac{-Q_1}{C_1} + \frac{-Q_5}{C_5} + \frac{-Q_2}{C_2} + 24 = 0##
2) ##\frac{-Q_3}{C_3} + \frac{-Q_5}{C_5} + \frac{-Q_4}{C_4} + 24 = 0##

It doesn't seem possible to solve those system of equations so I assume I messed up somewhere on creating the equations.Questions:

1) Can an open/close loop be created by the voltage connections on both sides of the capacitor, without necessarily returning to the original entrance of the voltage? Such as, if the voltage enters on the left side, can it leave through the right? I'm thinking enters from left (##V_L##) to ##C_1## through ##C_5## then ##C_4## then out the right (##V_R##) side, as an example?

2) How many equations would be needed for the system of equations? Does that depend on the number of junctions or the number of capacitors? So would I need 5 equations total before I can solve this?
 
Attachments
  • Circuit schematic of a 24.0V battery connected to five capacitors: C1=3.00μF and C3=2.00μF in series on the top branch, C2=1.00μF and C4=8.00μF in series on the bottom branch, with C5=4.00μF connected between the two branches.
    Z36ydF0.jpg
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Last edited:
Hello V, welcome to PF :smile: !

You either want to reduce the number of unknowns or increase the number of equations.

As to the first: if you pick the voltage at two smart points you can described all the voltage differences
As to the second: there's more loops to be found than you now mention -- that should also answer your first question: think loops, not paths