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I'm currently taking a course in electromagnetism, and I got stuck with this problem:

The Problem:

All of the resistors and capacitors have the same values, R and C respectively. The circuit is analyzed after a very long time. What is the voltage on the capacitors?

My Reasoning:

After a long time, the capacitors charge up and every loop that contains a capacitor has no current flowing.

C1 was charging through current I, I then was split and the rest of the capacitors where charging with current I1.

So the moment the current stopped flowing in the circuit, C1 built up Q1 charge and the other capacitors built up the same amount of charge Q2 (Same current flow through them for the same amount of time)

For some reason I get weird results, I would really appreciate it if someone could point to my mistake enter image description here

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  • \$\begingroup\$ You got correct results. But you actually don't need to write down all these charge equations. After infinite time capacitors stop conducting (charging) and become equivalent of open breaker. So in the end you have very simple circuit with resistors and open breakers. \$\endgroup\$ – AlexVB Mar 26 at 8:20
  • \$\begingroup\$ They request to find the voltage on the capacitors \$\endgroup\$ – primoshrimp Mar 26 at 8:22
  • \$\begingroup\$ What is the problem? You can find voltage across an open breaker. Even with your solution, zero charge means zero voltage. \$\endgroup\$ – AlexVB Mar 26 at 8:27
  • \$\begingroup\$ From the equation, you do get that the charge is zero. But what was the process to get there? it's unintuitive because you would guess there would be current flowing through the capacitors filling them up. Did the discharge somehow? \$\endgroup\$ – primoshrimp Mar 26 at 8:47
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    \$\begingroup\$ Yes, it charges while there is voltage across the vertical resistor and then discharges when the voltage starts to drop. \$\endgroup\$ – AlexVB Mar 26 at 8:56
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It's correct.

I would say to simplify that after long time the green current I is 0, so it is the voltage drop on the vertical resistor.

Knowing that in the second loop on the right also the current is 0 because we have at least 1 capacitor is series, the also the voltage drop across the horizontal resistor is 0.

The sum of voltages along the right loop is equal to 0 (Kirchhoff Voltage Law), so the only terms remaining are the voltage drops across the capacitors C2 ad C34, that must be 0. An objection might be that I cannot infer it, because they might be any value, one positive and the other negative, and they would compensate each other ad give a 0 anyway. But the flow of current was unidirectional, so they must have the same sign, and thus the only solution is VC2=0 ad VC34=0.

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  • \$\begingroup\$ Interesting. It's actually counterintuitive considering there was a current flow present. Did they even have a chance to build up charge and lose it somehow or simply they didn't even charge? \$\endgroup\$ – primoshrimp Mar 26 at 8:39
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Only C1 get charged up to emf of the battery. C2 gets no charged because it has not connected to a voltage deference. It is short circuited by 2 resistors. Even i don't want , but I made this for you. V= 10v in C1, which is my battery emf. V= 7 mV in C2, which is = 0 ( due to leakage) I simplified right side caps as C2.enter image description here![enter image description here]enter image description here![enter image description here]

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  • \$\begingroup\$ It's not a good idea to use car batteries to power your little prototypes. One small slip and the little wires vapourise causing injury. \$\endgroup\$ – Kartman Mar 26 at 9:37
  • \$\begingroup\$ It is safer. Only the wire may burn. But car battery wont blast. I have a 12 V lipo. But I am scared to use it. \$\endgroup\$ – upali Mar 26 at 9:51
  • \$\begingroup\$ "A car battery won't blast"? You should tell Prevent Blindness America because they state otherwise. \$\endgroup\$ – Elliot Alderson Mar 26 at 14:53
  • \$\begingroup\$ What do you propose to use please? \$\endgroup\$ – upali Mar 26 at 15:07
  • \$\begingroup\$ Hello every body thanks for advicing me. Now I use the car battery with a fuse. You saved my life. \$\endgroup\$ – upali Mar 27 at 9:38
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It's correct for the steady state analysis. For the transient analysis, here are the simulated voltages across all capacitors : Simulation results

Using this simulation circuit. The unrealistic component values are there to simplify RC time constants. The peak happens at ~756 ms. Simulation circuit

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  • \$\begingroup\$ I think it is a home work. He wants an answer with lot of equations! \$\endgroup\$ – upali Mar 26 at 11:07
  • \$\begingroup\$ Mr.Harnex, what is the name of your simulater please? \$\endgroup\$ – upali Mar 27 at 10:49

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