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Here I am trying to find the input current I(t) and Vc(t).

From my calculation, it's evident that Vc(0+)=Vc(0-)=0V and Vc(∞)= 10V but I am confused about the first capacitor. When t=0, it acts like a voltage source of 10V but does it supply current to the 2nd capacitor? The second capacitor was shorted when t=0+,then I(t) was flowing through it. Hence,I(0+)=10V/5kohm=2mA.also I am confused about the time constant of the circuit. According my assumption, C1 acts like an independent voltage source from t=0- to t=∞. Then would the time constant be (5*2)=10? Is my assumption correct?

enter image description here

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When t=0, it acts like a voltage source of 10V but does it supply current to the 2nd capacitor?

Yes it does and that current is infinite at t = 0 because connecting a discharged capacitor (C2) to a charged capacitor (C1) creates that possibility.

enter image description here

So, C2 and C1 instantly share/distribute the charge previously contained in C1 and, the resulting voltage across the connected pair is reflected by the charge equation Q = CV.

So, work out what the charge is on C1 before the switch closes and then you can calculate the voltage across both capacitors when the switch closes. That voltage will then start to rise exponentially with twice the time constant you mention because the capacitance is now 4 mF.

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This is the well known capacitor paradox. The normal conventions of circuit theory break when you close the switch, because you have two different voltages in parallel and neither of them can change instantaneously. To use normal circuit analysis techniques you will need to introduce some additional assumptions or circuit elements that are not shown in the problem.

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  • \$\begingroup\$ +1 for referring to the capacitor paradox and ensuing issues with the circuit as drawn, but do these reservations apply to the question as asked? The charge distribution and voltage question can be answered, esp. since it is given to be non-oscillating, e(-t/T)), and since the paradox is in the energy question? \$\endgroup\$ – P2000 Feb 5 at 18:22
  • \$\begingroup\$ I think the question should stipulate something about what happens to the two capacitors when the switch closes, just to prevent this sort of confusion. Actually, I don't think it is a very well designed question. \$\endgroup\$ – Elliot Alderson Feb 5 at 18:39
  • \$\begingroup\$ Not well designed no. That's the didactic mistake: raising unrelated questions that draw attention away from the lesson. \$\endgroup\$ – P2000 Feb 5 at 18:41
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This is a special case. Vc(0+)=Vc(0-)=0V doesn't stand here. What stands is conservation of charge. The charge of the two capacitors is the same before and after switching.

C1*V1(0-) + C2*V2(0-) = (C1+C2)*V(0+)

After figuring out the voltage on the two capacitors you will have the situation as if you are switching a pre-charged capacitor (two connected in parallel).

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ So if Vc(0+) does not equal Vc(0-) then what does happen? How can we get to a solution for the current through the resistor as a function of time? What happened to the energy that was lost when the two capacitors were connected together? \$\endgroup\$ – Elliot Alderson Feb 3 at 13:36
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    \$\begingroup\$ @ElliotAlderson Imagine what happens if there is another one very small resistor between the two capacitors. First capacitor charges second one very quickly; excess of energy is dissipated. Current through your big R remains zero between 0- an 0+. Then you start as if the capacitor (4mF) is already charged (halfway in your case). \$\endgroup\$ – AlexVB Feb 3 at 13:47
  • \$\begingroup\$ I understand this, but I think your answer skipped over it. The OP is trying to write a function of time for the currents in this circuit, and the reality is that your "very quickly" is really undefined. I'm not sure this answer leads easily to a function of time. \$\endgroup\$ – Elliot Alderson Feb 3 at 14:28
  • \$\begingroup\$ @ElliotAlderson I just didn't want to turn it into homework solving service. "Very quickly" bit is a way to understand what would happen if one used a real switch. In the idealized problem like this "very quickly" fits between 0- an 0+. Anyway, technically you solve the problem by applying "charge don't change" rule instead of "voltage don't change" rule and this is what I posted in my answer. \$\endgroup\$ – AlexVB Feb 3 at 14:52
  • \$\begingroup\$ Well, if capacitors share charge then QC1(0-)= 0.02C and VC2(0-)=0C( as C2 is initially uncharged). Then VC2(0+)= 5V. Then, the VC2 increase with time constant 20 and becomes 5e^(-t/20) then what happens to VC2 at infinite time?? And what's happen in I(t)? \$\endgroup\$ – Camila's voice Feb 3 at 15:14

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