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Yesterday I was browsing through some questions in preparation for my exam and I found this one:

Consider the circuit below consisting out of a capacitor C and two identical resistors R. For \$t<0\$ the switch is open and the capacitor is uncharged. At \$t=0\$ switch is shut and the circuit is connected to the voltage source with constant voltage U.

enter image description here

a) What's the total current in the circuit immediately after the switch is shut? What's the charge of the capacitor and the total current after a very long time?

b) Determine for \$t>0\$ the total current in the circuit and the charge of the capacitor as a function of time by setting up a suitable differential equation and solving it.

I didn't have an answer to that so I couldn't answer it and I don't have enough reputation yet to comment. There was answer but it wasn't explicit on b).

I had very similar problem in past exercises and I could never set up a differential equation and solve it. Could someone help me out here?

Edit: \$V_R=\frac{R}{2}\cdot I\$ and \$V_C=\frac{1}{C}\int Idt\$

With Kirchhoff's law it should be

\$\frac{R\cdot I}{2}+\frac{1}{C}\int Idt=\frac{U}{2}\$?

So differentiating with respect to t should be:

\$\frac{R}{2}\frac{dI}{dt}+\frac{I}{C}=0\$.

So that would give us \$I=\frac{2U}{R}e^{-\frac{2t}{RC}}\$, right?

But how do I get an expression for the charge on the capacitor?

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  • \$\begingroup\$ charge of capacitor: \$Q = V_C C\$ with \$V_C = U/2 - V_R\$. \$\endgroup\$ – Curd Jun 22 '15 at 8:32
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For t≥0 you can replace the voltage source with the two Rs (forming a voltage divider) by its Thevenin equivalent.

This simplifies your circuit (for t≥0) to:

enter image description here


EDIT:
Note: In the simplified but equivalent circuit the resistor's value is R/2 and the voltage of the voltage source is U/2.


You should be able to formulate the differential equation for this circuit.

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  • \$\begingroup\$ Could you look at my edit in the original post and see if I wrote it down correctly? And I also have trouble with the charge. \$\endgroup\$ – Fabian Henry Jun 22 '15 at 8:08
  • \$\begingroup\$ It looks good to me. \$\endgroup\$ – Curd Jun 22 '15 at 8:34
  • \$\begingroup\$ Thanks. Now I know how to deal with RC circuits and with voltage dividers. Although, now that I think about it, I'm kind of curious about a) of the exercise. What's the charge and the total current after a very long time? Isn't the total current just the \$I\$ we got from the differential equation? And since the current doesn't reach its maximum then the capacitor will not be fully charged? \$\endgroup\$ – Fabian Henry Jun 22 '15 at 8:39
  • \$\begingroup\$ Though the current never reaches a terminal value it asymtotically gets closer and closer to the "final" value 0. That value is meant (limit t--> infinity) by "after a very long time". The same is the case with voltage and charge. After several RC time intervals those values reach 0.9999...% of the asymtotic value. \$\endgroup\$ – Curd Jun 22 '15 at 8:46
  • \$\begingroup\$ @Chu: any linear circuit can by replaced ("combined" to) its Thevenin equivalent (which equals a voltage source with a serial resistor). If you don't know this, please look e.g. here en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem \$\endgroup\$ – Curd Jun 22 '15 at 12:33

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