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I have this circuit (picture below) and at first I had to calculate the eletric charge of the capacitors without the resistor which is parallel to C1. Which was not a problem at all. After that the resistor came up parallel to the C1 and I should calculate the electric charge again for both capacitors. The voltage is DC and the electric charge for t -> infinity.

I'm not sure how to calculate it because I think the resistor does not change anything.

Here the circuit:

enter image description here

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    \$\begingroup\$ Think about it. Initially uncharged (perfect) capacitors are shorts. Then, after an infinite amount of time they open. \$\endgroup\$ – st2000 Jun 22 at 23:28
  • \$\begingroup\$ R1 at t>infinity effectively nullifies C1. C2 then sees the full 800V. \$\endgroup\$ – vini_i Jun 22 at 23:31
  • \$\begingroup\$ The DC current of a capacitor is zero. Apply KCL and Ohms law to R1. \$\endgroup\$ – sstobbe Jun 22 at 23:34
  • \$\begingroup\$ So the Q = C*U of C1 is zero? At beginning when t=0 both capacitors charge and after a amount of time the C1 discharges through R1 and therefore the charge of C1 gets zero again? Is that right? \$\endgroup\$ – Manuel Jun 22 at 23:39
  • \$\begingroup\$ Stop! Answering! In! Comments! Gah. \$\endgroup\$ – pipe Jun 22 at 23:39
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No.

As long as there is a potential difference across R1, current will flow through it and into C2. This capacitor will eventually charge up to the point where no current can flow through R1, which implies that the voltage at both ends of the resistor is the same (V=R*I, if R!=0 and I=0 then V=0).

At this point, your capacitor C1 sees the voltage V1 at both terminals, i.e. the voltage across your C1 capacitor will be zero, hence it holds zero charge.

A easy way to visualize capacitors in DC circuits is to view them as open circuits as time tends towards infinity. Basically, if you remove C1 and C2 from your circuit, your R1 is left "dangling" in the air. If you now look at the terminals where C1 was, you have the terminal connected to V1, and another connected to your "dangling" resistor. Again, since no current flows through the resistor, the voltage in both terminals is V1 in reference to ground, hence no voltage across the resistor nor across the nodes from where you removed the capacitor.

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