Why Ca takes twice the time to charge as compare to Cx?

In following diagram there are two capacitors Cx and Ca, which are going to be charged in steady state. Now I know the method to calculate pole for simple circuits and after that take reciprocal to calculate time constant. By this method I know that charging of Ca will be slower than Cx. But I still don't get the intutive understanding of it.

The fact that it is still charging same size of capacitor and taking twice time confuses me, might sound like a stupid question but I have spent my two days to get a feel of why this would happen but still did not understand.

Please explain if my question makes sense to you. Charging of capacitor

  • 2
    \$\begingroup\$ CA and CB are virtual parallel via GND and you have 2pF to charge \$\endgroup\$
    – Jens
    May 28, 2022 at 23:16
  • 1
    \$\begingroup\$ While the question "what is the time constant" is still good, this test setup is flawed. If both capacitors start discharged, they are effectively a short circuit directly across a 1V source at time t=0. No algebraic solution exists for the state at time t=0, because current would be infinite, unless they both start with charges totaling 1V, and not even necessarily half each. Include a source resistance from Vin would help, but that would cause the capacitor voltages to rise at different rates, and would completely change the graph. Is it then appropriate to assign a time constant? \$\endgroup\$ May 29, 2022 at 4:11
  • \$\begingroup\$ @Jens, at steady state only Ca will be charged not Cb. Though at t =0+, both the capacitors will be charged in almost zero time(if everything is ideal) to 0.5V each, but as time passes Ca will continue to charge and Cb will discharge by resistor. Therefore, I don't thing we can say we have 2pF to charge. Please correct me if am wrong. \$\endgroup\$ May 29, 2022 at 9:17
  • \$\begingroup\$ @SimonFitch, Yes it is true that the capacitors will not be charged to 0.5V(at t = 0+) each if they are real capacitors, but if they are ideal and atleast how these simulators work, it is indeed 0.5V each. Also, can you propose a method to calculate time constant of the above two circuits(if I don't want to include source resistance)? \$\endgroup\$ May 29, 2022 at 9:24
  • \$\begingroup\$ @sumitasahu: The resistor changes the charge of both capacitors and both changes need time, it is symmetrically. Consider the VIn supply as an unlimited big capacitor and then CA and CB are effectively parallel. \$\endgroup\$
    – Jens
    May 29, 2022 at 13:52

1 Answer 1


Look at your circuit, Ca is more or less connected with Cb in parallel, although Ca is connected to a voltage. Find the equivalent capacitance seen by the resistor after shorting the voltage and you will see that the resistor sees 2 pF. Cb has to charge as well, when the input voltage is applied.


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