0
$\begingroup$

I have the following circuit

enter image description here

The switch has been open for a long time, and both capacitors are fully charged. After that, the switch is closed and the circuit is broken. I need to determine the change in charge $Q$ for each capacitor after a long period of time.

Why is that the change in charge through each capacitor is simply not the maximum charge each capacitor can hold? I see it like this. Each capacitor becomes fully charged and this charge can be determined as $Q = CV$. After the switch is closed, these capacitors begin to discharge through the circuit. However, after a long time has passed each capacitor completely discharges all of its charge as per $$Q(t) = VCe^{-\frac{t}{RC}}$$

$\endgroup$
2
  • $\begingroup$ What you said seems right, all the charge will go, the resistances just slow things down... $\endgroup$ Jun 16 at 12:52
  • $\begingroup$ No, there will be a voltage across each resistor, and hence across each capacitor. Do you know what a voltage divider is? $\endgroup$ Jun 16 at 13:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.