I've been struggling with this difficult RC circuit that asks for the voltage, Vo, across the R4 resistor.

Edit: Assume that the switch has been opened for a long time, before it is closed at time t = 0 s.

enter image description here I've already calculated the equivalent resistance for the circuit to be 0.75 ohms, under the condition of the switch closing at t=0 seconds. From there, I found the time constant to be 1.5 seconds. That being said, I am unsure as to how to proceed in solving for Vo. enter image description here I was under the assumption that, for t < 0, the left side of the switch would indicate what the initial voltage of the capacitor would be. Since the capacitor lies parallel to the V1 voltage source, I presumed that Vc(0-) = Vc(t=0) = 12 V. Now, I am unsure if this is correct, but since no current flows into the capacitor, it would represent as an open circuit. From that point, it's to my understanding that the final voltage, V_f (where t goes to infinity), needs to be found in order to solve for the voltage across capacitor, vc(t), which represents the following equation:

*vc(t) = V_f + [Vc(t=0) - V_f]e^(-t/RC)

Now I'm not sure how exactly does this relate to finding the voltage across resistor R4, but I do know that as time goes to infinity, the total current flows only through the short circuit of the switch. But is there any part where I'm wrong in my line of thinking? If I simplify the original circuit down to just the R4 resistor, the combined voltage sources, and another equivalent resistor, would that be the final step in solving for Vo?


1 Answer 1


Ignore the capacitor to calculate the output voltage just before the switch opens.

Ignore the capacitor to calculate the output voltage a very long time after the switch closes.

It goes from one voltage to the other with a time constant of 1.5 seconds- so you know there's a term \$e^{-t/\tau}\$ in the answer.

  • \$\begingroup\$ Apologies for not specifying the circuit, but the switch has been opened for a long time before it is closed at time t = 0. You're definitely right in that the term is included, but what I'm perplexed about is when you mentioned that the capacitor should be disregarded while the switch is opened. If that's the case, shouldn't R3, R4, and V2 be ignored as well, considering that they're along the path of the switch? \$\endgroup\$
    – Retrovert
    Dec 3, 2018 at 7:16
  • \$\begingroup\$ The in the first case, the switch is open and you can ignore R3 and V2 and the capacitor. That's the initial voltage. In the second case the switch is closed and only the capacitor can be ignored. That's the final voltage. \$\endgroup\$ Dec 3, 2018 at 7:18
  • \$\begingroup\$ +1, thank you for that clarification. So the first case is that much clearer that I can understand, however, I still am having trouble why the capacitor is ignored after the switch closes. Isn't the final voltage for t > 0 dependant on the voltage across the capacitor? Or is it simply just not needing to find that and just rather focus on finding the initial and final voltages for the R4 resistor? \$\endgroup\$
    – Retrovert
    Dec 3, 2018 at 7:41
  • \$\begingroup\$ The final R4 voltage (after many time constants) does not depend on the capacitor value. \$\endgroup\$ Dec 3, 2018 at 12:32
  • \$\begingroup\$ Okay, I see what you mean by that. If my understanding of your reasoning is right, then since the capacitor reaches its final voltage, its current will reach 0 such that it acts as an open circuit, thereby being independent of the final R4 voltage. And so, from that standpoint, both the initial and final R4 voltage can be utilized in setting up Vo(t) then? \$\endgroup\$
    – Retrovert
    Dec 3, 2018 at 13:20

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