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.
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.
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?
Best Answer
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.