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Most Rc circuits are not as severely difficult as this kind of circuit. I'm trying to solve for both the time constant and the voltage V2 across the R2 resistor, both of which will hold the necessary values needed to find the total energy for the capacitor. enter image description here However, despite knowing that the time constant is RC, I'm uncertain as to whether I ended up with the right equivalent resistance, Req, that's used to find the time constant. I was under the assumption that with the modified approach I took to find Req, it came out to be 5000 ohms, leading to a time constant of 25 ms. I'm almost certain this part is done right. enter image description here For (b), I understood that because a discharging capacitor is involved, the initial voltage for the capacitor would be zero: vc(0-) = 0, such that it would equal vc(t=0), based on the given assumption. Unfortunately, beyond this, I am stuck on where to proceed next in finding vc(0+), V0 (initial voltage) and other key variables needed to solve for V2. My instinct tells me that finding the voltage across the capacitor, Vc(t) is likely to be needed for (c), but I could be wrong about that. For an RC circuit with a discharged capacitor, I am still confused as to how to approach this circuit. And then there is finding the total energy of the capacitor, which I believe is dependent on the voltage across capacitor too?

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  • \$\begingroup\$ You need to find the Thevenin equivalent resistance with the switch closed. It's obviously higher than 3K because there is 3K in series with the capacitor. Short out the voltage source and redraw it- the calculation can be done in your head. \$\endgroup\$ Dec 2 '18 at 18:57
  • \$\begingroup\$ Don't be distracted by R3 being drawn diagonally - it's in parallel with R2. When you've redrawn the circuit, simplifying the R4/R5/R6 configuration is easy. \$\endgroup\$
    – Chu
    Dec 2 '18 at 19:55
  • \$\begingroup\$ Indeed, I believe this to be true, upon redrawing the circuit diagram. The R4 || (R5 + R6) configuration turned out to be in parallel to (R2 || R3), leaving out R1 as being in series with all of that. However, considering that the circuit is redrawn, wouldn't I have to re-draw the original circuit again (after the switch is closed) to find out the voltage for the resistor? Because while I did end up with a Req ~5k, the capacitor would become an open circuit as t goes to infinity, right? \$\endgroup\$
    – Heavywake
    Dec 3 '18 at 4:54
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Apply a sanity check, the cap is in series with a resistor R1, so Req clearly cannot be less then the 3k series resistance....

I would start by rolling R4,5,6 up into one resistor, and then combining R2,3 into a single resistor, just to simplify the thing (Resistors in series and parallel).

Then for B (and maybe it is just me) I would convert the 1V voltage source in series with a resistor to a current source in parallel with a resistor, which at T0+ ends up in parallel with R2.... Combine with R2 and convert back to voltage source in series with resistor, at which point you have 1 loop of three series connected components (plus voltage source), should be soluble from there I would have thought?

Hint, once you have that little series network, solve for current then calculate the relevant voltage.

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  • \$\begingroup\$ Yes, I came to understand that the Req is definitely more than 3k ohms, and after practicing your configuration of the R4/R5/R6 resistors and the R2/R3 parallel resistors, that confirmed it. \$\endgroup\$
    – Heavywake
    Dec 3 '18 at 7:59
  • \$\begingroup\$ However, I am confused about the approach you mentioned for (b). From what I understand, you're suggesting on building the Thevenin equivalent of the circuit for t = 0+, because with including the voltage source and the resistors, Kirchoff voltage law can be used to find the current source, which lead to the voltage across resistor R2. But doesn't this only account for the t > 0 case? Is the voltage across the capacitor not relevant? \$\endgroup\$
    – Heavywake
    Dec 3 '18 at 8:07
  • \$\begingroup\$ Try this reasoning: At T0, the cap has zero voltage across it, so R1 is in parallel with R2, making Vr2(0) = 1.5k/7.5k = 0.2V. After many time constants the cap is fully charged so you can ignore R1, making the voltage across R2 a simple potential divider problem, between these two conditions you have an exponential rise given by the timeconstant of the circuit. Where I was going with the Thevenin and then Norton transforms was ending up with a series circuit but whatever works. \$\endgroup\$
    – Dan Mills
    Dec 3 '18 at 10:42
  • \$\begingroup\$ Ah, I see where you're going with that: So I do get the fact that the cap, having 0 terminal voltage, acts as a short-circuit, making R1||R2. But I'm puzzled as to how you ended up with Vr2(0) = 0.2 V? Did you perhaps simplify the circuit down by combining all the resistors via parallel/series into a single loop to find that? Also, as you said, at t=0+, the cap meets is fully charged (1V) and becomes an open circuit. So from that point, is that the reason why R1 is neglected? \$\endgroup\$
    – Heavywake
    Dec 3 '18 at 11:05
  • \$\begingroup\$ Fully charged is NOT 1V because you have a series combination of R2 || R3 and that mess on the right (R4/5/6). I got 0.2V from R1 || R2 || R3 = 1.5k, R4,5,6 simplifies to 6k, so 1V * 1.5k/(1.5k + 6k) = 1/5 = 0.2V. For the fully charged case you can ignore R1, as there is no current flowing in it if dV/dT~=0, so you get R2 || R3 = 3k, hence T0+ gets you 1V * 3k/(3k+ 6k) = 1/3V as the terminal condition. You know the timeconstant so just write the expression for Vr2(t). Pay attention to the polarity of that voltage source. \$\endgroup\$
    – Dan Mills
    Dec 3 '18 at 12:12

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