RC Circuit w/ discharged Capacitor: How to find total energy delivered?

Thread Starter

Vertigone

Joined Dec 5, 2018
3
I'm currently brushing up on RLC circuits for an upcoming final exam, but I haven't encountered one as difficult as this kind of circuit. It asks to solve for RC, the voltage V2 across the R2 resistor, and the the total energy delivered to the capacitor.

Despite knowing that the time constant (for RC circuits) 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 by using the condition for t > 0 (or @ infinity) to find Req, I knew that the cap is in series with the resistor R1, suggesting that Req > 3k ohms... Upon further calculation, the Req came out to be 5000 ohms, leading to a time constant of 25 ms. I'm not certain if this was done right however.

For the voltage across resistor: by presuming that at for t<0 and t=0, the cap has 0 V, meaning that R1 should be parallel to R2, leading to Vr2(0) (the voltage across R2) = 1.5k/7.5k = 0.2V. After several time constants, I thought that the cap is fully charged at t = inf. Here is where I'm really confused: do I ignore R1 at this point, since the cap acts as a open circuit? I felt that this would make Vr2(t=0+) a potential divider problem. As such, since there is no current flowing in it (if dV/dT~=0), this would lead to get R2 || R3 = 3k, hence t=T0+ leads to 1V * 3k/(3k+ 6k) = 1/3V as the steady state condition! From there, I believed that between the 2 steady states, we have an exponential rise given by the time constant of the circuit. Can anyone verify whether I took a correct method to finding the voltage across resistor R2?

Finally, the real deal is solving for the total energy delivered to capacitor. My knowledge tells me that for the initial voltage of the cap, it is 0 V, since it is assumed as a fully discharged capacitor. However, I have no idea what is the condition for it at t = 0+. I suspect that that while the capacitor would usually reach steady state at t = inf, the voltage of capacitor is not 1 V. Beyond that, I have no idea how to actually solve for the energy delivered at this condition. Does anyone have ideas, suggestions, or advice as to how to do this?
 

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wayneh

Joined Sep 9, 2010
16,399
I’d start by simplifying the drawing by collapsing the resistors into their equivalents. Then I’d scratch my head and wonder whether the question is asking for the energy stored on the cap, which would be easier to calculate, or the energy provided by the source to provide that charge, which seems to be what’s being asked. I don’t think they’re the same.
 

Thread Starter

Vertigone

Joined Dec 5, 2018
3
I’d start by simplifying the drawing by collapsing the resistors into their equivalents. Then I’d scratch my head and wonder whether the question is asking for the energy stored on the cap, which would be easier to calculate, or the energy provided by the source to provide that charge, which seems to be what’s being asked. I don’t think they’re the same.
Wait, so the energy provided by the source? What do you mean exactly by that? I did take the opportunity to simplify the circuit down to one equivalent resistor of 2k ohms (this excludes the R1 since it's ignored). However, what I'm left with is the voltage source, the said equivalent 2k resistor, and the open circuit portion of the capacitor. I recall the equation of the energy being: w = (1/2)(C)*(V)^2, and I keep thinking the V is referring to the voltage of the capacitor at t = infinity. Is this perhaps where I'm wrong in my line of thinking?
 

wayneh

Joined Sep 9, 2010
16,399
Wait, so the energy provided by the source? What do you mean exactly by that? I did take the opportunity to simplify the circuit down to one equivalent resistor of 2k ohms (this excludes the R1 since it's ignored). However, what I'm left with is the voltage source, the said equivalent 2k resistor, and the open circuit portion of the capacitor. I recall the equation of the energy being: w = (1/2)(C)*(V)^2, and I keep thinking the V is referring to the voltage of the capacitor at t = infinity. Is this perhaps where I'm wrong in my line of thinking?
That describes the stored energy in the capacitor and is always (and instantaneously) correct for any voltage. It’s not necessarily the same as the energy supplied by the source. That depends on the path, not just the thermodynamic endpoints.
 

Thread Starter

Vertigone

Joined Dec 5, 2018
3
Hmm, so if both of those are not the same, then considering that the energy provided by the source is what is being looked for, that energy must be affected by the equivalent resistor along the path too then, right?


 

MrAl

Joined Jun 17, 2014
7,806
Hello,

The total energy delivered to the cap is a reasonable question i think.
Since the response will be exponential only, you could solve for Vc(infinity) and then E=(1/2)*C*V^2.
That's what it looks like anyway and should not be that hard to calculate using for example Nodal Analysis.
 
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