Circuit analysis - series capacitors

1. Apr 1, 2014

Steve13579

I am hoping that general physics is the correct location for this problem, but if I am wrong please say so. I have a problem where two capacitors of different value are in series and are attached to a DC source.

1. The problem statement, all variables and given/known data

Image is a bit blurry but that is how our professor scanned it. This is practice he said we should do, and I solved it for V total but not individually.

relevant formulas: C(dv/dt) tou=RC

attempt at solution:
at t=0; v=0
at t=∞; v=12

(v-12)/100k + c(dv/dt) = 0
(dv/dt) + 5v = 60
than solving the differential equation I get
v = c1e-5t + 12
solving for c1 I got the final solution
v(t) = -12e-5t + 12

How would I go about solving individually? And did I even get the correct solution? It is logical for t=0 being 0v and at t=∞ getting 12v

Thanks,
Steve

Last edited: Apr 1, 2014
2. Apr 1, 2014

Steve13579

I figured it out, sort of.
You set up:
C(dv/dt) + (v-4)/(100k) = 0 and
C(dv/dt) + (v-8)/(100k) = 0
where C = 2uF
The part I still don't understand is why capacitance is the same for both equations but voltage is not.

3. Apr 2, 2014

Maylis

I'm working on this problem as well. It turns out that problem you posted comes out of the textbook that my professor wrote , funny enough. You should use voltage division to find the voltages at t=infinity

I'm stuck how how the time constant is the same both of the capacitors, isn't it just R*C but C is different?

4. Apr 3, 2014

Staff: Mentor

How do capacitors in series "add"? What capacitance does the rest of the circuit "see" that determines the time constant? In a series circuit can there be different currents in different components? Can current be flowing in one part of a series circuit but not in another?

5. Apr 3, 2014

Maylis

The phrase ''what capacitance does the rest of the circuit see'' is just too vague for me to be able to make anything useful out of it. Anything related to sight and I don't know, I can't see any circuit components vision. I thought that the time constant is the thevenin resistance times the capacitance. It is actually the thevenin capacitance times the thevenin resistance?

The way I understand the time constant conceptually, its the time required for the capacitor to reach ~63% of its max charge. If the capacitance is different, then it stands to reason that time to reach that 63% is different, no?

What if I kept the capacitors separate instead of combining them? Then one capacitor ''sees'' the other one and vice versa.

This is part of why I have issues with these problems. I drew up another one to contrast this problem, where the time constant is different for each inductor. To me, the problems look different and I cannot make the connection why in this problem in the time constant is different, but in the original problem they are the same. I think they should be solved in a similar manner. I am missing something fundamental I think that is underlying the concept regarding thevenin equivalents but not discussed in great enough detail in the book to know it without discovering it myself or having outside help. I definitely haven't discovered it myself.

For the problem I posted, I don't see what inductance and resistance each inductor ''sees''

Attached Files:

• problem 5.50 attempt 1.pdf
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Last edited: Apr 3, 2014
6. Apr 3, 2014

Staff: Mentor

"Thevenin Capacitance" doesn't have an official definition that I'm aware of, but I think I understand what you're driving at. A first order circuit (one containing just one type of reactive component) will have resistance and capacitance, or resistance and inductance. Reducing the circuit to a simple equivalent circuit of a lumped resistance and lumped reactance ill tell you how the whole thing will behave in terms of time.

If the circuit is simple enough that inductances or capacitances can be combined and reduced using the usual rules for parallel or serial components then that's the way to go. When things are complicated by having resistances in the way so that such simple combinations are not viable then one can resort to performing a circuit analysis (by whatever method) in the Laplace domain to derive the differential equation or transfer function as desired.

Yes, if the net capacitance is changed then the time constant will change. The net capacitance does not change for a fixed circuit. A given first order circuit can have only one time constant that applies to all voltage and current dynamics in the circuit.

The net capacitance of the fully reduced circuit model will still be the same. Overall circuit behavior depends on the actions of all components working in combination. Cutting out a component changes the circuit. If you cut out a component and reduce what's left, then tack the component back on, you must then reduce the new circuit again. You'll end up with the same equivalent model.

Your example is a bit misleading because the short circuit created by the switch at t = 0 effectively isolates the two inductive branches into separate circuits. The short circuit will not allow any current from one branch to flow through the other branch, and the short permits no potential difference from one branch to affect the other. Short circuits and fixed voltage sources can effectively isolate loops from interacting with each other.

7. Apr 3, 2014

Maylis

Thanks for the lengthy response. I was walking on the quad about 10 mins and your argument about the current clicked in my head. So the current stops in both capacitors at the same time. Hence the time constants must be equal.

I also took a look at my inductor circuit and I see what you are saying. My professor loves putting short circuits in exam problems to make complicated circuits at first site reduce to something more manageable as long as you notice the short circuit

8. Apr 3, 2014

Staff: Mentor

Yup

Yes, it's one of those sneaky tricks that one has to beware of, or perhaps more constructively, be ready to take advantage of

9. Apr 3, 2014

Steve13579

Thanks guys! My exam actually had two series capacitors, but it wanted voltage on one after a given time knowing current. I used C calculated from both capacitors and found overall voltage using 1/C integral i(x)dt, than used a ratio between the two capacitors. Is there a more direct root? And charge is same on both capacitors right, despite different capacitance values?

10. Apr 4, 2014

Staff: Mentor

If you know the current with respect to time for the particular capacitor you can directly perform the integration to find its charge and hence voltage.