Electrical Engineering-- Capacitor Resistor Circuit

n00bcake22

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

Hello all. First time posting so excuse any fourm faux pas I may make. I was wondering if anyone could lend some help with the circuit below (ignore the *, they are just space holders to prevent reformatting of the circuit):

|------) |-----) |--------\/\/\/\/\----------|
|*****C*****C*********R********** |
|********************************|
|--------------------/*** -----------------|
****************Switch

Here are the details of my question:

C1=C2=C (value unknown, used as the energy source, and both charged to 2.5V initially)
R=500 Ohms

Question: Find the minimum value of capacitance such that when the switch is closed at t0 = 0 to t=1ms, the voltage across the resistor (R) does not drop below 4.9 Volts. Please show/explain equations and work.

2. Relevant equations



3. The attempt at a solution

Below is what I have (let me know if any of it is wrong):

Kirchhoff Voltage Law
- v_c - v_c = v_r
2 * v_c = v_r

Inserting voltage equations for v_c and v_r
2 * ( 1/C * Integrate( i_c(t) dt, t0, t) + v_c0 ) = i_r(t) * R

Series circuit so i_c(t) = i_r(t) = i(t)
2 * ( 1/C * Integrate( i(t) dt, t0, t) + v_c0 ) = i(t) * R

Now I'm stuck; how can I express i(t) as a function of knowns and/or C.

Thanks again for all the help!!!

Zryn

I've attached a paint picture of what I think you were aiming for, which came out looking weird on my screen.

Essentially, you need to find a value of capacitance for the circuit so that when the switch is closed, the RC time constant is sufficiently large so that less than 0.1V is discharged, i.e. more than 4.9V across the resistor at all times during the 1ms that the switch is open.

This forum has a knowledge repository which is quite useful, here is a link to the section on capacitors:

http://www.physicsforums.com/library...tem&itemid=112

Specifically, if you check out the voltage and current charging/discharging exponential equations, you may be able to rearrange and solve. Additionally, keep in mind how voltages in series combine, and how capacitors in series combine.

Attached Thumbnails

 

n00bcake22

I appreciate the help. This question is actually Problem 3.26 in Electrical Engineering: Principles and Applications, 3rd (Hambley). Up to this point, transients (and, therefore, time constants) have not been discussed (covered in the next chapter) so I was wondering how to come up with the correct answer using what has been covered thus far. If anyone else could offer some advice I would really appreciate it. Below are the "primary" equations for capacitors offered so far by the text (note: i will use _ to indicate subscript, i.e. v_0 would stand for initial voltage):

q = C*v : (q= charge, C= capacitance, v= voltage)
i = C* dv/dt : (i = current)
q(t) = integrate(i(t) dt, t_0, t_f) + q(t_0)
v(t) = 1/C * integrate(i(t) dt, t_0, t_f) + v(t_0)
w(t) = 1/2 * C * (v(t))^2 : (w(t) = stored energy)
w(t) = 1/2 * v(t) * q(t)
w(t) = (q(t))^2 / (2*C)

The book states that the correct value for C is 198 micro-Farads. If anyone could help me out with how to arrive at this answer using the above equations and basic circuit laws (KVL, KCL, etc.) I would really appreciate it. Thanks again everyone!

Zryn

Well, lets not let a little thing like having not been taught get in the way! How are you with algebra and calculus?

Using Kirchoffs Voltage Law around a closed loop (assuming the two capacitors voltages are adding together to create one source for now), your original equation is slightly wrong.

After you correct it, see if you can rearrange those voltages into currents and resistances.

Next, see if you can solve the differential equation.

Next, rearrange the result of this differential equation algebraicly to get C as your target.

Next, stick in the values and solve.

Keep in mind the way capacitors combine in series, and presto chango you should see ~197.99uF. This is a very 'first principles' and mathematical way of doing things, and hopefully you won't have to do it too often, because its difficult and time consuming and really only needs to be done to see how much nicer it is when you can jump to the final equation instead of doing it all from scratch.

n00bcake22

Hello again. Well, I was able to get a look at the solution and they did some simplifying of the problem. They "assumed" a constant current discharge instead of time variant (which makes sense, I suppose, as they haven't yet discussed transients) and that's where I ran into my little conundrum-- knowing that, in reality, that is not the case.

Thanks again for all your help. I am a mech. eng. so I get to have a super crash course in EE with little time to digest each topic before being rushed onto the next. Sigh... time to dig up my old electromagnatism text. Yahoo!

Zryn

So that was just rearranging [tex]I\ =\ C\frac{dV}{dt}[/tex] for C and plugging in the numbers? Turns out that that is a reasonable approximation in this case.

Sorry for overly complicating things, I thought I was helping an EE! Good luck with the next topic too.

n00bcake22

Yeah that is exactly what they did, but in a rather strange way (they fixed both current and voltage throughout the delivery pulse--which makes no sense for a capacitor... right?). In my opinion, this was a terrible problem to have without first discussing transients (they should have saved it for the following chapter). It was an unnecessary, oddly worded problem.

Anyways, thanks again Zryn for all your assistance; btw, I actually enjoy the "complexity" of engineering. It's amazing how much better you understand the material once you understand all of the underlying theory and mathematics.

All the best!

p.s. I'm going out on a limb here. This website... best find in 2010!!!

Zryn

Well, any model representing something (capacitors for example) can be "correct" depending on the assumptions that govern it. Not defining those assumptions makes the understand part a bit difficult though.

In reality both current and voltage are not fixed throughout the delivery pulse indeed. In theory they discharge as a time dependant exponential ... which is asymptotical around zero and thus takes infinite time to reach ... and yet in practice they do reach zero within five (ish) time constants. Theres always assumptions.

And I concur, this website is a strong contender for my best find in 2010 too!