Question

Please help with problem #2 of the attached file. I'm asked to solve for Vc(t). Do you think that is supposed to say solve for Vc(s)? If not, how do I go about this problem?

Thanks

Answer 1

it's an exponential decay, I think.

Answer 2

I think ...
The circuit will oscillate or "ring" as the voltage goes to zero. So, underneath the curve I've tried and failed to draw, there could very well be damped oscillations. R damps the behaviour. There's an equation for this which is based on DAMPED HARMONIC MOTION as opposed to SIMPLE HARMONIC MOTION, and FORCED HARMONIC MOTION, and FORCED DAMPED HARMONIC MOTION. Am i boring anyone ?
http://perendis.webs.com

Answer 3

I don't understand what that definite integral notation was supposed to indicate so I'm assuming that the t0 as the only border of the integral is some infinitesimal moment after t=0 because otherwise that would just evaluate to 0 which doesn't make sense.

If we assume that's the case it's basically the same as saying that the integral is not meant to evaluate to 0 and instead is meant to evaluate to some constant number, then the task seems to be silly (therefore I don't know if I understood the question right).

All you have to do is set up the circuit and realize that you're dealing with a normal case of an discharging RC circuit that just has some fancy looking initial condition. The general setup for the problem is:
\[R \frac{dQ}{dt} - \frac{Q(t)}{C} = 0 \]
which is just
\[RI(t) - V_c(t) = 0 \]
if it helps you see it better. This is a linear ordinary differential equation of first order which has a known solution in the form of:
\[
[\frac{dQ}{dt} = - \frac{1}{RC}Q(t)
V_c(t) = V_0 e^(-\frac{t}{RC}
\]

which means all you'd have to do is put that fancy looking V0 given to you inplace of the V0 of the solution. Provided that is what that integral means. Once it's stated that for an initial moment t=0 the voltage across the capacitor is *expression that uses t0* it's assumed that the t0 is 0 because it said so a second earlier. But solving a definite integral where the borders are the same yields 0 as a result as stated by first fundamental theorem of calculus:
\[\int^a_b f(x) dx = F(b) -F(a) \]

Answer 4

oh ffs, why is there no edit button the mashed up equation is meant to say:
\[\frac{dQ}{dt} = -\frac{1}{RC}Q(t) \\
V_c(t) = V_0e^{-\frac{t}{RC}}
\]

Answer 5

@ljetibo the "technical graphics" site are dreadful, if my browser is anything to go by. I'm not convinced by the equation in your last post ... as I remember it, the circuit originally posted was an RLC circuit, and the posted equation doesn't contain the L bit. So, it can't "ring" or oscillate. But the RLC will oscillate, which is one reason why it has uses in radio reception and transmission.

Answer 6

@osprey
OP: "help with problem #2 of the attached file"
Problem #2 has no inductor drawn therefore it's not an RLC circuit. It contains a resistor and a capacitor which makes it an RC circuit.

I don't mind the graphics although I prefer them simple. I just like edit and history buttons and notifications on edit to be able to sound coherent and keep good track of the discussion.

Answer 7

@Babyslapmafro this is a RC circuit. what is "s" in your mind.

that is the real question, right?

Answer 8

@IrishBoy123 The only "s" that I can think of is one of the transforms - Is it laplace ?
@osprey I've just looked at the circuits again in relation to the post, and realised that I've got qs 2 and 3 muddled up. 2 is an RC circuit which wouldn't oscillate, just "time constant exponentially decay". It's 3 that would oscillate. So, apart from punching myself in the face, or custard pie, I guess I can only say that I think that the lousy sketch graph drawn earlier is an approximation to the answer. If the poster wanted to solve the differential equation using Laplace transforms - s - they probably could.

Answer 9

s being in units of time (seconds) ......is that what you're thinking @Babyslapmafro?

Answer 10

@ljetibo

Why do you subtract Q(t)/c

Answer 11

Why isn't it...

\[R \frac{ dQ }{ dt }+\frac{ Q(t) }{ C }=0\]

Answer 12

https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws#Kirchhoff.27s_voltage_law_.28KVL.29

The total voltage/emf availible in that circuit is the one that's in the capacitor. Capacitor empties (ergo minus) to create the current in that circle. See the little note under that circuit diagram to the right:

"The sum of all the voltages around a loop is equal to zero. v1 + v2 + v3 - v4 = 0"

where the v4 is the battery in that diagram. Basically the voltage losses on the 3 resistors must be equal to the voltage across the battery. Except in this case battery is the capacitor.

Answer 13

Ok I got it

Answer 14

@ljetibo

where does e^-(t/RC) come from?

Answer 15

nvm, discharging voltage for RC circuit, right?

Answer 16

you have to solve the differential equation. That depends on how your prof. wants you to do them. In general most of them have general solutions which you can use and just substitute whatever letters they used for your letters.
Googling "linear ordinary differential equation of first order" will help you out here.

Answer 17

http://tutorial.math.lamar.edu/Classes/DE/IntroFirstOrder.aspx

click on the first link: "linear" that's the diff. eq. you with proof for a general case solution.

Answer 18

@ljetibo

i got...

\[Vc(t)=R(1-e ^{\frac{ t }{ RC }})\]

Answer 19

doesn't look quite right. Not sure where you got R from. Initial conditions are what you should usually put in. That diff eq. just looks like your standard exponential growth solution.

$$\frac{dQ}{dt} = \frac{1}{RC}Q \\
dQ = \frac{1}{RC}Qdt $$
Which looks like the diff. eq. of the shape:
$$\frac{dy}{dt} = p(t)y + g(t)$$
where
$$p(t) = \frac{1}{RC} \\
g(t) = 0$$
The link says to assume that the answer is in the form (below):
$$\mu = e^{\int-p(t)dt}$$
From which point you can do it the "official way" or you can try to do it over variable separation and see if it goes. Results should be the same. I'll try to separate the variables since it's faster (hopefully). Starting from eq 2.
$$\frac{dQ}{Q} = \frac{dt}{RC} \\
\int_{Q_0}^{Q_1} \frac{dQ}{Q} = \int_0^t\frac{dt}{RC}\\
\ln\left( \frac{Q_1}{Q_0}\right) = \frac{t}{RC}$$
Since R and C are constants they go infront of the integral and you jsut do the empty one which is t. On the other side you have an integral of charge Q, that starts off with initial charge Q0 and as it empties over time t ends up with charge Q1; where it's mandatory that Q0>Q1. You actually get the difference of logarithms, but by the rules of handling logs and ln's: ln1-ln2 = ln(1/2). We raise everything by an exponential function to get rid of the natural logarithm:
$$\frac{Q_1}{Q_0} = e^{\frac{t}{RC}} \\
Q_1 = Q_0 e^{\frac{t}{RC}}$$

And that's the answer. Except it's obviously wrong somewhere because the exponent is positive which means that as time goes on you'll reach infinite charge. I missed a minus somewhere.

Answer 20

oh, yeah, you get the potential over time from this by dividing the equation by C because by definition C = Q/V so V = Q/C, then:
$$\frac{Q_1}{C} = \frac{Q_0}{C}e^{-\frac{t}{RC}}\\
V(t) = V_0e^{-\frac{t}{RC}}$$