Help on First order Diff Eq problem

rude man

You're right, sorry. I meant R.

pghaffari

Voltage is somehow still in the Resistor?

rude man

No, a resistor can't store voltage. What componernt can?

pghaffari

The capacitor

rude man

Right! Now, when the switch was closed for a long time, what voltage do you think might be stored on C?

pghaffari

entire 2 Volts? ..what about the v(t) ? Is that just the voltage drop across the resistor? SO it doesn't matter to the Capacitor?

pghaffari

Wait wouldnt it just be 1/c integral i dt

rude man

Yes. After a long time, C is fully charged and i = 0, so no drop across R₁.

So now we open the switch, and what do you think i(0+) has to be?

I have to take a break, will be back in 3-4 hrs. Think some more about what the circuit is really doing once the switch is opened, what the source of i(t) is, what i(0+) has to be. The result will be your value for R.

pghaffari

Source of i(t) is the Capacitor.. where Vc = 1/c integral (i dt)

We set that equal to 2 because thats how much voltage is stored in the capacitor at before switch is open for a very long time. If we do that we get ::

2 = 1/c integral idt

2c = integral idt

take derivative of both sides:

i(t) = 0.

This makes sense I think because I(0-)has to be I(0+) = 0 .. ?

butthen how would i find R??? R = V/I .. if I = 0, then I cant solve for R,

Not sure about this :| Ill think some more> Let me know if any of this makes sense when you get back.

Thank you

rude man

∫i(t)dt from -∞ to 0 is a constant, it's a definite integral and its time-derivative is thus zero. You can't come up with the voltage on C at t=0+ that way.

Instead, just look at the circuit and what you have after the switch has been on a long time: +2V sitting to the left of R₁, charging up C to what value?

pghaffari

Charges up to 2V? So C = 2? How many Farads would that be ?

IF C = 2, Then we can just plug it into the euqation


C = (-2*10^-6)/(R+1) and solve for R?

Can you explain to me why entire 2V is stored into the Capacitor?

pghaffari

Source of i(t) after switch is open: The Conductor
if i(0-) = 0, then i(0+) = 0

Right?

rude man

Why would C=2 just because its voltage is 2V???

Take the 2V source, connect via R1 to an uncharged C, and write the equation for i(t). Tne voltage on C will be (1/C)∫i(t)dt integrated from t= 0 to t = ∞. That'll give you your initial voltage on C for your particular problem, where the switch is opened at t = 0.

Have to go to bed. Will probably not get much of a chance to help further until tomorrow late afternoon.

pghaffari

Ok that makes alot more sense..


So basically I think i figured it out. Ths is what I did:


After opening the switch: I applied 2V to the right hand lop like you said and i get THIS equation:



2 + R1 + i(t) + 1/c integral(0-> infinity) i(t) dt = 0

BUT WE KNOW that i(T) = e^-t/R(C+1)

So we plug that in for i(t) and integrate from 0->inifnity and we end up geting

2 + R1 + i(t) - R(c+1)/c = 0

BUT because this is AFTER we open switch, we know that i(t) has to EQUAL 0, because i(0-) = 0.

SO equation becomes:

2 + R1 - R(c+1)/c = 0


We can solve the top equation and we end up getting:
2C = R


Now we have a R-C relationship. We plug thisback into our old equation:

Eq 1) R(c+1) = 2/10^-6
Eq 2) 2C = R

Plug both of them in:

We end up getting a quadratic equation and I end up finding C and R as:


C = 1/10 (5 +- sqrt(26)

and since R = 2C

R = 1/ (5 +- sqrt(26))

But which do we take? the Plus orthe Minus? Im assumnig Positive value because we want it to be postive/higher?

Not sure if this is correct, but it feels right.

Let me know!
THX FOR UR HELP AGAIN@!!

jrive

As you have now figured out with the help Rude Man, after the switch has been closed for a long time, there are 2V stored in the capacitor. As soon as the switch is opened up, the capacitor discharges through R and R1. With this you have all the information you need to solve for R.

You had obtained the equation for the current through the circuit earlier: i(t)=K*e^(-t/RC), where R is R+R1. Knowing the initial voltage across the cap, and the current flowing through te circuit at t=0+, you can determine K, and therefore the value of R.

With the second piece of information given to you in the problem, you can now solve for C. You are given the resistance of R1 and you have the equation of the curren tflowing through it as function of time. Since V_acrossR1=I*R1, V_acrossR1(t)=K*e^(-t/RC)*R1. you can now solve this equation for t=2usecs.

hope this helps....

pghaffari

my value for i(t) is actually K*e^(-t/(R1(C+1)) where K = 1.

Not sure how you got C(R1+R) under the divider?

ANyways, can you look at my previous post and see if I did it correctly. thanks

jrive

K has to have some physical meaning.....in this case the current at t=0+.
The only way the circuit can produce the 1A at t=0, given that the capacitor has been charged up to 2V, is to divide that voltage by the total resistance in the circuit (after the switch is open) through which the capacitor discharges.

so, K=Vcap_{at t=0}/(R1+R). You're right that K must equal 1....but the 1 has to have physical meaning -- it is not just a unitless number -- it is 1amp.

By the way, your equation should be : i(t)=K*e^-t/((R1+R)C), for t>0...since the capacitor discharges through both resistors.

Keep at it....it will come to you...

sorry if i'm unclear or too cryptic, I'm struggling a little bit between providing responses that help you think through the problem so you can understand what's going on and figure it out, and just "giving it away" in trying to explain the solution.