
#19
Feb1312, 07:45 PM

P: 35





#20
Feb1312, 07:46 PM

HW Helper
Thanks
PF Gold
P: 4,433

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




#21
Feb1312, 07:47 PM

P: 35

The capacitor




#22
Feb1312, 07:48 PM

HW Helper
Thanks
PF Gold
P: 4,433

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




#23
Feb1312, 07:50 PM

P: 35

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?




#24
Feb1312, 07:54 PM

P: 35

Wait wouldnt it just be 1/c integral i dt




#25
Feb1312, 07:54 PM

HW Helper
Thanks
PF Gold
P: 4,433

Yes. After a long time, C is fully charged and i = 0, so no drop across R_{1}.
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 34 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. 



#26
Feb1312, 08:03 PM

P: 35

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 



#27
Feb1312, 11:02 PM

HW Helper
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PF Gold
P: 4,433

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 R_{1}, charging up C to what value? 



#28
Feb1312, 11:32 PM

P: 35

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? 



#29
Feb1312, 11:58 PM

P: 35

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



#30
Feb1412, 12:40 AM

HW Helper
Thanks
PF Gold
P: 4,433

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. 



#31
Feb1412, 03:31 AM

P: 35

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 RC 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@!! 



#32
Feb1412, 08:39 AM

P: 37

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.... 



#33
Feb1412, 11:36 AM

P: 35

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 



#34
Feb1412, 12:23 PM

P: 37

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. 



#35
Feb1412, 01:28 PM

P: 35

You are right, it should be C(R+R1), i had a typo. But also we are given that R = 1 ohm, so I just replaced it in the equation to get C(R1+1).
So now that we have i(t), we need another equation to relate R and C after switch is open. We know that before switch is open, 2V is stored in the capacitor and i(0) is 0. So rudeman told me that we can basically connect the 2V to the right circut @ t =0 to get a new equation. So an equation would be: 2 + R1 + i(t) + 1/c integral (0 to infinity) of i(t) and the i(t) on the left is essentially 0 because i(0)= 0 = i(0+) is this correct? then we solve 2 equations 2 unknowns for R,C ? 



#36
Feb1412, 02:48 PM

P: 37

The problem stated that the switch had been previously closed for a while (at rest). So, the cap has been charged to 2V (the same as the source), and there is no more current flowing through the cap.
Then, the switch was opened. The circuit became a "source free" circuit analysis problem, and if you recall, you wrote the equation for i(t) flowing through the circuit at this time using KVL ("the sum of all the voltage drops equals the sum of all the voltage rises in the loop"). Ie. 1/c∫idt +v(0)i(R+R1)=0. (notice how all the terms in the equation above a in units of volts, term 1= volts across capacitor, v(0)=initial voltage across capacitor, and term 3=voltage drop across resistors always good to verify units are consistent in any equation). You then solved this differential equation which gave you the result in the form of i(t)=K*e^{t/((R1+R)C)}. At this point, you have an equation that describes the flow of current in the circuit from the time the switch is open to eternity(while the switch remains open) This is all you need. From the initial conditions, you can solve for R as I stated in a previous post, since you know the voltage stored in the cap (and the voltage in a capacitor cannot change instantaneously) and you know the value of R1, and the measured current in the loop, 1 amp. Then, from i(t)=K*e^{t/((R1+R)C)} (you know the value of K (1amp),and now the value of R as computed from the initial conditions), you can solve or C at t=2usecs. 


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