
#1
Oct2906, 02:30 AM

P: 5

hey everyone, this is an extra credit problem which i really need to get in order to pass this class! haha i did horrible on 1 of my 3 tests!
But yeah any help would be MUCH appreciated, thanks! here's a link to the problem http://www.physics.odu.edu/~hoy/232Bonus%20Problem.pdf 



#2
Oct2906, 03:51 AM

Mentor
P: 4,499

Use your knowledge of circuits to write out as many equations as possible that have I_{1,2,3} in them. You should be able to set up enough equations to solve for the I's relatively easily.... once you have all the equations, if you still need more help, post what you have here. Keep in mind I = dq/dt




#3
Oct2906, 09:33 AM

HW Helper
P: 1,446

I would think that the potential over the capacitor need to be the same as that over R2.




#4
Oct2906, 03:45 PM

HW Helper
P: 1,446

REALLY NEED A PHYSICS EXPERT IN CIRCUITS, if you can help!!!
When the switch is closed the capacitor in effect shorts R2 out. This means that the current in the circuit is determined by R1 giving I1 = I2.
At the other end the current Iq will eventully be zero. This means that we again will have that I1 = I2 with the same potential over the capacitor and the potential over R2 the same as the capacitor is fully charged up. At this point the current will be Io = I1 = I2 = E/(R1+R2). Giving for the potential over the fully charged capacitor Vo = Io x R2. 



#5
Oct2906, 03:59 PM

Mentor
P: 4,499

Look again andrevdh... at the beginning, I_{1} = I_{q} and at the end I_{1} = I_{2}




#6
Oct2906, 04:45 PM

P: 5

oh man im so confused, does anyone have an aim sn?




#7
Oct2906, 04:47 PM

Mentor
P: 4,499

jevan, if you had a circuit with just a capacitor on it, would you know how to solve that?




#8
Oct3006, 12:13 PM

P: 5

im getting confused at the part where the current moves in 2 directions. all we were taught in the class were the loop rule and junction rule. since there are no junctions in this problem, im assuming we use the loop rule. also another part im getting confused with is the switch that is open on the bottom left. could you give me a run through on how i would go about to solve for the equations?




#9
Oct3006, 12:19 PM

HW Helper
P: 1,446

So at [tex]t = 0[/tex] we have that [tex]I_o = I_1 = I_q = \frac{\mathcal{E}}{R_1}[/tex] and after a very long time (infinity) we have that [tex]I_q = 0[/tex] so that [tex]I_i = I_1 = I_2 = \frac{\mathcal{E}}{R_1 + R_2}[/tex] In between we know that [tex]I_1 = I_2 + I_q[/tex] and that [tex]V_{R2} = V_C[/tex] 



#10
Oct3006, 04:49 PM

P: 5

so where do we use diff eq to solve?




#11
Oct3106, 03:25 AM

HW Helper
P: 1,446

With the switch open the capacitor will discharge through R2. So when the switch is closed the capacitor presents no "resistance" to the flow of current. That is it shorts R2 out by absorbing the flow of current into its plates. This can also be understood via the fact that the potential over the capacitor is zero when the switch is closed (it has been discharged). This means that the only resistance in the circuit at that point is R1. So the current through the circuit will be
[tex]I_o = \frac{E}{R_1} = I_1 = I_q[/tex] The function of R1 is therefore to limit the intial current flowing in the circuit when the switch is closed. Since there is no potential over R2 at this point (discharged capacitor) we have that [tex]I_2 = 0[/tex] when the switch is thrown. 



#12
Oct3106, 04:16 AM

HW Helper
P: 1,446

The capacitor will charge up to the "final voltage over R2".
At this point [tex]I_q = 0[/tex] so we have that [tex]V_{2i} = I_i R_2[/tex] with [tex]I_i[/tex] as in post #9 (i referst to when time = infinity) The cap will therefore charge up through R1 according to [tex]V_C = V_{2i} \left( 1  e^{\frac{ t }{\tau}\right) = V_2[/tex] where [tex]\tau = R_1 C[/tex] Use the loop rule to set up the equations. The charge, [tex]I_q[/tex], will flow into the capacitor as it charges up to [tex]V_{2i}[/tex]. 



#13
Oct3106, 07:42 AM

Mentor
P: 7,290

Can you shed any light on the DiffEq? 



#14
Nov106, 07:30 AM

HW Helper
P: 1,446

Note that
[tex]V_2 = V_C[/tex] which gives us that [tex]I_2 R_2 = \frac{q}{C}[/tex] from which it follows that [tex]\frac{dq_2}{dt} = \frac{q}{R_2C}[/tex] using [tex]I_1 = I_2 + I_q[/tex] we get that [tex]\frac{dq_1}{dt} = \frac{q}{R_2C} + \frac{dq}{dt}[/tex] applying the loop rule over R1 and C gives [tex]\mathcal{E} = R_1 I_1 + \frac{q}{C}[/tex] using the above this comes to [tex]\frac{\mathcal{E}}{R_1} = \left( \frac{1}{R_2 C} + \frac{1}{R_1 C} \right)q +\frac{dq}{dt}[/tex] 



#15
Nov106, 09:14 PM

P: 5

Thank You Very Much



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