Recent content by juicec

Unfortunately I have to sign off for the evening, I will look forward to continuing this tomorrow hopefully. I am almost there :) Perhaps with a fresh mind and a bit of scrap paper I can finish this off and report the findings to you tomorrow. Thanks again for all the help. I have been trying...

I'm not 100% sure how to answer this.

ah OK, interesting. So with that said now, we've got 2 time constants, and also 2 circuits? One for discharging and one for charging. Or in fact we have 4 circuits, 2 when charging (capacitor empty/full) and 2 when discharging (capacitor full/empty).

Initially only R1 is limiting what is going through the capacitor right? When the capacitor is fully charged, eh, I should probably know these answers but it would simply be I going through it since there is no resistance when its fully charged? -- Or does the capacitor "close" when its fully...

Thinking about it now... would it be a matter of subbing in our time constant into a charging/discharing equation and calling it a day?

OK, I'll take that as a, "awsome! perfect!" ;) lol OK so now with the time constants in toe, how do we attack the "explicit time dependence for all currents" ?

Yes, OK, so if you read the problem one more time... I would have 2 time constants, one when the switch is closed which we have now found above, and one when it is open where the capacitor is discharging into R2 which I would venture a guess at it simply being: tao = (R2)(C) ?

ok so that's the time constant of the circuit without the battery and the switch closed right? Does the time constant not change? For example, would the time constant be different if the switch was open and the capacitor was discharging? Because in that case it would just be discharging across...

Woops... So its, R(eff) = (1/R1 + 1/R2)^-1(C) which in turn gives us ((R2+R1)/R1R2)^-1(C) which in turn gives us [(R1R2)/(R1+R2)](C)?

OK so with resisters in parallel we simply find: R(eff) = R1 + R2. So for the time constant of discharging with the battery removed at 100% charge, tao = (R1 + R2)(C) ?

lol I feel like we are beating this to death here, but still so foggy :( The resisters are now in parallel, the capacitor will have its current split across both?

well this I think is my problem, and perhaps after you help me out here I will be good to go. I thought that R2 and C are in parallel, but R1 and R2C are in series. So perhaps I just answered my own question, the effective resistance is therefore in series? But the current still undergoes a...

Well I'm not to sure, don't we separate into different loops? So it sees R1 in one situation and R2 in another, I am not sure how to do the effective resistance in this situation.

Oh your not saying open the switch, so the capacitor also discharges across R1, eh, again confused.

Using the same diagram, the capacitor would just discharge through R2 and C. So the time constant when discharging would simply be tao = R2C ?