Very Hard Question

[Dean820]

Here's a question I got for homework and I really do not have a clue on how to do it. Even if somebody could do pieces of it, I would be most grateful.

A capacitor(C) of 10 μF is connected in series with a resistance(R) of 5kΩ across a 20 V
d.c supply (E). Calculate
(i) The initial charging current.
(ii) The initial voltage across the capacitor.
(iii) The Time Constant (T) of the circuit.
(iv) The voltage and current at the Time Constant.
(v) The voltage across the capacitor after 75 milliseconds
(vi) The time for the current to reach 2 mA.
(vii) The time for the voltage to reach 18 V.
(viii) The voltage on the capacitor after a long time period.

Note : The charging current in the circuit is given by i(t) = (E/R) e^-t/CR.

The charging voltage across the capacitor is given by v(t) = E(1- e^-t/CR).

[jemidiah]

I believe the following is all correct. Normally I wouldn't have done the entire homework problem, but it's most likely a college assignment, so you'll need to understand the material to pass the test (and the class) anyway.

At first I thought this was going to actually involve calculus which scared me off, but then I got to looking at the individual problems and they're all just problems with functions (except the last one, which has a hint of calculus).

I assumed that you were given the two functions for i(t) and v(t). If you were asked to derive these functions, the problem would be considerably harder (i.e. a semester/a year of difficulty higher), and I don't remember which Physicist's laws you have to apply in that case.

Since this problem is pretty straight-forward after you delve into it, I would suggest you seek help from your professor or someone who is doing well in the class if you think you need it after looking over my answers.

Finally, you might want to put more descriptive titles on your posts in the future--it'll get you a faster answer. I would probably have used "RC Circuits" or something here.

Welcome to the forums!


(i) Isn't this i(0) = E/R = 4 mA?
(ii) Isn't this v(0) = E(1- e^0) = 0 V?
(iii) From http://www.tpub.com/neets/book2/3d.htm the Time Constant is apparently your resistance * capacitance, or 10 μF * 5kΩ = 50 ms
(iv) e^(-50 ms / [10 μF * 5kΩ]) = e^(-50 ms / 50 ms) = e^-1 = 1/e = ~0.368. So, i(50 ms) = (E/R) / e = ~1.47 mA and v(50 ms) = 20V * (1-1/e) = ~12.6 V
(v) This is simply v(75 ms) which I won't plug in, seeing as I've plugged in two values already
(vi) Just set i(t) = 2 mA and solve for t; just algebra, though you might have to use logarithms
(vii) Similarly just set v(t) = 18 V and solve for t. Again you'll probably use logarithms
(viii) "A long time" means "as time goes to infinity". So, just take the limit as t goes to infinity of v(t) [I got 20 V]

[Dean820]

Thanks for the reply. I always come up with crazy answers for parts (vi) and (vii). Could you show the best way to go about getting the answer. Algebra and logs have always been my weak point. Sorry for causing you any trouble.

[jemidiah]

Alrighty, here's my algebra for simplification:

(vi) i(t) = (E/R) e^-t/CR -> ln(i(t)) = ln((E/R) e^-t/CR) = ln(E/R) + ln(e^-t/CR) = ln(E/R) - t/CR * ln(e) = ln(E/R) - t/CR -> t/CR = ln(E/R) - ln(i(t)) = ln(E/(R i(t))) -> t = CR ln(E/(R i(t)))

(vii) v(t) = E(1- e^-t/CR) -> v(t)/E = 1 - e^-t/CR -> v(t)/E - 1 = -e^-t/CR -> 1 - v(t)/E = e^-t/CR -> ln(1-v(t)/E) = -t/CR*ln(e) = -t/CR -> -CR ln(1-v(t)/E) = t


Checking my work with dimensional analysis: the arguments to the logarithms are both unitless; good. The units of C*R is time, as it should be in both cases.

Hope this helps

[gary223]

Thanks for the help.