RC Circuit Problem

1. Apr 18, 2014

astro2525

[Mentor's note: this thread does not use the standard homework-help template because it originated in a non-homework forum. It was moved here instead of being deleted because it had already gotten significant help.]

"A capacitor is being charged from a battery and through a resistor of 20 KOhms. It is observed the the voltage on the capacitor rises from 10% of its final value to 90%, in 15 seconds. Calculate the capacitor's capacitance."

I know we are supposed to show an attempt at a solution, but I'm not even sure where to start with this one :/. I know I need to be able to find the time constant from the data given, and from there tau = RC, but I don't know how to incorporate the percentage difference. I just need a push in the right direction

Last edited by a moderator: Apr 21, 2014
2. Apr 18, 2014

Simon Bridge

Do you know any equations that show voltage as a function of time for charging a capacitor?
(Check your course notes, or look online.)

Call the max voltage V, and the initial time t.
The initial voltage is V/10 at time t and the final voltage is 9V/10 at time t+15s

Write down the equations for each of the times.
What are the unknowns?

You can also ask - what is the percentage change in voltage in time $\Delta t= \tau$
What is the percentage change in voltage in $\Deta t = 15\text{s}$?
How many times $\tau$ to get that percentage change?

Last edited: Apr 18, 2014
3. Apr 19, 2014

Simon Bridge

Correction: that should be $\Delta t = 15\text{s}$

4. Apr 19, 2014

astro2525

well, i know V(t) = V0 * e-t/RC

so would my equations be:

V(t) = (V/10) * et/RC

V(t+15) = (9V/10) * e(t+15)/RC

which leaves me two unknowns, V & C. sorry i don't really understand that last part, my notes are super confusing. ;x

5. Apr 19, 2014

Simon Bridge

Your equations have five unknowns: V(t), V(t+15), V, t, C.

Also: look at the equation you used:
As time increases, what happens to the voltage? (i.e. does it increase or decrease?)

Now look at the problem description:
In the problem, as time increases, what happens to the voltage?

If your notes are too confusing for you - have a look online:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capchg.html
http://www.electronics-tutorials.ws/rc/rc_1.html

Take special care over what each variable means when you do the substitutions.

Last edited: Apr 19, 2014
6. Apr 19, 2014

astro2525

well, as time increases, voltage on the capacitor increases and voltage on the battery decreases, so V decreases and V(t) increases

if there are five unknowns i need more equations before i do substitutions right? what am i missing?

7. Apr 19, 2014

Simon Bridge

The voltage across the battery is always the same!

The equation you used - was that for the voltage across the capacitor or something else?
Which voltage does the problem want to know about?

Thinking about the votage across the capacitor:

Look at the equation you used:
As time increases, what happens to the voltage? (i.e. does it increase or decrease?)

Now look at the problem description:
In the problem, as time increases, what happens to the voltage?

You used the wrong equation and you put the numbers in the wrong place in that equation.

8. Apr 20, 2014

astro2525

ok, voltage across capacitor:
Vc = capacitor voltage
Vs = voltage on battery
initially:
0.1 * Vc = Vs(1 - e-t/RC)
0.9 * Vc = Vs(1 - e(-t+15)/RC)
are these right??
as time goes on, voltage on the capacitor is increasing

9. Apr 20, 2014

Simon Bridge

You have the right equation now: Vc=Vs(1-exp[-t/RC]) ... well done.

- but still the wrong substitutions.

At time t, Vc = ?
At time t+15, Vc = ?

(Vc is given as a percentage of another value - what is it in terms of the equation you are now using?)

10. Apr 20, 2014

astro2525

so its maximum voltage is the voltage of the battery?

at time t, Vc = 0.1 * Vs
at time t+15, Vc = 0.9 * Vs
?
in which case Vs would cancel leaving something like 0.1 = 1-e-t/RC & 0.9 = 1-e(-t+15)/RC ?

thanks so much for all this help btw, i really appreciate it

11. Apr 20, 2014

Simon Bridge

Now you are thinking.

If you look at the equation Vc=Vs(1-exp(-t/RC)) - what value does Vc tend towards as t gets very big?
There's your maximum value - you don't need me to tell you ;)

So - you end up with two equations with two unknowns - just solve for the value you need.

12. Apr 21, 2014

astro2525

well as t gets bigger, e-t/RC goes to 0 i think .. so Vc does tend towards Vs? so my two equations are right? now i just substitute and solve?

13. Apr 21, 2014

Simon Bridge

If you are uncertain, you can always just put some numbers in and see.
OR work it out

note: exp[-t/RC] = e^(-t/RC) = 1/e^(t/RC) [the "exp[]" form is nicer to write down]
so when t=10RC, 100RC, 1000RC
... what happens to exp[t/RC] ?
... what happens to 1/exp[t/RC] ?
... see? You don't need me to reassure you ;)

Did you try it and see?

You need to think of ways you can be more confident in your own work.

14. Apr 22, 2014

astro2525

thank you so much for all your help.

one more quick question if i may, say i had the capacitor being discharged through a resistor, from 90% of it's max value to 10% in 20 seconds. i believe the formula i would use is Vc = Voe-t/RC. the steps would basically be identical i presume?

0.9 * Vo = Voe-t/RC
0.1 * Vo = Voe-(t + 20)/RC

and just substitute and solve from there

15. Apr 22, 2014

Simon Bridge

Exactly.

The underlying problem you have is "how can I know I have the right approach, without asking?"

It is important for you to develop a set of skills and habits to deal with this issue, because, at some point, you will have to work on problems where nobody knows the right approach - so there is nobody to ask.

One of the important habits is to "play" with mathematical relations - you could have checked out the last few questions by yourself just by plugging sample numbers into equations and watching what happens.

A great tool for this is a computer graphing program of some kind.
Then the computer can test your ideas numerically for thousands of sample numbers in less than a second. Very useful.

Physics is not about knowing the right formulas - it's about understanding the relationships between phenomena. I, personally, have never memorized the formula for charging and discharging capacitors.

16. Apr 22, 2014

astro2525

yeah I know, i'm not usually this cautious. currently a comp. science major so most of my time goes there, and this question is part of an assignment that's rather important grade wise, wanted to be as sure as possible that i was doing the right stuff

jus wanted to say thanks again you were such a huge help

17. Apr 23, 2014

Simon Bridge

No worries - enjoy.