# Homework Help: RC circuit capacitor problem

1. Mar 15, 2015

### vysero

1. The problem statement, all variables and given/known data

With a 12V battery, a 5.00µF capacitor and a 8x105Ω resistor determine the following:
a) The time constant of the circuit
b) The maximum charge on the capacitor
c) The maximum current in the circuit
d) The charge on the capacitor as a function of time, q(t)
e) The current in the circuit as a function of time, I(t)
f) The time until the charge on the capacitor is 75% of it’s maximum value

A long time later the capacitor starts fully charged. At this new t=0, starting with a fully charged capacitor, the switch is moved to position b. Determine the following:
g) The charge on the capacitor as a function of time, q(t)
h) The current in the circuit as a function of time, I(t)
i) The time for the capacitor to reach 15% of it’s maximum value

2. Relevant equations

τ = RC
q = CV
V = IR

Charging:
q = CV(1-e^(-t/RC))
i = (V/R)e^(-t/RC)
V = (q/C) = V(1-e^(-t/RC))

Discharging:
q = qₒe^(-t/RC)
i = -(qₒ/RC)e^(-t/RC)

3. The attempt at a solution

a) RC = (5x10^-6)(8x10^5) = 4 seconds

b) I am not sure what the question is asking. Obviously it wants the maximum charge but does it want that maximum when t = 0 or at some other point? I said the answer was zero but I am not sure.

c) Once again it depends on the time. Right after t = 0 all the current is on the resistor but before that at t = 0 there is no current in the circuit. So for t = 0 the answer is zero but just after zero say like .01 sec the current is all on the resistor: R = V/I = 12/(8x10^5) = 1.5x10^-5 A.

d) Same problem not sure what time I should be looking at. At t = 0 the answer is zero.
e) At t = 0 it is zero again.

f) Not sure how to do this part.

The next few questions I think will be better if left alone until I get the previous set. Any help would be appreciated, mostly I need to know if what I am assuming the questions are asking is correct. Also, any other help would be appreciated.

2. Mar 15, 2015

### Staff: Mentor

(b) maximum means at any time, just whenever the maximum occurs

(c) you need to consider time of 0+, as you have

3. Mar 15, 2015

### vysero

So for b should I be using q = CV so q = (5x10^-6)(12) = 6x10^-5 C? Can I assuming I was correct about d,e both being zero? Still not sure how to do f.

EDIT: An update on part f. I want to say I need to solve for t when .75 = e ^(-t/RC) is this correct? I came up with t = -RCln(.75) = 1.151 sec.

Last edited: Mar 15, 2015
4. Mar 15, 2015

### Staff: Mentor

(d) asks for q(t), i.e., how capacitor charge varies as a function of time. It will involve an exponential.

Something similar applies to (e).

5. Mar 15, 2015

### vysero

So for d it asks for q(t) isn't that going to be an equation and not like a number? Is it looking for: q = CV(1-e^(-t/RC))?

6. Mar 15, 2015

### Staff: Mentor

That looks right, and you know values for most of those pronumerals.

7. Mar 15, 2015

### vysero

Maybe I am making a math error then because I tried saying t = -RCln(1-(1/CV)
Okay so solving for t from that equation you get: t = -RCln(1-(q/CV)). At this point do I assuming RC = 4, q = 5x10^-6, and for CV do I use CV(.75)? If I do that I come up with: t = .4711 seconds for f.

8. Mar 16, 2015

### vysero

Not feeling to confident in my answers atm...

9. Mar 16, 2015

### BvU

So far I've seen
$\tau =$ 4 sec,
60 $\mu C$,
15 $\mu A$,
$q = CV(1-e^{-t/\tau})$ (or q = CV(1-e^(-t/RC)) ),
i = (V/R)e^(-t/RC).​

All good. No reason for lack of confidence.

Maybe yes. You have an expression for q(t) and one for qmax.
An easy way to check that it's not the correct expression:
you can only take a logarithm of a number.
So whatever you take the logarithm of, it can not have a dimension (like 1/Coulomb)
same thing for exponential, and things like sine etc. -- where you need an angle, but angles are dimensionless too.​

But you correct yourself a little further down :

t = -RCln(1-(q/CV)) is good.
And you want to know the time when q = 0.75 qmax,
so q/CV = .... ?

And there's a simple check: at t = $\tau$, q is at 1 - e^{-1} = 63% of qmax, so the correct answer must be more than 4 seconds...

10. Mar 16, 2015

### vysero

Okay so what I come up with is 5.45 sec for f and 7.59 sec for i. Which I am fairly confident in :D thanks for your help guys. One more thing does voltage work the same way? Like I have an equation for voltage Vc = V(1-e^(-t/RC)). Does that mean that voltage and charge will both take the same amount of time?

Last edited: Mar 16, 2015
11. Mar 16, 2015

### Staff: Mentor

For a capacitor, its charge is related to its voltage as: Q = CV
or, more formally, q(t) = C.v(t), where C is a constant here.

So, there's a direct proportionality between v(t) and q(t). Exactly the same waveshape, just with a different scaling factor.

12. Mar 16, 2015

Well done !