# RC Circuit with switch. Find (discharging?) current

1. Mar 2, 2013

### Color_of_Cyan

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

http://imageshack.us/a/img831/8798/homeworkprobsg31.jpg [Broken]

Assuming that the switch has been in position #1 for a long time; at t = 0 the switch is moved to position #2.

Calculate the current i(t) for t>0.

2. Relevant equations

Q = CV

V = IR

V(t) = -V0*e-t/RC

(V0 is the capacitor voltage at time t = 0)

some formula I am probably forgetting

3. The attempt at a solution

I did not get that well in to RC circuit problems before so please bare with me, even though I sort of know how they work.

I think the capacitor becomes fully charged, so no current goes across it, but charge is not given, so not sure how to determine Vc(t).

At t=0, is the drop across the capacitor the same as the drop across the 3kΩ, since they are parallel elements?

So doing KVL with the 6k and 3k only

12V - 6kΩ*i - 3kΩ*i = 0

then

12V = 9kΩ*i

i = 0.00133A (1.33 mA) through the 3k resistor at t = 0

so if V = IR then

V = (0.00133A)*(3000Ω)

V = 4v across 3kΩ at t = 0

so then

Vc(t = 0) = 4V ??

Not sure where to go from here, (if this is all is right so far). I can calculate the Q on the capacitor but not sure where I would use it then.

I know the capacitor will then discharge current through the 6k and 3k if the switch goes to #2 but not sure what the formulas are for it.

Sorry if there's something that's supposed to be right under my nose but I just can't seem to find / derive the proper formula anywhere.

Last edited by a moderator: May 6, 2017
2. Mar 2, 2013

### Saitama

Right!
At t=0, the capacitor acts as the battery so it starts discharging. The current flowing at t=0 is the maximum current and then goes on decreasing. What's the current flowing through the circuit at t=0?

3. Mar 2, 2013

### Color_of_Cyan

Well you got me, but I'm still not sure how that would help either.

I'm going to assume that the V through the capacitor / 3k was correct so the current through the resistors with the capacitor charged and the switch still at the first position and t = 0 was 0.00133A.

Am I missing a formula?

Last edited: Mar 2, 2013
4. Mar 2, 2013

### Saitama

Yep, your thinking is correct and you found the correct potential difference across the 3k resistor or the capacitor.

You need to find the current as a function of time. Apply KCL at any instant of time and solve the differential equation.

5. Mar 3, 2013

### Color_of_Cyan

That seems a bit tedious.

How would I use KCL at any time if there's no current through the capacitor at t = 0?

Would the equation for the current simply be

V(t) = (-V0*e-t/RC)

and then I = V(t)/R ?

I mean, even if you're doing a differential equation, don't you have to take in to account the voltage source being cut off ?

Last edited: Mar 3, 2013
6. Mar 4, 2013

### Color_of_Cyan

Okay wait, I can do this.

But just to be sure, you completely disregard current through the 12V when writing the KCL equation when t>0, right?

The charge on the (fully) charged capacitor was 0.0004C

I finally found out the formula for a discharging capacitor current too.

I did NOT really know that the formula for the charge ON a discharging capacitor is

Q = CVbe-(t/RC)

and then the same for current is apparently the derivative with respect to time which is

i = dQ/dt = (Vb/R)e(-1/RC)

So if I understand now, I'm supposed to write KCL with this current now, right?

And then set the above equal to the equation standing for (V(t) = (-V0*e-t/RC))/(resistance) for each path with the resistances, right?

7. Mar 4, 2013

### Saitama

Well, after t>0, the 12V battery isn't a part of closed circuit so there is no current in that branch.

What I meant was that if you apply KCL at any instant of time, you get
$$\frac{q}{C}=iR$$
You can replace $i$ with $dq/dt$ and solve the differential equation.
The formula you quoted for the charge as a function of time is correct but you differentiated it wrong.

8. Mar 4, 2013

### CWatters

Thats correct but it's not an equation I would remember. I'd be thinking back to....

Q = VC (which in England we sometimes remember as Queen = Victoria Cross)

then

I = dQ/dt
so
I = dQ/dt = CdV/dt

You have an equation for the capacitor voltage...

Vc = Vbe-(t/RC)

so you can differentiate that to get an equation for dV/dt and substitute.

9. Mar 5, 2013

### Color_of_Cyan

I think my problem is just with the calculus here (not really my strong point :( ).

Not really any problem with Q = VC, V = IR, etc.

@Pranav would you say then that I'm solving first for the initial charge Q on the capacitor then, instead of the current? (Even though current IS dQ/dt)

@CWatters So if you have any equations varying with time, are you supposed to replace ALL variables with differentials (ie do you replace both I and V with dQ/dt and dV/dt because they both change with time) ?

Sorry for also getting the derivative wrong, was it

I = dQ/dt = C(dV/dt) = C(d(V0*e(-t/RC))/t)

= -V0*e(-t/RC)/RC

= -V0*e(-t/RC)/R ?

But this would just be the current from the capacitor, correct? And not the current through the 3k, right?

Last edited: Mar 5, 2013
10. Mar 6, 2013

### CWatters

No it's just that I prefer to work from first principles rather than remember a complicated equation. Current _is_ the rate at which charge is flowing.

Correct, but it's not difficult to calculate one from the other. Think ratios.

11. Mar 6, 2013

### Color_of_Cyan

But you also replaced V with dV/dt

So it's really the case then that dQ/dt = dV/dt, as voltage is the only other variable that can also change?

Ok I seem to have finally the got the answer now, thanks.:

R tot = 2000Ω

V knot through capacitor = 4V

Current through capacitor = -V0*e(-t/RC)

= (4V/2000Ω) * e(-t/((100x10-6μ)*2000Ω

= 0.002*e(-5t) A

Then you do current division for it so:

I = 0.002*e(-5t) A * (6kΩ/9kΩ)

I = 1.33*e(-5t) mA