RC Circuit with switch. Find (discharging?) current

[Color_of_Cyan]

Homework Statement

http://imageshack.us/a/img831/8798/homeworkprobsg31.jpg 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.

Homework 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

The Attempt at a Solution

I did not get that well into 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*ii = 0.00133A (1.33 mA) through the 3k resistor at t = 0so 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.

 

[Saitama]

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

Right!

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.

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?

 

[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?

 

[Saitama]

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.

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

Am I missing a formula?

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

 

[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 into account the voltage source being cut off ?

 

[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 = CV_be^-(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?

 

[Saitama]

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

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

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 = CV_be^-(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?

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.

 

[CWatters]

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

Q = CVbe-(t/RC)

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...

V_c = V_be^-(t/RC)

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

 

[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?

 

[CWatters]

@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) ?

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.

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

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

 

[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 = 4VCurrent 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