Determining current in a transient RC circuit

[i_am_stupid]

Homework Statement

For the below circuit, the switch was open for a long time. Find the current Ia(t) for t>0. All capacitors are initially uncharged.

Homework Equations

Complete response of a capacitor: v(t) = Voc + (v(0) - Voc)e^(-t/τ)
Combining parallel capacitors: summation
Current through a capacitor: i = Cdv/dt

The Attempt at a Solution

I began by combining the three capacitors on the right side into a single 500μF capacitor connected in series with the 3K resistor.

Because the switch was open for a long time, we can assume the circuit has reached steady state, at which point the capacitor behaves as an open circuit. Since no current is flowing through this branch, the 3K resistor has no voltage, and the initial capacitor (open circuit) voltage is 12V by KVL. v(0-) = v(0+)

However, opening the switch does not seem to change anything, because after the circuit has once again reached steady state the capacitor acts like an open circuit once again, with a voltage of 12V by the same reasoning as before.

So, my equation would work out to be v(t) = 12 + (12 - 12)e^(-t/τ) = 12. Is this correct, or am I missing something?

Also, the question asks for the current through the 3K resistor, so would this be found by [12 - v(t)] / 3K?

Thanks.

 

[berkeman]

Homework Statement

For the below circuit, the switch was open for a long time. Find the current Ia(t) for t>0. All capacitors are initially uncharged.

Homework Equations

Complete response of a capacitor: v(t) = Voc + (v(0) - Voc)e^(-t/τ)
Combining parallel capacitors: summation
Current through a capacitor: i = Cdv/dt

The Attempt at a Solution

I began by combining the three capacitors on the right side into a single 500μF capacitor connected in series with the 3K resistor.

Because the switch was open for a long time, we can assume the circuit has reached steady state, at which point the capacitor behaves as an open circuit. Since no current is flowing through this branch, the 3K resistor has no voltage, and the initial capacitor (open circuit) voltage is 12V by KVL. v(0-) = v(0+)

However, opening the switch does not seem to change anything, because after the circuit has once again reached steady state the capacitor acts like an open circuit once again, with a voltage of 12V by the same reasoning as before.

So, my equation would work out to be v(t) = 12 + (12 - 12)e^(-t/τ) = 12. Is this correct, or am I missing something?

Also, the question asks for the current through the 3K resistor, so would this be found by [12 - v(t)] / 3K?

Thanks.

Welcome to the PF.

I don't understand the part of the problem statement that says the caps are initially uncharged. They sure look like they would have some charge with the switch open or closed...

 

[i_am_stupid]

Wouldn't it just mean that they start at 0V?

 

[berkeman]

Wouldn't it just mean that they start at 0V?

Not given the problem statement. If the switch is left open for a long time, what voltage is on the caps? You can figure that out by inspection...

 

[i_am_stupid]

Not given the problem statement. If the switch is left open for a long time, what voltage is on the caps? You can figure that out by inspection...

Well, eventually it would end up being 12, but I was thinking that the rate at which it would reach 12 would change given an initial charge.

 

[berkeman]

Well, eventually it would end up being 12,

That is correct. So it makes no sense that they would say assume the caps are discharged, unless they mean at time t = negative infinity. Because at t=0-, they are all at 12V.

So maybe double-check the question statement that you were given, to see if it's worded differently. Anyway, assume the caps are charged to 12V at t=0-, and figure out the rest of the circuit the way you were doing it in your original post (OP).

Actually, I see now that you correctly said the cap voltages are 12V at t=0 in your OP. I think you are also right that closing the switch (you said "opening" in your text, but I think you meant closing as shown in the figure) does not affect the voltage on the caps. There is a power supply on one side of the feed resistor, and then just the caps to ground. I'm not seeing now how the stuff to the left will change any of that...

So it would seem that Ia(t) is zero after the caps are initially charged up...weird...

 

[i_am_stupid]

Yeah, I'm not sure what the point of including that was. The problem statement I posted originally is exactly the same as given - furthermore, this is an old exam question, so I doubt it's a mistake. I guess the main problem here is figuring out what the switch changes (though I've even put it through a circuit simulator in case I was missing something; nothing). In any case, thanks for your help!