# Circuit help - resistor and capacitors... time too!?!?

#### Supervisor

Joined Jan 17, 2009
32
The redistribution of charge across the two capacitors happens very rapidly so that it is settled long before the time delay measured in seconds. The voltage across the two capacitors in paralled settles to 40V soon after the switches are thrown.

hgmjr
Alright, so if I did understand that above (-.-) then the answer would be 40 Volts across each?

EDIT: No! I still don't get it to be 40 Volts across each... There is a ten second time lapse between the switch being closed so it HAS to be different for each!!!

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#### hgmjr

Joined Jan 28, 2005
9,029
Alright, so if I did understand that above (-.-) then the answer would be 40 Volts across each?

EDIT: No! I still don't get it to be 40 Volts across each... There is a ten second time lapse between the switch being closed so it HAS to be different for each!!!
The two capacitors are in parallel. Do you agree with this statement? If not, explain what you believe to be the case.

If two components are in parallel, the voltage across one must equal the voltage across the other in the steady-state. Do you agree with this statement? If not, explain what you believe to be the case.

There is no voltage feeding the two caps in parallel. Agree or disagree?

These are ideal caps. Agree or disagree?

hgmjr

#### shteii01

Joined Feb 19, 2010
4,644
Alright, so if I did understand that above (-.-) then the answer would be 40 Volts across each?

EDIT: No! I still don't get it to be 40 Volts across each... There is a ten second time lapse between the switch being closed so it HAS to be different for each!!!
It has been my experience in various lab assignments that capacitors charge and discharge in milliseconds. In the lives of the capacitors 1 second is a very very long time.

#### JoeJester

Joined Apr 26, 2005
4,390
It matters not if the capacitor charges in milliseconds. Once charged, they remain charged until discharged. If yuu ever working in a high RF enviroment, you would short the capacitors while they were in storage.

I've seen an antenna, with top loading elements, that charged from the wind blowing. The antenna would normally be a hot antenna, but the transmitter wasn't connected, so it acted as a capacitor till I shorted it out with a nut between the johnny balls on the tower lighting transfromer.

The point of the ten seconds is so it is well beyond the 5 time constants normally used to achive 100 percent charge or discharge.

Here is a simulation of the problem. I purposely changed the timing on the switches so the whole sequence can happen in a shorter time for illustrative purposes.

I zoomed in on the critical events, when sw1 closed, when switch 2 opened and switch 3 closed.

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#### Supervisor

Joined Jan 17, 2009
32
After much thought and self reflection...
Nah, not like that at all - just me thinking.
Well, either way, I looked at the circuit again and then just told myself that the 0.00036 will redistribute itself to the 3 and 6 micro farad capacitors, splitting itself in half.

Thus, 0.00018 / (3*10^-6) = 60 V
and
0.00018/ (6*10^-6) = 30 V

I'm saying this because the capacitor will start to leak out and since it can't go back to the battery and get recharged, I guess what happens is that the whole thing cycles through that circuit (as you can see, when switch two is opened and switch three closed, there's no way for it to return back).

Hopefully this makes sense and passes with my teacher as well as all of you who have worked so hard to get this done.

Thanks for all your help and I will be back with more questions later when I start to become an engineer!

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#### Supervisor

Joined Jan 17, 2009
32
It matters not if the capacitor charges in milliseconds. Once charged, they remain charged until discharged. If yuu ever working in a high RF enviroment, you would short the capacitors while they were in storage.

I've seen an antenna, with top loading elements, that charged from the wind blowing. The antenna would normally be a hot antenna, but the transmitter wasn't connected, so it acted as a capacitor till I shorted it out with a nut between the johnny balls on the tower lighting transfromer.

The point of the ten seconds is so it is well beyond the 5 time constants normally used to achive 100 percent charge or discharge.

Here is a simulation of the problem. I purposely changed the timing on the switches so the whole sequence can happen in a shorter time for illustrative purposes.

I zoomed in on the critical events, when sw1 closed, when switch 2 opened and switch 3 closed.
Wow. That is just epic...
I can't believe you actually did that! Good report.

EDIT: Looks like there was one key bit of information that was missing from all of this -- switch two is opened while switch three is closed.
Also, switch one and two are closed in the beginning for the first three problems and then at 20 seconds, switch 2 is opened and switch three is closed.
Though, 40 V would have been the correct response if switch two were closed.

EDIT2: Oh, I see you do have it like that... Why do I get 60 and 30 Volts...

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#### hgmjr

Joined Jan 28, 2005
9,029
EDIT: Looks like there was one key bit of information that was missing from all of this -- switch two is opened while switch three is closed.
Also, switch one and two are closed in the beginning for the first three problems and then at 20 seconds, switch 2 is opened and switch three is closed.
Though, 40 V would have been the correct response if switch two were closed.
Below is an excerpt from post #16 of this thread in which you clearly indicated the state of switches 2 and 3 associated with the question in part D.

d) At twenty seconds, switch two is opened and switch three is closed; at thirty seconds, what will the voltage be on each of the capacitors?
EDIT2: Oh, I see you do have it like that... Why do I get 60 and 30 Volts...
The only thing I can figure is that you have different circuit configuration in mind than the one.

hgmjr

#### JoeJester

Joined Apr 26, 2005
4,390
EDIT2: Oh, I see you do have it like that... Why do I get 60 and 30 Volts...
Did you not read hgmjr's post #38 in this thread? I thought the formula there explained why there was 40 volts on both capacitors instantaneously when switch 3 closed.

The simulation also showed that C1's voltage went from 60 to 40 and C2's voltage went from 0 to 40 instanteously. Without any resistance slowing down the charging current (from C1), it happens instanteously.

You split the 0.00036 joules in half, causing your error. Qc / (C1 + C2), as illustrated by hgmjr's post #38, calculates out to be 40 V. The parallel capacitors must have the same voltage else the Kirchoff's Voltage po'lease will arrest you.

The only thing I left out was that sw 1 closed at t = 100uS

At t = 700, I opened both Sw1 and Sw2 to illustrate the charge remained on the capacitor with those two switches open. It also illustrated that there was no current flowing in the resistor. I could have used the exact numbers you did

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