Homework Help: Independent circuits

1. Aug 27, 2011

jaumzaum

What will happen to the following circuits immediately after T is closed?

Obs:
A) 2 generators of voltage 3V and V, 2 resistors of resistance R and 2R
B) 2 generators of voltage 3V and V (second inverted), 2 resistors of resistance R and 2R
C) 1 generator of voltage 3V, 1 capacitor initially carged with capacitance C, charge Q and voltage V, 2 resistors of resistance R and 2R

[PLAIN]http://img189.imageshack.us/img189/5149/34532wtgedgsdg.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Aug 27, 2011

lewando

This is a weakly posed problem. Are you required to simply redraw the circuits? If so, you have already described the new circuit configurations with your words. I say it is a weakly posed problem because it is not asking for anything specific, like a voltage across a component or a current through a component. Is the question requiring you to to find all voltages/currents across/through all components?

3. Aug 28, 2011

jaumzaum

The question is if there will be current from one circuit to another or if they are all independent themselves (it means that even with T closed down, one circuit will not communicate to the o ther, and there won't be current in T). The number (1V, 3V) is only to express the way of the current (first circuit to second or second circcuit to first)

4. Aug 28, 2011

lewando

Okay, then what are your thoughts? Have you tried to determine the currents?

5. Aug 28, 2011

jaumzaum

This question isn't homework (I think I've posted in the wrong topic). But anyway,

In Brazil we have a system to go in the university, the best universities are public and for you to go in you have to do a test calledd "vestibular". The 2 most difficult "vestibulares" of the country are ITA and IME (the first is the only vestibular that has "Aerospace Engineer" couse, that's what I want to be). Anyyway, in ITA and IME the only subsjects are Math, Physics and Chemistry. In ITA Physics is MUCH harder than the others and in IME math is the most difficult. For you to have an idea the most common vestibular is ENEM (that is a unified vestibular). 180 question and a wording. 1st day 90 question 4:30 hours, second day 90 question and a wording 5 hours. In IME you have 5 days of vestibular. Math, Physics and Chemistry are one in each day. Each day has 10 questions and the same 5 hours to solve ( IT'S REALLY DIFFICULT).

Anyway. I was doing an exercise of ITA 2003, analytical (with alternatives) part, the easier part (easier compared to written part), which was doing a lot of statements about the following circuit.

[PLAIN]http://img88.imageshack.us/img88/5564/imagemxeo.jpg [Broken]

And the right one was saying that the current in resistor g was 3.75 A IMMEDIATELLY after T is opened. And that would only be true if there was no current between the 2 circuits (the one with the 2 generators, and resistors a, b and c, and the one with the 2 capacitor and resi stors g and e). The template says the two circuits are independent, and there is no current between them. That's the part I didn't understand (I've calculated all potential differences/currents in the closed circuit, but I don't understand why there is no current when T is opened).

So I've drawn these circuits to ask if they are independent or not (as my learning material does not say anything related to that).

Last edited by a moderator: May 5, 2017
6. Aug 28, 2011

lewando

For now, let's focus on this fragment:
At equilibrium (before switch T is opened), how much current is flowing through the two capacitors?

At the moment switch T is opened, how much current is flowing through the two capacitors?

7. Aug 28, 2011

jaumzaum

You can calculate it.

When T and S are closed:

Current in the generators -> 10A
Current in c: 2,5A
Current in g: 7,5A
Current in f: 3,75A
Current in d: 3,75A
Current between the 2 capacitors: 0 (of course)
Current in e: 0
Current in a/b: 10A
Voltage in each capacitor: 7,5V

When T is open (by the template):

Current in the generators -> 5A
Current in c: 5A
Current in g: 3,75A
Current in f: 0
Current in d: 0
Current in the 2 capacitors:0
Current in e: 3,75A
Current in a/b: 5A
Voltage in each capacitor: 7,5V

8. Aug 28, 2011

lewando

Good. If no current is going through the capacitors, then would you agree that all of the current going through resistor g is going through switch T when it is closed?

9. Aug 28, 2011

jaumzaum

I think I've found out why, it couldbe because of the short circuit between g and e, but short circuit is done when a ideal thread with no resistance is put between 2 points of a circuit that has resistance making the current not to pass through it. But there we have a SINGLE thread, not 2 .
By the way it doesn't say if there will or not be current in the circuits I did.

10. Aug 28, 2011

lewando

?

11. Aug 28, 2011

jaumzaum

Sure. As the capacitors are charged, and B is closed, All the current is from the generators.

12. Aug 28, 2011

lewando

Okay, we are almost there. So when you say "Sure", you are acknowledging that all of the current that goes through resistor g is also going through switch T and only switch T (since you have also acknowledged that there is no current going through the capacitors before or at the moment switch T is opened).

So, if we agree that, at the moment when switch T is opened, there will be no current going through switch T, what must the current in resistor g be?

13. Aug 28, 2011

jaumzaum

After T is opened, the 2 capacitors ( which never has current passing through it because it's made by a dieletric material inside) will act like a generator, generating 15V of voltage that makes g and e has a current of 15/(2+2) = 3,75A

14. Aug 28, 2011

lewando

This is what started our discussion. The key word is "IMMEDIATELY". We are discussing the current through resistor g at the moment switch T is opened.

This may be true for the new equilibrium (well after switch T is opened-- t = ∞) Side note: capacitors can have current passing through them--when they are charging and discharging, for example.

To quote myself, this question still stands:

15. Aug 28, 2011

jaumzaum

All the feedback made by the vestibular (from all learning materials and also from the vestibular) have the same answer (that was the one I said). The immediately means that the current is 3.75A at first, then it will decrease as the time passes and capacitor uncharge. So the time to this current goes out the capacitor and reach g is like 0.000001s (wich is NOT well after). We also know that a capacitor can charge/uncharge millions of time per second (so the time is not Infinity).

Also, if you search for

A Capacitor NEVER HAS current inside it, even when it's charging/uncharging . What happens is that the electrons of the initial neutral metalic part of the capacitor (plate) , goes to the positive part of the battery, making the plate become positive, which creates a voltage and makes the electrons from the negative part of the battery goes to the other side of the capacitor, that becomes negative .

After the capacitor have reached its maximum voltage (which is very fast ), the current is stopped, and there is no current in the circuit. Why? Because in the capacitor we have a dieletric material that makes NO current pass, so even when the capacitor is charging/uncharging there is no current in the capacitor, only in the wire.

Last edited: Aug 28, 2011
16. Aug 28, 2011

lewando

Current through a capacitor is defined by this equation:

Ic = C*dV/dt

Let's try it this way. From your prior post, this represents the switch T closed equilibrium:
This represent the switch T open equilibrium:
Somewhere in between the two equilibriums (the moment that switch T is opened), an interesting thing happens to the current through resistor g. It is not 7,5A. It is not 3,75A. What is it? It would be helpful if you could draw a graph of this current versus time during this transition between the two equilibriums.

17. Aug 28, 2011

jaumzaum

I still don't understand this second equilibrum.

An equilibrum is made when, independently of the time, the charge/voltage /current of each piece of circuit is constant. Immediatelly after T is closed, there will be current induced by the capacitor voltaage, creating a current, that's direction is to the other plate of the capacitor, making it to uncharge (in a very little time), and then it will reach a equilibrum.

Anyway, couldn't you give only the anwer? I think it's the only thing I don't know in the question. What would be your graphic of the current?

18. Aug 28, 2011

lewando

Please standby while I figure out if I have been misleading myself and you. Need a moment.

19. Aug 28, 2011

lewando

Profound apologies are offered. I was wrong to suggest that the current through the capacitors, initially zero with switch T closed, would remain at zero when the switch was opened. Resistors e, and g, along with the capacitors form a circular "stub" hanging off the parallel combination of resistors d and f. The capacitors act like a voltage source and the resistors e and g draw current initially (at the moment the switch is opened) according to i=v/r = 3,75A. The final, open-switch equilibrium will be reached when all the energy from the capacitor is dissipated through resistors e and g and no current will be going through these resistors. I think I just violated the policy on giving out the answer, but I think its even worse violation to waste a person's time.