# RC Circuits, capacitor charges and voltages

1. Jan 18, 2014

### lovelyrwwr

These are 2 stand alone problems, however the answers disagree with one another and I am trying to reconcile their differences. Any insight would be appreciated! :)

View attachment untitled2333.bmp
The answer is C. The answer provided with this question is that "there is no dependence on resistance of any kind".

View attachment untitled23.bmp
View attachment untitled233.bmp

Following along the same logic as question 24, I assumed the answer was B. However the answer for problem 20 is supposedly D and the the following explanation was provided:

When switch 1 has been closed for a long time, the capacitor becomes fully charged and has a potential difference across the capacitor Vc that is equal to the potential difference V3.

Since the capacitor is fully charged, no more current should flow towards the capacitor, so the circuit can be thought of as having just R1 and R3.
View attachment 444.bmp

The equivalent resistance of R1 and R3 is the sum so Req = R1 + R3.

EMF = I(Req)
I = EMF / Req

V3 = IR3 = (EMF/Req)R3

Taking V3 to equal the potential difference across the capacitor, we can find the charge on the capacitor:
Q=CV = CR3(EMF/R1+R2), which is answer D.

I just don't see why the answer for problem 20 doesn't follow the same rule as problem 24. I expect that all capacitors - when fully charged - to have the same potential as the emf source. And that resistance does not affect the maximum amount of charge plates can hold as problem 23 suggests.

Last edited: Jan 18, 2014
2. Jan 18, 2014

### Staff: Mentor

You'll have to provide your own attempt at solutions for the questions. What are your thoughts on each? What is confusing you?

Question 23 appears to be associated with some particular circuit configuration which is not shown. Question 24 may also have a particular configuration assumed. Can you clarify the question context?

3. Jan 18, 2014

### lovelyrwwr

No - questions 23 and 24 are stand alone questions and neither has a figure provided. I will edit this post again with my thoughts and further clarification of why I am confused, please check again gneill in a few minutes. Thanks so much.

Last edited: Jan 18, 2014
4. Jan 18, 2014

### BvU

The capacitor never gets to 'see' more than the voltage over R3 during and after charging. So battery EMF minus the voltage drop over R1.
If you want to convince yourself, imagine the situation with the capacitor charged to the battery EMF. What would happen ?

5. Jan 18, 2014

### lovelyrwwr

Ok I have edited my question, hopefully this explains it better. I am still so confused.

6. Jan 18, 2014

### lovelyrwwr

BvU hi. that still doesn't explain problem 24 though. Problem 24 is answer C. The answer provided states "there is no dependence on resistance of any kind".

Last edited: Jan 18, 2014
7. Jan 18, 2014

### lovelyrwwr

--------------

8. Jan 18, 2014

### Staff: Mentor

If the questions are posed exactly as you've shown and are from the same section of the book, then they are badly posed questions.

The ultimate potential that a capacitor "sees" always depends upon the circuit that is providing the potential difference; if the battery voltage is reduced by a resistor network, then the capacitor can never "see" the full emf of the battery. It seems that you've come to the same conclusion in your edits to post #1 (I would ask that you not make drastic changes to initial posts, it tends to make the follow up responses look confusing and out of context; Make any changes obvious with appropriate indications or better yet, just add a post with corrections/explanations).

Question 24 appears to be without context of a particular circuit configuration so strictly speaking it is not an answerable question.

9. Jan 18, 2014

### lovelyrwwr

Whoops sorry about that! I was trying to make it easier for you to understand haha.

Ok...sooooo basically, question 24 is just an awful question? And so, when a current comes across a resistor before reaching the capacitor, that resistor will zap some of the voltage that would otherwise get stored as charge on the capacitor? Is that right, gneill?

Thanks for all your help!!! I am a student working minimum wage and could never afford a 25\$/hour physics tutor!!!! You are appreciated!

10. Jan 18, 2014

### BvU

I don't think they want you to keep the circuit in figure 20 in mind when answering 24.
After all, if someone is dumb enough to short-circuit the capacitor (i.e. R3=0 in fig. 20) he won't be able to get any charge on the capacitor at all, but that doesn't mean the maximum charge on the capacitor is zero.

If you insist on a deeper layer: 24 is not a good question. What battery? Real capacitors have a maximum voltage (breakdown voltage).

11. Jan 18, 2014

### Staff: Mentor

Question 24 is not a well posed question without an accompanying circuit for context.

A series resistance (so that battery, resistor and capacitor are all in series) will not affect the ultimate potential that the capacitor reaches -- it will eventually charge up to the same potential as that of the battery and current will cease to flow. The resistance will only affect the time it takes to charge up to full potential. But a resistor network that divides the battery potential, as in the circuit of question 20, WILL limit the ultimate potential developed across the capacitor.

12. Jan 18, 2014

### BvU

1. It's a bit simplistic, with the risk that smart students get sidetracked....

2. You want to be a bit more careful in your expressions here to avoid misunderstandings in the future: The zapping is correct. The voltage is not what is getting stored: current is moving charge; that charge can accumulate on a capacitor, thus building up voltage. Once this voltage (Q/C) equals the externally applied voltage, the charging stops.

Now the zapping as you describe it so nicely: If you charge an empty capacitor (V=0) from a battery via a resistor R1 (as in 20, with R3=∞) the resistor will zap the full EMF at time zero, because it sees EMF - V = EMF (assume everything ideal, no self-inductance, that's for a later chapter). Current at time zero is EMF/R1. But that current charges the capacitor, so the driving force EMF-V diminishes, etc. until V=EMF. No more voltage drop over the resistor, no more current, no more zapping. Who knows, the next chapter might even present you with a differential equation that gives the voltage (charge, current) as a function of time !

Happy studyinig!

13. Jan 18, 2014

### lovelyrwwr

Ok I think I got...so when they're in series maximum charge q = emf x C but when you have resistors in parallel with the capacitor, it's another story?

14. Jan 18, 2014

### lovelyrwwr

BvU: that was an amazing explanation! ....Why can't textbooks just put it that simply! I wish I could transplant y'all's brains into my head for this test. Yikes! :)

15. Jan 18, 2014

### Staff: Mentor

That's pretty close to the story. The circuit between the voltage source (or sources) and a capacitor can get fairly complicated, so simply saying "resistors in parallel with the capacitor" doesn't cover all the possibilities (There might even be current sources rather than voltage sources involved).

Eventually in your studies you will come across a couple of very helpful theorems called Thevenin's Theorem and Norton's Theorem which will simplify analyzing the more complicated cases. But for now, a practical approach is to just "remove" the capacitor from the circuit and determine the potential difference that would appear across the open terminals where the capacitor was connected. That will give you the potential that the capacitor would eventually reach when charging finishes.

16. Jan 18, 2014

### lovelyrwwr

0_o I've taken both physics I and II in college and I'm brushing up for the MCAT. I most definitely never came across either theorems :X

THank you for all of your insight! It help tremendously! :)