1. Jan 10, 2012

### Jay9313

1. The problem statement, all variables and given/known data
I have this circuit here.. http://s1167.photobucket.com/albums/q639/jay9313/?action=view&current=EM.jpg
The question says that when switch S is open in the circuit shown above, the potential difference between points a and b would be..
A) 0V
B) 6V
C) 12V
D) 24V
E) 36V

Part 2 says when switch S is closed, the carge on the 3 microfarad capacitor would be..
A) 0 μF
B) 12 μF
C) 18 μF
D) 36 μF
E) 72 μF
2. Relevant equations

3. The attempt at a solution

My professor ways that the answer is A. I need somebody to help me understand why.

Last edited: Jan 10, 2012
2. Jan 10, 2012

### SammyS

Staff Emeritus
Where's the switch ?

Also, the rules for Homework Help in this Forum state:
You MUST show that you have attempted to answer your question in order to receive help. ​

So, what do you think the answer should be --- and why ?

3. Jan 10, 2012

### Jay9313

I redrew it and added the switch, and I don't know how he got 0V. My logic was that all of the current would go through the capacitors, so there would be a potential difference of 36V, the same as the EMF source.

4. Jan 10, 2012

### Curious3141

You may assume that the capacitors charge up very quickly and both capacitors are fully charged.

Take the negative terminal of the power source as being ground (0V).

What is the potential at a? Hint: voltage divider.

What is the potential at b? Hint: what do you know about capacitors in series? Compare the charges across the plates of each capacitor. Now compare the voltages across the plates of each capacitor.

5. Jan 10, 2012

### Jay9313

Oh, ok. It's 0v because both drop 24V in the first resistor or capacitor! Nnow what about the second part, where do I even start for that?

6. Jan 10, 2012

### Jay9313

Is it just 72 μC?

7. Jan 10, 2012

### Curious3141

Yes, since a and b are equipotentials. Closing that switch makes no difference. The capacitors don't discharge and the charges (and voltages) across both of them don't change.

8. Jan 10, 2012

### Jay9313

Oh thank you so much, you helped me understand it so much =D