How Do Capacitors Affect Potential Differences in a Circuit?

[Dell]

in the following diagram the numbers above the capacitators indicate μF,

http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE&pli=1&gsessionid=ABXE5awGNEiCKaJR5eBwnA#5333028795415941746

i am asked to find
I)the capacitance between A and B
II)if we know that the charge on the 5μF capacitator is 120μC, what is the potential difference between A and C

I)
i add up all the capacitators in the following manner

i add the 3 from the top left hand corner as parallel capacitators
and get C=12μF
then add the 12μF to the top right hand capacitator is capacitators in a row,
and get 3μF
then do the same for the bottom and eventually add up the total from the top and total from the bottom as parallels,
finally getting

C=5μF


II)
if i am looking for the potential difference from A to C can i look only at the bottom left hand capacitator and say that the potential difference on that capacitator is the same as the potential diff from A-C?

therefore
C=Q/V
V=Q/C
V=120μC/5μF

V=24V

does this look correct?

 

[PalashD]

check your calculations for the bottom cpacitors

 

[rl.bhat]

then do the same for the bottom and eventually add up the total from the top and total from the bottom as parallels,
finally getting

C=5μF
This is wrong. Check the calculation.

 

[Dell]

why? my bottom calculations seem right to me,, for the top 4 i get 3μF, for the bottom 3 i get 2μF, for the whole system that comes to 5μF, which is the correct answer to the 1st part

for II) i didnt take the calculations from I) into account at all, just looked at the bottom left hand capaciitator since it is connected directly to A and C

 

[rl.bhat]

why? my bottom calculations seem right to me,, for the top 4 i get 3μF, for the bottom 3 i get 2μF, for the whole system that comes to 5μF, which is the correct answer to the 1st part

It is possible if you have 3μF instead of 5μF.

 

[Cyosis]

How did you get 2μF for the bottom three? On the right hand side we have two capacitors parallel to each other, thus c_eq1=4+2=6μF. This equivalent capacitor is in series with the left hand side capacitor giving, 1/c_eq2=1/6+1/5=11/30 μF. Therefore c_eq2=2.73μF.

 

[Dell]

when i worked out the FIRST part i get that the top comes to 3, the bottom comes to 2 all in all comes to 5,

but the second part is another question, they tell me that the single capacitator with the 5μF capacitance (ie the top-left-middle and bottom left capacitators in the diagram) have a charge of 120μC on them, now find the potential difference from A-C,

the correct answer is meant to come to 64V according o the book but mine comes to 24V, where have i gone wrong?

 

[Cyosis]

when i worked out the FIRST part i get that the top comes to 3, the bottom comes to 2 all in all comes to 5,

Three people have told you that is wrong, perhaps pay attention to what they say. Check the picture you've linked to us, does it display the correct capacitance values as r.l.bhat suggested?

 

[Dell]

o, man, you are all, right, i copied this totally wrong, the bottom 5 is meant to be a 3, and the only 5 is at the top,

thanks

 

[Cyosis]

II)
if i am looking for the potential difference from A to C can i look only at the bottom left hand capacitator and say that the potential difference on that capacitator is the same as the potential diff from A-C?

therefore
C=Q/V
V=Q/C
V=120μC/5μF

V=24V

does this look correct?

Why did you use 5μF for the bottom left capacitor after all? We had just determined it was 3μF. You don't know the charge of the bottom left capacitor at all, but we do know the charge on the upper left capacitor the 5 μF one. With this you can calculate the charge on the equivalent capacitor (upper left again). From this you can calculate the charge on the entire upper equivalent capacitor. Then you can calculate the potential difference A-B, which you will need to calculate the potential difference A-C. So that you can check if you've done it correctly V_AB=96V.

 

[Dell]

lets number the capacitators,
C1-top left top
C2-top left-middle
C3-top left-bottom
C4-top right
C5-bottom left
C6-bottom right-top
C7-bottom right-bottom

C1 C2 C3all have the same potential and since i know the capactiance and charge on C2 i can find that potential, which comes to V=24V
now using that V i can find the total chrage of (C1+C2+C4) which comes to 288, now i know that that charge is the same charge as the charge of C4. now i can find the potential difference of the whole top capacitator, which comes to V=96V,

from here all i need to know is the potentiasl diff of the bottom right capacitators (which have the same potential difference) and i can add that to the 96 i found here and that will give me the total difference from A-C,

i have not managed to find this though,

 

[Cyosis]

Those calculations are correct so far. Now you can calculate q on the C5 capacitor and on the equivalent capacitor at the bottom right. Then you know, using the same argument as for the top capacitor that the equivalent capacitor (the entire thing at the bottom) has charge q. So you can calculate q=C5 V_AC=C_eqbottom V_AB.