Capacitance for series and parallel connected capacitors; C=Q/delta_V, derivations of last formula
The Attempt at a Solution
C1 and C2 appear to be connected in series. Therefore, use appropriate formulas.
(1/Cs)=(1/3C)+(1/C) = 3C/4 (answer I got is "flipped" here due to the way the formula is stated)
Alright, so we have the equivalent capacitance for C1 and C2. Now move on to C3.
(C1+C2) + C3 appear to be parallel-connected. So,
Cp=(3C/4) + 5C = 23C/4
I also tried 23/4C in case it couldn't read my entry. A preview confirms it is what I intend, and yet it is wrong. Tried different combos, all wrong. I'm certain I did this correct... what did I do wrong here?
I'm not sure how to even solve this part. I actually guessed on this. At first I thought higher capacitance = more charge storage, but that was wrong.
Anyway, Capacitance = Q/delta_v. So they want charge, so I derive Q=C(delta_V).
delta_V is not provided, so I don't understand what they want. I'm not even sure how I'm supposed to input it... obviously, integer values are what they want. I am clueless on this problem :(
I understand V2=V3 because they are connected in parallel. Why is V1 bigger than the other two, though?
As far as I can derive, delta_v=Q/C. So, shouldn't a bigger C mean a smaller delta_V? Am I supposed to use the answers I supposedly got in Part B?
I don't understand. C3 is the last to be charged (am I thinking of this the wrong way?), and so I am thinking it shouldn't have be affecting the other caps at all. Plus, shouldn't the charges on C1 and C2 be equal, because they are connected in series?
EDIT: I thought about it a little more. C2 and C3, connected in parallel, so therefore they "share" a charge. Because of this, If C3 increases, C2 has no choice but to decrease in charge. Since one could argue that C1 is connected to C3 via a series connection, then that would mean charge is the same - thus meaning that if C3's Q increases, C1's charge MUST increase as well. Is that right?
I would really appreciate any assistance! thanks This is also my first post here.
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