What I meant was in a series, the Q's in Q=CV are the same for each capacitor in that series and in a parallel series the V's in Q=CV are the same.
Thanks for explaining the differences, why the book couldn't make it that easy I'll never understand
I was distinguishing between capacitor C2 and C, but since they are equal it would just be C&C. Does that clear it up? If not I did this C*C/C+C=C/2
current, charge, voltage...I still can't get the differenct between them and it's really hard becuase I can't picture the difference in my head...
I'm sorry, I made a huge mathematical mistake from the beginning, I edited it so it should make better sense now.
when I say the capacitors in the series, I'm talking about the one labled C2 in my picture and the one right below it.
Then the one that is parallel to those 2 would be the...
Capacitance across the series is C2*C/C+C or C2*C/2 which I assume I can make into just C^2/2Csince all the C's are equal
Capacitance across all three would be 3C/2
Then if I make it all into just 1 capactior it would be 3C/5
The current across it would be q=CV so q=10(3*10)/(5) since...
Parallel capacitors is q/V so C1+C2+C3...
Series capacitors are 1/(1/C1+1/c2+1/C3)
Yes all the Cs are equal. I think I had my series equation wrong but I just don't understand it. I know I have to use the voltage across the capactiors but I'm also not sure how to find that.
Capacitance -- How handle multiple caps in series and parallel?
I can't draw the circut for the cuestion so I attached it.
I am having trouble finding the charge across capacitor 2. I don't understand what to do with the 3 capactiors right in that group and how to figure the change in voltage
I am having trouble with this derivation: f(q)=q(Q-q). We want to find the values that maximize the vaule of q so the book sets the derivative equal to 0 and gets Q-2q=0
When I try to do the derivative of f(q)=q(Q-q), I used the product rule
1(Q-q) + q(Q-1)=0
Q-q+Qq-q=0
Q+Qq-2q=0
I...