Capacitator Circuit and determining unknown charges voltages

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The discussion revolves around solving a capacitor circuit problem involving four capacitors with given capacitance values and charge. The user initially miscalculated the charges on the capacitors, particularly misunderstanding the series and parallel configurations. After receiving feedback, they redrew the circuit and correctly identified the relationships between the charges and voltages. The final calculations yielded Q1 = Q2 = Q4 = 37.4 μC and Q3 = 14.175 μC, confirming the correct approach to the problem. The user expressed satisfaction upon resolving their confusion about the circuit's behavior.
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Homework Statement



Suppose in the figure (Figure 1), that

C1 = C2 = C3 = 18.0μF and C4 = 28.2μF.

If the charge on C2 is Q2 = 37.4μC, determine the charge on each of the other capacitors
Determine the voltage across each capacitor.
Determine the Voltage Vab across the entire combination.

https://dl.dropbox.com/u/47465778/MP24.24.PNG


Homework Equations



Q=CV

Rules for Series:
Q is the same across a series
\frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + . . .
Vtotal = V1 + V2 . . .

Rules for Parallel:
Qtotal = Q1 + Q2 . . .
C_(eq) = C1 + C2 . . .
V is same across parallel

The Attempt at a Solution



Let 1 and 2 be in Series. Then Capacitance can be added as reciprocals
\frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18}
Ceq = 9μF for 1 and 2

Q1=Q2...because they are in series, Q is same. Q1 = Q2 = 37.4 μC

Q = CV. So,
37.4 = 9*V
V = 4.155 V

I figured that because Voltage is the same in parallel, then this voltage is the same across the board for 3 and 4. So I found C for 3 and 4

3 and 4 are in series with each other, so
\frac{1}{Ct} = \frac{1}{18} +\frac{1}{28.2}
Ct = 10.987...
Then Q = CtV
= 10.987*4.155 = 45.65 μC

Q3 = Q4 because Q's are equal in series.

Final Answer: Q1=Q2 = 37.4 and Q3 = Q4 = 45.65 all in microColoumbs (Wrong)

Although I'm fairly certain Q1 does equal Q2, but since I have to enter all of the answers at once, if one answer is wrong then the rest is wrong.

Where did I go wrong? Please help, I did my best but I'm pretty shaky on this stuff. Also this is my first time posting so sorry if I did anything weird with the formatting.
 
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welcome to pf!

hi liliacam! welcome to pf! :smile:
liliacam said:
Q3 = Q4 because Q's are equal in series.

no, they're not in series, there's a junction between them!

also, what happens to the electrons that leave the top plate of Q4 ?

they have to spread over the left plates of both Q1 and Q3

so how could they all be on Q3 ? :wink:
 
Hi! Thanks for responding so quickly

So I tried it again...I guess the diagram confused me. I redrew it here:

https://dl.dropbox.com/u/47465778/MP24Redrawn.PNG

With 3 still in parallel and 1, 2, 4 in series with each other. My current teacher still hasn't taught us how to redraw capacitors but I hope this is on the right track at least. So the electrons from 4 spread over 1 and 2, then 3?

If this drawing is correct, then:

Q1 = Q2 = Q4 = 37.4μC
Q3 = ?

\frac{1}{Ceq} = \frac{1}{C1} + \frac{1}{C2} + \frac{1}{C4}
\frac{1}{Ceq} = \frac{1}{18} + \frac{1}{18} + \frac{1}{28.2}
Ceq for 1,2,4= 6.822

New Capacitance:
Ceq total = Ceq for 1,2,4 + C3
Ceq total = 6.822 + 18 = 24.822 μF

New Voltage:
Q = CV
Q1 = C1V1
V1 = 2.077 V
If 1, 2, 4 is in parallel with 3, then V is the same across

V3 = 2.077 V

Q total = Ctotal*Vtotal
Qt = 24.822*2.077
Qt = 51.575 μC

Qt = Q3 + Q(1,2,4)
51.574μC = Q3 + 37.4
Q3 = 14.175 μC


Final Answers: Q1 = Q2 = Q4 = 37.4, Q3 = 14.75 all μC


Is this correct? I haven't had an official lecture on this yet, so it's just hard for me to visualize what exactly is going on.
 
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Nevermind, figured it out.
 
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