# Capacitator Circuit and determining unknown charges voltages

1. Feb 24, 2013

### liliacam

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant 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

3. 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.

Last edited by a moderator: May 6, 2017
2. Feb 24, 2013

### tiny-tim

welcome to pf!

hi liliacam! welcome to pf!
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 ?

3. Feb 24, 2013

### liliacam

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 [Broken]

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.

Last edited by a moderator: May 6, 2017
4. Feb 27, 2013

### liliacam

Nevermind, figured it out.