Capacitator Circuit and determining unknown charges voltages

[liliacam]

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.

 

[tiny-tim]

welcome to pf!

hi liliacam! welcome to pf!

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 ? ​

 

[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

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.

 

[liliacam]

Nevermind, figured it out.

 

[Nickie]

I would first like to commend you for attempting to solve this problem on your own and providing a detailed explanation of your thought process. That being said, there are a few errors in your solution.

Firstly, when capacitors are in series, the equivalent capacitance is calculated as follows:

\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...

In your solution, you have only added the reciprocals of two capacitors in series, which is incorrect. The correct calculation would be:

\frac{1}{C_{eq}} = \frac{1}{18} + \frac{1}{18} + \frac{1}{18} = \frac{3}{18} = \frac{1}{6}

Therefore, the equivalent capacitance for capacitors 1, 2, and 3 in series would be 6 μF, not 9 μF as you have calculated.

Next, when capacitors are in parallel, the equivalent capacitance is calculated as follows:

C_{eq} = C_1 + C_2 + C_3 + ...

In your solution, you have added the capacitances for capacitors 3 and 4 in series, which is incorrect. The correct calculation would be:

C_{eq} = 18 + 28.2 = 46.2 μF

Now, to determine the charge on each capacitor, we can use the equation Q = CV. Since we know the charge on capacitor 2 (Q2 = 37.4 μC) and the equivalent capacitance for capacitors 1, 2, and 3 in series (Ceq = 6 μF), we can calculate the voltage across this series combination as follows:

Q2 = Ceq * V
V = \frac{Q2}{Ceq} = \frac{37.4}{6} = 6.23 V

Now, using this voltage, we can calculate the charge on each capacitor as follows:

Q1 = C1 * V = 18 * 6.23 = 111.84 μC
Q2 = C2 * V = 18 * 6.23 = 111.84 μC
Q3 = C3 * V = 18 * 6.23 =