# Intuitive understanding of capacitors in series and parallel

1. Jun 29, 2013

### psilocybin

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

I am trying to determine the charge on C2 which is a 3 μF cap in parallel with a 2 μF cap that are both in series with a 4 μF capacitor. The voltage source is 12V

(diagram attached)

2. Relevant equations

Q=CV
1/Ceq= (1/C1)+(1/C23)

3. The attempt at a solution

(work attached as img)

Hi,
So I have been attempting to develop an intuitive understanding of capacitance and capacitors for almost 2 weeks now, and am driving myself crazy. I get the basics of what happens to cause capacitance, and what a dielectric does, but once there are multiple capacitors in a circuit it is getting very hectic for me. I have been working on this problem all day:

I am trying to determine the charge on C2 which is a 3 μF cap in parallel with a 2 μF cap that are both in series with a 4 μF capacitor. The voltage source is 12V. I initially thought it was as easy as (1/4)+(1/5)=9/20 which makes the total capacitance 5.4 μF

After that I thought it was as simple as 5.4(3μF) to get an answer of 16.2, but I don't believe this to be correct. The voltage across C1 needs to be taken into consideration right? I have been attempting to solve this on my own, but after reading every page I can google on capacitance and series and parallel circuits, I am so confused that I think I am doing more damage at this point.

My text doesn't provide any examples other than formulas and the basic idea. Any help figuring out this problem and any intuitive assistance would be greatly appreciated. If anyone would be willing to help me with understanding what is happening in each capacitor that would really help. Thank you
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### cap_qc2.jpg
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2. Jun 29, 2013

### technician

I think you have made a common mistake in this sort of calculation.....
You have 1/C = 1/4 + 1/5 = 9/20
This means that C = 20/9 microfarads

3. Jun 29, 2013

### barryj

The equivalent capacitor for a 3 and a 2 in parallel is indeed 5. Now you have a 4 and a 5 in series.

When you put 12 V across series capacitors, each capacitor has the same charge as curent doesn't go through the capacitor from one plate to the other. So lets assume when you apply 12 volts you have a charge of Q on both capacitors. So the voltage on the 4 capacitor = V = Q/C = Q/4. Likewise the voltage on the 5 capacitor is Q/5. Now, the voltages must add to 12 volts, right so Q/4 + Q/5 = 12. Solving for Q you get Q = 26.67. Now that you have the Q, the voltage on the 4 cap is Q/C = 26.67/4 = 6.67 volts and the voltage on the 5 C is 26.667/5 = 5.33. Add them up and you get 12 volts. Does this help?

4. Jun 29, 2013

### psilocybin

Ah, I do see the error with the reciprocal for Ceq, thank you. So then since both C2 and C3 must have the same voltage, 5.33 V,

Q2/3μF=5.33V

3 μF(5.33 V)=Q

Q2= 15.99 μC

correct?

Taking this farther, I could do the following calculation to see that the charge on the other capacitor in parallel is the remaining amount to verify my result. So does everything seem to be what it should?

Q3=CV
(2μF)(5.33 V)=10.66 μC

Q2+Q3=Q
15.99+10.66≈ 26.67μC

Thank you for all the help, the intuition still needs a little work, but it is making more sense.