Intuitive understanding of capacitors in series and parallel

[psilocybin]

Homework Statement

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)

Homework Equations

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

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

 

[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

 

[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 current doesn't go through the capacitor from one plate to the other. So let's 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?

 

[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.

 

[Nickie]

for your time.

Dear student,

Firstly, let's break down the problem into smaller parts to better understand it. We have three capacitors in series, C1, C2, and C3. Let's label the charge on each capacitor as Q1, Q2, and Q3 respectively.

In a series circuit, the same current flows through each component. This means that the charge on each capacitor will be the same. So, Q1 = Q2 = Q3.

Next, we can use the equation Q=CV to find the charge on each capacitor. We know the voltage source is 12V, and the total capacitance is 5.4μF. So, Q = (5.4μF)(12V) = 64.8μC.

Now, we can use the equation 1/Ceq = (1/C1) + (1/C2) + (1/C3) to find the equivalent capacitance of the three capacitors in series. Plugging in the values, we get 1/Ceq = (1/4μF) + (1/5μF) + (1/3μF) = 23/60μF. Therefore, Ceq = 60/23μF.

Finally, we can use the equation Q=CV to find the charge on C2. Since the charge on each capacitor is the same, we can use the total charge we found earlier, 64.8μC, and the equivalent capacitance we just calculated, 60/23μF, to get Q2 = (60/23μF)(64.8μC) = 168.7μC.

I hope this explanation helps you understand the problem better. In general, capacitors in series have a smaller equivalent capacitance than each individual capacitor, while capacitors in parallel have a larger equivalent capacitance. This is because in series, the total charge is the same but the voltage is divided, while in parallel, the voltage is the same but the charges are added together.

As for the intuitive understanding, think of capacitors as tanks that can store charge. In series, the charge has to pass through each tank, so the total charge that can be stored is limited by the smallest tank. In parallel, the charge can be divided among multiple tanks, so the total charge that can be stored is larger.

I hope this helps. Keep practicing and don't