# Capacitors charging capacitors

1. Feb 24, 2005

### biigturkey

I'm completely boggled by this problem:

"A 12.0V battery is used to charge a 4.00μF capacitor. A switch is then thrown disconnecting that capacitor from the battery and connecting it to a circuit with two capacitors (3.00uF and 6.00μF) connected in series. What are the final charges on the capacitors?"

I'm not looking for the answer to the problem, just some nudges in the correct direction.

The initial charge on the 4.00μF capacitor is easy enough...

q = C*V = (4.00*10^-6)(12.0) = 48μC

Next I assume that I'll need the voltage in the new circuit (the sans-battery one). The voltage can be obtained by assuming the charge from the old system will carry over and be shared with each capacitor in the new circuit:

V(eq) = q/(C1 + C2 + C3)
V(eq) = 0.48μC/(4.00μF + 3.00μF + 6.00μF) = 3.69V

Now since the capacitors are all running in series:

1/C(eq) = 1/C1 + 1/C2 + 1/C3 = 1.33μF

Which makes sense, since the capacitance of a series is less than the weakest capacitor in that series.

Now this is where I run into troubles. Referencing the numbers I get for charges using a variety of methods do not agree with the answers in the back of the book. My methodology is flawed here in the final step and I'm not sure how.

I figured that since I know the capacitance of each capacitor in the series now (1.33μF) and the voltage of each (3.69V) that multiplying the two should give me charge on each capacitor:

q = CV
q = 1.33μF * 3.69V = 4.92μC

If anyone can possibly nudge me in the correct direction here without violating any of the forum rules I'd be much obliged.

2. Feb 24, 2005

### vincentchan

combine your last two capacitor so that your question will looks simpler
1/C_4 = 1/C_2 + 1/C_3

the first capacitor(C_1) has charge 48 and connected to the C_4

first, you know after they were connected, the voltage on C_1 and C_4 are same, therefore, Q_1/C_1 = Q_4/C_4 ---------(1)

the charge Q_1 and Q_4 add up to 48, because there charges is came from the C_1 capacitor
Q_1+Q_4 = 48-----------(2)

you have 2 equation and 2 unknown now....

3. Feb 24, 2005

### biigturkey

4. Feb 24, 2005

### CartoonKid

I noticed a weird thing in Vincent's solution, using his method, I got $$32\mu C$$ for Q1 and $$16\mu C$$ for Q4. After finishing all the calculation, how come C1 which is only $$4\mu F$$ has more charges than the one which is $$6\mu F$$ if they are all in series and have same potential??

5. Feb 24, 2005

### CartoonKid

I think eventually they will all have the same amount of charges because they are connected in series. What will actually change is the final voltage.
My approach is as follow:
$$Q = C_{1}V_1 = 48\mu C$$
$$C_4 = (C_{1}^{-1} + C_{2}^{-1} +C_{3}^{-1})^{-1}$$
$$C_4 = \frac{4}{3} \mu F$$
$$C_1 V_1 = C_4 V_4$$
$$V_4 = 36 V$$
To further check, I do:
$$V_1 = \frac{Q_1}{C_1} = 12 V$$
$$V_2 = \frac{Q_2}{C_2} = 16 V$$
$$V_3 = \frac{Q_3}{C_3} = 8 V$$
$$V_1 + V_2 + V_3 = 36 V = V_4$$
These are what I thought.
Can somebody please verify which one is the correct approach...Thank you.

6. Feb 24, 2005

### vincentchan

your calculation is completely wrong (both in #4 and #5)

in #4, you make some algebra mistake (or arithmetic mistake), go back and check...
in #5, your concept is totally messed up, C_4 is a LOOP of capacitor, not series....... you don't get any equivalence capacitance for a LOOP of capacitor....

7. Feb 24, 2005

### CartoonKid

Ok, now I work out what I understood from your solution,
$$C_1 = 4\mu F, C_2 = 3\mu F, C_3 = 6\mu F$$
So, $$C_4 = (C_{2}^{-1} + C_{3}^{-1})^{-1} = 2\mu F$$
$$\frac{Q_1}{C_1} = \frac{Q_4}{C_4}$$
$$\Rightarrow Q_1 = 2Q_4 ---- 1$$
$$Q_1 + Q_4 = 48\mu C ---- 2$$
$$2Q_4 + Q_4 = 48\mu C$$
$$Q_4 = 16\mu C$$
$$Q_1 = 48 - 16 = 32\mu C$$
Right up to here, the $$C_1$$ can be concluded as having the most charges in the circuit. This is the point I don't understand.
Can you please pointed out my mistake. Thank you. I am a bit lost.

8. Feb 24, 2005

### CartoonKid

Vincent, I think I know my problem. The circuit is considered as if $$C_1$$ is connected parallelly to $$C_2$$ and $$C_3$$ which are in series. Therefore, it is reasonable for the $$6\mu F$$ capacitor to have smaller charge compare to the $$4\mu F$$ capacitor because it is in a series which is parallel to the $$4\mu F$$ capacitor. Am I right this time? Thank you.

9. Feb 24, 2005

### vincentchan

in a circuit, the voltage of each element on a loop add up to zero..... therefore, C1 and C4 has the same voltage...
$$Q=VC$$ or $$V=\frac{Q}{C}$$
if the voltages are same, the higher the capacitance, the higher the charge, therefore, C_1 has more charge than C_4

I didn't check your arithmetics carefully but I assume you did every right, why surprise C_1 has more charge than C_3? the charge is originally came from C_1!

just wanna notice that the because the charge is came from C_1, the C_2 and C_3 must have same charge , (hopefully you can see why)

10. Feb 24, 2005

### CartoonKid

Thank you very much, Vincent. Now I have a clear concept of it already.