Capacitor in series voltage problem

1. Mar 2, 2017

terryds

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

There are three capacitors C1 = 2 uF, C2 = 4 uF, C3 = 6 uF. Each of these capacitors were connected to 200-V voltage source so every capacitor has been fully charged. Then, the three capacitors are connected like the image above. When S1 and S2 are closed, but S3 is opened, determine the voltage across each resistors!

2. Relevant equations

1/C = 1/C1 + 1/C2 (for series)

3. The attempt at a solution

I wonder if the voltage across the resisors will be equal one to another since there'll be no current flowing after they achieve equilibrium condition. So, I think it is V_1 = V_2 = V_3 and q1+q2+q3 = q_total where q_total is 200(C1+C2+C3) because they have all been charged with 200V source.

But, I remember in determining the substitute capacitor for capacitors in series, the charges on each resistor is equal to one another which means q_1 = q_2 = q_3 and V_1+V_2+V_3 = 200V (maybe?)

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2. Mar 2, 2017

ehild

I do not see any resistors in the problem.
The situation is not the same as capacitors initially uncharged are connected series or parallel and then connected to a voltage source. Can the charge on the left plate of C1 go anywhere? Can the charge on the right plate of C3 go anywhere? Do the initial charges on the plate change when switches S1 and S2 get closed?

3. Mar 2, 2017

terryds

I mean capacitors not resistors sorry.
Initial charge do not change.
Charge on left capacitor has nowhere to go, I think.
So, V1 = Q1*C1 = 200V
But what about c2 and c3? Is the voltage equal?

4. Mar 2, 2017

ehild

W
What do you think? can the charge on C3 go anywhere?

5. Mar 2, 2017

terryds

Of course, it can go to C2, right?

6. Mar 2, 2017

ehild

From the right plate???

7. Mar 2, 2017

terryds

It can go to C2 (going left) because the switch is closed, right?

8. Mar 2, 2017

ehild

C3 was charged with Q3= 200*6E-4=1.2 E-3 C. That means 1.2E-3 C on the left plate and - 1.2 E-3 C on the right plate. The negative charge on the right plate can not go anywhere, and it attracts the positive charges on the left plate more than the negative charge on the right plate of C2, so they stay there.

9. Mar 2, 2017

terryds

So,
V1 = V2 = V3 = 200 V
Is it right?

10. Mar 2, 2017

Staff: Mentor

If you mean when S1 and S2 are closed and S3 is still open, then yes.

In circuits, charge cannot move without a there being closed path (a "circuit").

11. Mar 2, 2017

terryds

But, if S3 is closed, is it correct to use
V1 = V2 = V3 and q1+q2+q3 = 200 (C1+C2+C3) ?

12. Mar 2, 2017

Staff: Mentor

No. Once charge starts moving the voltages on the capacitors will definitely no longer be equal.

When charge moves in a series circuit, the same current must flow through each component. That means each of the capacitors experiences the same change in charge. Since the capacitors all have different values, the same change in charge will cause different changes in voltage for each.

Knowing that the change in charge will be the same for each capacitor, you'll need to come up with an expression to solve for that change. I'd suggest using KVL around the closed loop.

13. Mar 2, 2017

ehild

When S3 gets closed, a closed loop forms, and the sum of voltages across the capacitors becomes zero, according to KVL.
The charges on the connected plates can move from one plate to the other, but the sum stays the same as it was. You can write one equation for the voltages and also equations for the new charges on the capacitors (q1, q2, q3) .

14. Mar 3, 2017

terryds

$dq_1 = dq_2 = dq_3 \\ C_1 dV_1 = C_2 dV_2 = C_3 dV_3 \\ dV_1 + dV_2 + dV_3 = 0 \\ dq_1/C_1 + dq_2/C_2 + dq_3/C_3 = 0$

How to get the voltage??
Is my equation correct?

15. Mar 3, 2017

ehild

What are the dq-s? It is better to calculate with the charges and voltages instead of their changes.
The original charges were Q1, Q2, Q3, and the voltage across each capacitor was 200 V. What were the individual charges?
Originally, there were positive charges on the left plates, and negative charges on the right plates. So it was -Q1 charge on the right plate of capacitor C1 and Q2 charge on the left plate of capacitor C2, -Q1+Q2 is the charge on the two plates together. The sum stays the same when the capacitors are connected, so -Q1+Q2=-q1+q2. You have similar equations for each pair of capacitors.[/QUOTE]

16. Mar 3, 2017

terryds

[/QUOTE]

What's the difference between -Q1 and -q1?

17. Mar 3, 2017

ehild

What's the difference between -Q1 and -q1?[/QUOTE]
Q1 is the charge on the left plate of capacitor C1 before connecting all capacitors, and q1 is the charge after connecting them. They are quite different!

18. Mar 3, 2017

terryds

Q1 is the charge on the left plate of capacitor C1 before connecting all capacitors, and q1 is the charge after connecting them. They are quite different![/QUOTE]

$-Q1+Q2 = -q1+q2 \\ -Q2+Q1 = -q2+q1 \\ -Q2+Q3 = -q2+q3 \\ -Q3+Q2 = -q3+q2 \\ -Q1+Q3 = -q1+q3 \\ -Q3+Q1 = -q3+q1 \\ \\ V_1 + V_2 + V_3 = 0 \\ -C_1V_1 + C_2V_2 = -q1+q2 \\ q2 - q1 = 4*10^{-4} \\ q3 - q1 = 8*10^{-4} \\ q3 - q2 = 4*10^{-4} \\$

I still can't get q1, q2, and q3 values

19. Mar 3, 2017

ehild

Write V1, V2 , V3 as q1/C1, q2/C2, q3/C3. And write q2 and q3 in terms of q1 as q2 = 4*10-4+q1 ....

20. Mar 3, 2017

Staff: Mentor

Since the change in charge is identical for every capacitor in a series circuit you only need one variable to represent it. Call it Δq. Each capacitor will experience a change in voltage according to its capacitance and Δq: $ΔV = \frac{Δq}{C}$ .

Now, in this problem every capacitor starts out with the same voltage, $V_o = 200~V$.

The Δq that moves when S3 is closed affects each of the capacitor's charges equally, and so each experiences a change in voltage according to this Δq and the capacitor value as described above. So a given capacitor will end up with a voltage:

$V_C = \left( 200 - \frac{Δq}{C} \right)$

Use this to write your KVL for the final state and solve for Δq. One equation, one unknown.