# Capacitor in series voltage problem

• terryds
In summary, when S3 is closed in the given circuit, the voltages across the capacitors will no longer be equal and can be determined using Kirchhoff's Voltage Law (KVL). The sum of the charges on the connected plates will remain the same, but each capacitor will experience a different change in charge. Equations can be written for the voltages and charges on each capacitor to solve for the voltage across each resistor.
terryds

## Homework Statement

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!

## Homework Equations

1/C = 1/C1 + 1/C2 (for series)[/B]

## 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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terryds said:

## Homework Statement

View attachment 113980

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!

I do not see any resistors in the problem.
terryds said:

## Homework Equations

1/C = 1/C1 + 1/C2 (for series)[/B]

## 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?)

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?

ehild said:
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?

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?

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

ehild said:
W

What do you think? can the charge on C3 go anywhere?
Of course, it can go to C2, right?

From the right plate?

ehild said:
From the right plate?

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

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

terryds
ehild said:
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.

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

terryds said:
So,
V1 = V2 = V3 = 200 V
Is it right?
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").

gneill said:
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").

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

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

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

terryds
ehild said:
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) .
View attachment 114006
gneill said:
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.

##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?

terryds said:
##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?

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]

ehild said:
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]

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

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!

ehild said:
What's the difference between -Q1 and -q1?
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

terryds said:
##
-Q1+Q2 = -q1+q2 \\
-Q1+Q3 = -q1+q3 \\
\\
q2 - q1 = 4*10^{-4} \\ q3 - q1 = 8*10^{-4} \\ q3 - q2 = 4*10^{-4} \\
V_1 + V_2 + V_3 = 0 \\##I still can't get q1, q2, and q3 values
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 ...

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.

## 1. How do I calculate the total voltage in a series circuit with capacitors?

To calculate the total voltage in a series circuit with capacitors, you need to first find the equivalent capacitance of the circuit by adding the individual capacitances together. Then, you can use the formula V = Q / C, where V is the total voltage, Q is the total charge, and C is the equivalent capacitance. This will give you the total voltage across the entire series circuit.

## 2. What is the role of a capacitor in a series circuit?

In a series circuit, a capacitor acts as a charge storage device. It stores energy in the form of an electric field between its two plates. This stored energy can then be released when needed, such as during a sudden change in voltage or when the circuit is turned off.

## 3. How does the total capacitance change when capacitors are connected in series?

When capacitors are connected in series, the total capacitance decreases. This is because the equivalent capacitance of a series circuit is equal to the reciprocal of the sum of the reciprocals of each individual capacitance. Therefore, as more capacitors are added in series, the equivalent capacitance decreases.

## 4. Can capacitors in series cause a voltage drop?

Yes, in a series circuit, the total voltage is divided among the individual components. This means that each capacitor will receive a portion of the total voltage, resulting in a voltage drop across each capacitor. The sum of these voltage drops will equal the total voltage of the circuit.

## 5. What happens to the total voltage in a series circuit when a capacitor is removed?

If a capacitor is removed from a series circuit, the total voltage across the remaining capacitors will increase. This is because the equivalent capacitance of the circuit decreases, resulting in a higher voltage across the remaining capacitors. However, the total voltage across the entire circuit will remain the same, as it is determined by the voltage source and the sum of all the individual voltage drops.

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