Is this capacitor association in series or in parallel?

[pc2-brazil]

I found this problem while self-studying "Physics" by Resnick, Halliday and Krane.

Homework Statement

When the switch of the attached figure is turned to the left, the plates of the capacitor of capacitance C₁ acquire a potential difference V₀. Initially, C₂ and C₃ are discharged. The switch S is now turned to the right. What are the values of the final charges q₁, q₂ and q₃ of the respective capacitors?

Homework Equations

Capacitance:
C = \frac{Q}{V}

The Attempt at a Solution

When the switch is turned to the left, C₁ gets a charge q_0 = C_1V_0.
After the switch S is turned to the right, it initially seems that the capacitors C₁, C₂ and C₃ should be in series. Then, the charges of C₁, C₂ and C₃ would stay equal. But, according to a solution I found, the potential difference across C₁ stays the same as the potential difference across the combination of C₂ and C₃ (which are in series). Why? This seems to imply that C₁ is in parallel with the combination of C₂ and C₃.

Thank you in advance.

 

[gneill]

When components are connected in series around a closed loop as is the case with these three capacitors, there are several ways of looking at it. You can isolate any single component and treat the rest as being in parallel with it. As long as KVL is satisfied going around the loop, all is well.

In some cases it may be easier (or at least more obvious!) to analyze a circuit by isolating a particular component in such a fashion. In this case, only C1 is starting out with a charge on it, and that charge is going to end up distributed over all three components. It just happens that the logic behind distributing a charge across two parallel capacitors, which must share their final potential difference, is pretty straightforward. So, isolate C1 with its initial charge, and add C2 and C3 to find its equivalent capacitance. Since neither C2 nor C3 had any initial charge there's nothing to be done with charge distribution for this pair when they are so combined into an equivalent capacitance. Proceed to do the charge distribution over the resulting pair, and then determine the individual charges.

 

[pc2-brazil]

When components are connected in series around a closed loop as is the case with these three capacitors, there are several ways of looking at it. You can isolate any single component and treat the rest as being in parallel with it. As long as KVL is satisfied going around the loop, all is well.

In some cases it may be easier (or at least more obvious!) to analyze a circuit by isolating a particular component in such a fashion. In this case, only C1 is starting out with a charge on it, and that charge is going to end up distributed over all three components. It just happens that the logic behind distributing a charge across two parallel capacitors, which must share their final potential difference, is pretty straightforward. So, isolate C1 with its initial charge, and add C2 and C3 to find its equivalent capacitance. Since neither C2 nor C3 had any initial charge there's nothing to be done with charge distribution for this pair when they are so combined into an equivalent capacitance. Proceed to do the charge distribution over the resulting pair, and then determine the individual charges.

Thank you for the answer.
What guarantees that the potential difference of C₁ will be the same as the potential difference across C₂ and C₃? Is it the fact that this is a closed loop?
I thought that the three capacitors should be analyzed in series, and, thus, the charge on each capacitor should be equal.

 

[gneill]

Thank you for the answer.
What guarantees that the potential difference of C₁ will be the same as the potential difference across C₂ and C₃? Is it the fact that this is a closed loop?
I thought that the three capacitors should be analyzed in series, and, thus, the charge on each capacitor should be equal.

The potential difference across the series combination of C2 and C3 will equal the potential difference across C1, because doing KVL around the loop, V1 - V2 - V3 = 0, or rearranged, V1 = V2 + V3.

You have to be a bit careful when you're dealing with circuits where some components are 'carrying over' some initial conditions from previous activity. In this case C1 comes with some charge q1 = C1*V already on it. Charge will flow from C1 onto the equivalent capacitor comprising C2 and C3 until the voltages balance (the voltage on C1 equals the sum of the voltages on C2 and C3).

While C2 and C3 must have the same charge on each (since they must experience exactly the same current, being in series), C1 is not so constrained because it already had a charge (mind you, the CHANGE in charge in C1 will equal the charge that ends up on both C1 and C2).

 

[pc2-brazil]

The potential difference across the series combination of C2 and C3 will equal the potential difference across C1, because doing KVL around the loop, V1 - V2 - V3 = 0, or rearranged, V1 = V2 + V3.

You have to be a bit careful when you're dealing with circuits where some components are 'carrying over' some initial conditions from previous activity. In this case C1 comes with some charge q1 = C1*V already on it. Charge will flow from C1 onto the equivalent capacitor comprising C2 and C3 until the voltages balance (the voltage on C1 equals the sum of the voltages on C2 and C3).

While C2 and C3 must have the same charge on each (since they must experience exactly the same current, being in series), C1 is not so constrained because it already had a charge (mind you, the CHANGE in charge in C1 will equal the charge that ends up on both C1 and C2).

Thank you for the clarification. This is very interesting. I understand it better now.
I will probably understand it more consistently when I learn about Kirchhoff's laws (in the book I'm reading, the part on capacitors comes before the part on current and resistance).