Capacitors in series or in parallel? - calculating CR

In summary: when the total capacitance of two components is greater than the capacity of a single component, they will be in series.
  • #1
nwmyname
8
0

Homework Statement


Screen_Shot_2017_06_28_at_19_11_44.png


I have the circuit above.
When I have the switch at A, the capacitor C1 is charging at 100μF is at 10V.
I then discharge the capacitor by connecting the switch to B.

I was asked to calculate the time constant.
I thought the capacitors were in parallel but according to the answer, C1 and C2 are in series.
I calculated the time constant to be 20 seconds, but the answer was apparently 5 seconds as C1 and C2 were in series.

I attempted a similar question as shown below.
Screen_Shot_2017_06_28_at_19_21_26.png


but they said that the total capacitance was 6.0μF where the capacitors were in parallel for part (i) which I can understand.

Screen_Shot_2017_06_28_at_19_32_20.png


So who is correct for the first question?

Homework Equations


T = CR

The Attempt at a Solution


Total C = 100 + 100 = 200μF
So T = 100x103 x 200x10-6 = 20 seconds.
 
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  • #2
The capacitors are in series. The inclusion of the 100 kΩ between their connections means that they cannot be treated as a parallel connection. In order to be parallel-connected, both leads of both capacitors must be paired with no other circuit element to be traversed.
 
  • #3
gneill said:
The capacitors are in series. The inclusion of the 100 kΩ between their connections means that they cannot be treated as a parallel connection. In order to be parallel-connected, both leads of both capacitors must be paired with no other circuit element to be traversed.
how is that then different from the second question I showed where S1 was open and S2 was closed?
 
  • #4
nwmyname said:
how is that then different from the second question I showed where S1 was open and S2 was closed?
The second problem is of a different type (Note that you were not asked to find a time constant there -- in fact you would be hard pressed to do so as there is no defined resistive element at all involved). In the second problem the capacitors are directly connected to each other, no components separating them. The method of solution would also be different, likely relying on conservation of charge.
 
  • #5
gneill said:
The second problem is of a different type (Note that you were not asked to find a time constant there -- in fact you would be hard pressed to do so as there is no defined resistive element at all involved). In the second problem the capacitors are directly connected to each other, no components separating them. The method of solution would also be different, likely relying on conservation of charge.

so in the second problem, they are in parallel but in the first one they are in series?
 
  • #6
nwmyname said:
how is that then different from the second question I showed where S1 was open and S2 was closed?

With switch S2 closed the capacitors are in parallel and series with each other at the same time. Consider...

Just before S2 is closed the 4.5uF charged and the 1.5uF not. So when S2 is closed current will flow from the 4.5uF to the 1.5uF circulating around the right hand loop. In this state the capacitors look like they are in series. The circuit behaves like the first one except the resistance is very small (R->0) instead of 100k .

Eventually (rather quickly!) the voltage on both capacitors will become the same. From then on the capacitors look more like they are in parallel.
 
  • #7
CWatters said:
With switch S2 closed the capacitors are in parallel and series with each other at the same time. Consider...

Just before S2 is closed the 4.5uF charged and the 1.5uF not. So when S2 is closed current will flow from the 4.5uF to the 1.5uF circulating around the right hand loop. In this state the capacitors look like they are in series. The circuit behaves like the first one except the resistance is very small (R->0) instead of 100k .

Eventually (rather quickly!) the voltage on both capacitors will become the same. From then on the capacitors look more like they are in parallel.

So in the first example, when I move the switch to B, the capacitors are in series.
In the second example, however, the total capacitance is in parallel... so what is the difference?
 
  • #8
nwmyname said:
So in the first example, when I move the switch to B, the capacitors are in series.
In the second example, however, the total capacitance is in parallel... so what is the difference?
Hint: Current through series elements is same and voltage across parallel elements is same.
Can you see the difference now?
 
  • #9
When you have just two components connected in a closed circuit, whether to treat them as series or parallel connected can be a bit ambiguous. In fact, such a connection satisfies both the definitions of series connection and parallel connection! That means that you can treat them as either series-connected or parallel connected, whichever is most convenient for your analysis at the time.

One caveat that you should be aware of is that sometimes a given problem will implying that you need to look at a circuit from a particular point of view, that is, you should judge the circuit from the point of view of a certain pair of nodes (connection points). In such a case that may force your interpretation of the connection as being series or parallel. In your second problem the implication of wanting to know the potential across both capacitors implies that you should construe the circuit in the following way:

upload_2017-6-28_15-57-14.png


so that the parallel connection of the capacitors is the preferred interpretation. You could also treat them series-connected and solve for the potentials across the capacitors separately (and find out in the end that they are equal), but it's a lot more work to do it that way!

One way to identify whether two components are in parallel is whether you are able to trace a closed path through those components without having to pass through any other circuit component. If you can, then they are parallel-connected. If you cannot, they are not parallel-connected.
 

Related to Capacitors in series or in parallel? - calculating CR

1. What is the difference between capacitors in series and in parallel?

The main difference between capacitors in series and in parallel is the way they are connected. In series, the capacitors are connected end to end, while in parallel, they are connected side by side. This results in different total capacitances and charge distributions.

2. How do I calculate the total capacitance of capacitors in series?

To calculate the total capacitance of capacitors in series, you can use the formula: 1/Ct = 1/C1 + 1/C2 + ... + 1/Cn, where Ct is the total capacitance and C1, C2, ..., Cn are the individual capacitances of each capacitor.

3. How do I calculate the total capacitance of capacitors in parallel?

The total capacitance of capacitors in parallel can be calculated using the formula: Ct = C1 + C2 + ... + Cn, where Ct is the total capacitance and C1, C2, ..., Cn are the individual capacitances of each capacitor.

4. Can I mix capacitors in series and in parallel in the same circuit?

Yes, you can mix capacitors in series and in parallel in the same circuit. However, it is important to calculate the total capacitance of each configuration separately before combining them to ensure proper functioning of the circuit.

5. How do I calculate the time constant (CR) for a circuit with capacitors in series or in parallel?

The time constant (CR) for a circuit with capacitors in series or in parallel can be calculated using the formula: CR = Ct * R, where Ct is the total capacitance and R is the resistance in the circuit. This value represents the amount of time it takes for the capacitor to charge or discharge to approximately 63% of its maximum charge.

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