Energy and Voltage of Capacitor Circuits

In summary, a capacitor of 4.0uF charged to a potential difference of 20V has an initial energy of 0.8 mJ. When an uncharged capacitor of 2.0uF is connected in parallel, the potential difference across the initial capacitor decreases to 13.3V. The total energy stored in both capacitors after connection is 0.53mJ. While the capacitors can be considered to be in series or parallel, the final result is the same regardless of how they are connected. The discrepancy in energy can be attributed to energy loss through electromagnetic radiation.
  • #1
Jimmy87
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Homework Statement


A capacitor (A) of capacitance 4.0uF is charged to a potential difference of 20V. An uncharged capacior (B) is connected in parallel to capacitor A. What is:
(a) the initial energy stored on capacitor A
(b) the p.d. across A after B has been connected
(c) the final energy stored in A and B

Homework Equations

The Attempt at a Solution


Answers in back of my book are (a) 0.8mJ, (b) 14V (c) 0.53 mJ
I am ok with part (a) - I did E = 1/2 CV^2 = 0.8 mJ (milli Joules) as per the answer
(b) I really don't get this whole circuit problem. For one thing the answer to (b) must be wrong because if you use 14V you don't get answer (c) right. I got 13.3V for (b) which leads to the correct answer they give for (c). The way we were shown these for parallel capacitors where you have one initially charged then add another is to work out the total voltage by dividing the total charge by the total capacitance. The total charge to start with is 8uC (micro coulombs) using Q = CV for the charged capacitor. This gives 13.3V. Since they are in parallel they both get 13.3V. But this is what I really don't get. It implies the 4uF capacitor is charged initally (i.e. there is no power source when B is added. therefore there are only 2 capacitors in the circuit and no power source so how can they be in parallel - surely it is just one loop (i.e. series)? Also the answer of both A and B getting 13.3V adds up to more than the 20V we had to start with. If we do use 13.3V for both capacitors you end up with the correct energy for part (c). But this energy is less than the energy we started with in part (a). Its so confusing, please help!

Thanks
 
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  • #2
You don't mention the size of capacitor B, nor do you show the details of your work so we can't obtain its value from there. So that makes it difficult to comment on your observations about the correctness or lack of correctness of the book's supplied answers.

When there are only two components connected as a circuit you can treat them as either series-connected or parallel-connected. The connections satisfy the requirements for both topologies. You can take advantage of the characteristics of either or both topologies to analyze such a circuit.

Connecting a charged capacitor to an uncharged capacitor in the scenario presented is somewhat analogous to an inelastic collision between two masses. Some things are conserved while others are not. In the case of the mass collision, momentum is conserved but energy is not (there's some loss of kinetic energy). In the case of capacitors, charge is conserved but electric potential energy is not.
 
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  • #3
You don't tell us the value of capacitor B, but it appears to be 2uF if your 13.3V is correct, as I think it is.
Then, as you say, it all works out ok.

As for the energy discrepancy, yes I used to find this strange once. Maybe if you thought of the wires as having resistance, you would see some energy loss to heat as the current is transferred. But that resistance plays no part in your calculations! You get the same answers whatever (finite) value of resistance the wire has. That seems strange, doesn't it? The answer seems to be electromagnetic radiation. So you could think of the energy being lost in either Ohmic resistance in the wire (if that resistance is large) or in radiation resistance (when the Ohmic resistance is small.) Either way it's lost from the caps.
If you consider the gravitational PE of a tank of water, which is the connected to another tank so that the water gets shared, you'll find a similar effect I think. But there all the lost energy ends up as heat of course (well maybe some sound as well.)

Serial parallel, again I can see your point. Draw some diagrams, that always helps.
With the series combination, only one end of the caps (or batteries) is joined and there is always some other component between the other ends in any sensible circuit. Here, if the charging circuit were reinstated, how would you say the capacitors were connected to that?
At the end of the day (IMO) series and parallel are just (Edit: cancel "ways of" subst. " aids in " ) describing how components are connected. Here, I would say parallel, but if you said series, you'd get the same result.
 
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  • #4
Merlin3189 said:
You don't tell us the value of capacitor B, but it appears to be 2uF if your 13.3V is correct, as I think it is.
Then, as you say, it all works out ok.

As for the energy discrepancy, yes I used to find this strange once. Maybe if you thought of the wires as having resistance, you would see some energy loss to heat as the current is transferred. But that resistance plays no part in your calculations! You get the same answers whatever (finite) value of resistance the wire has. That seems strange, doesn't it? The answer seems to be electromagnetic radiation. So you could think of the energy being lost in either Ohmic resistance in the wire (if that resistance is large) or in radiation resistance (when the Ohmic resistance is small.) Either way it's lost from the caps.
If you consider the gravitational PE of a tank of water, which is the connected to another tank so that the water gets shared, you'll find a similar effect I think. But there all the lost energy ends up as heat of course (well maybe some sound as well.)

Serial parallel, again I can see your point. Draw some diagrams, that always helps.
With the series combination, only one end of the caps (or batteries) is joined and there is always some other component between the other ends in any sensible circuit. Here, if the charging circuit were reinstated, how would you say the capacitors were connected to that?
At the end of the day (IMO) series and parallel are just (Edit: cancel "ways of" subst. " aids in " ) describing how components are connected. Here, I would say parallel, but if you said series, you'd get the same result.

Whoops sorry guys, yes the capacitance of B is 2.0uF. Thank you both for your help, it all makes sense now. A parallel arrangement just means that the negative side of one capacitor is connected to the negative side of the other. I get your points you both made about energy going into heat as the circuit has resistance. However, I is really strange that the equation automatically takes it into account, why is this? For example, if you consider a circuit with a 6V battery and two 100 ohm resistors, calculations would tell you they each get 3V. However, in reality they won't as some of the voltage is lost as heat but we didn't factor this into the equation (i.e. the resistance of the connecting wires we assumed to be zero). It seems weird that the capacitor equations somehow have the energy loss already built into them even though we didn't consider the heat loss through circuit resistance.
 
  • #5
Jimmy87 said:
It seems weird that the capacitor equations somehow have the energy loss already built into them even though we didn't consider the heat loss through circuit resistance.
It's on the same level of weird as collision equations "knowing" how much KE is lost despite not specifying the loss mechanism.

Quantities like potential energy that are intimately related to the physical geometry of their storage tend to have this quality. There's probably a fundamental theorem that explains this but I can't recall it at the moment.
 
  • #6
As I said, the resistance of the wires doesn't matter. The energy is lost even if the wires were perfect conductors. (The lower the resistance the greater the rate of change of current and the greater the EM radiation.
In your 2x 100 R eg, you are consciously neglecting wire resistance and capacitance (of both wire and resistors). If you take the extra resistance into account, you get a different result. But the capacitance makes no difference. In the capacitor question, you again neglect resistance and capacitance of the wires. Here the resistance makes no difference, but the capacitance of the wires would. Energy is a different kettle of fish in this situation, as no particular amount of energy has been defined.Thinking further about your series parallel, if you had an LC tank circuit in isolation, traditionally viewed as a parallel LC circuit, it has the same properties (ideally) as a series LC circuit with the ends connected together - a high current in each component, a high voltage across each component and zero voltage / impedance between the joined ends and between the other ends.
Perhaps we have to use context to give meaning to the S/P terms? If you were told to connect two components in parallel, you'd make two connections, but if you were told to connect them in series, only one. For polarised components, you'd also connect different ends in each case. And you'd probably quibble if told to connect batteries, LEDs, Zeners, bulbs etc. of different voltage ratings in parallel. But you might be happy to connect some of them in "inverse parallel".
Maybe it's like I first thought, they're just hints or clues, rather clear definitions?
 

Related to Energy and Voltage of Capacitor Circuits

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.

How does a capacitor store energy?

When a voltage is applied to a capacitor, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, storing energy in the form of potential energy.

What is the relationship between capacitance and voltage?

The capacitance of a capacitor is directly proportional to the voltage across it. This means that increasing the voltage will result in an increase in the amount of charge that can be stored in the capacitor.

How is the energy stored in a capacitor calculated?

The energy stored in a capacitor can be calculated using the formula E = 1/2 * CV^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.

What is the role of a capacitor in electrical circuits?

Capacitors are commonly used in electrical circuits to store energy, filter out unwanted frequencies, and regulate voltage levels. They can also be used to provide a temporary power source in the event of a power outage.

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