Effect of Leakage on Capacitor Energy Storage

In summary, the conversation discusses a problem with two capacitors in series and the calculation of the voltage across each capacitor. The key concept is that the two capacitors store equal charge and the voltage across each can be calculated using the equation V = Q/C. However, the initial conditions of the circuit are not given, making the analysis more complex. It is important to carefully consider the topology of the circuit and any possible initial conditions in order to accurately solve the problem.
  • #1
wcjy
73
10
Homework Statement
Consider a RC circuit shown in Figure 6(b) with V1 = 4 V, V2 = 2 V, R1 = 1.4 Ω,
R2 = 1 Ω and R3 = 1.5 Ω. The capacitors have capacitances C1 = 6 uF and C2 = 3 uF
and the circuit has been connected for a long time.
(i) Calculate the energy stored in C2.
Relevant Equations
E = 0.5cv^2
1606192695084.png


$$R_{eff} = R_1 + \frac{1}{\frac{1}{R_2}+\frac{1}{R_3}} = 1.4 + 0.6 = 2$$

PD of R3 = 4 / 2 * 0.6 = 1.2
Net PD of the capacitor = 2- 1.2 = 0.8V
$$ E = \frac {1}{2} CV^2$$
E = 0.8 ^2 * ( 3*10^-6) / 2
E = 9.6 * 10^-7

Correct answer is: 4.27 * 10^-7

was thinking if i should find the Effective capacitance or something.
 
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  • #2
You have correctly calculated the voltage across the two capacitors as 0.8V. But part of this voltage is dropped across C1 and part across C2, so the voltage across C2 is less than 0.8. Do you know how to calculate how much of the voltage is dropped across C1 and how much is dropped across C2?
 
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  • #3
Is it find the effective capacitance,
then use Q = CV to find the charge,
then using the charge to find the pd of C2?
 
  • #4
yep got it with the ^
THANKS!
 
  • #5
The easiest way is to realize that the two capacitors will store equal charge. Knowing that the two charges are equal, and knowing the total voltage, you can calculate the voltage across each capacitor.
 
  • #6
phyzguy said:
The easiest way is to realize that the two capacitors will store equal charge. Knowing that the two charges are equal, and knowing the total voltage, you can calculate the voltage across each capacitor.
so basically by taking the ratio of the capacitance will do?
 
  • #7
Go ahead and write it down. There should be two equations and two unknowns to give you the two voltages.
 
  • #8
Without any specification of the voltage distribution on C1 and C2 (like initial conditions), this is an unsolvable problem.
 
  • #9
I am getting really tired of trying to help students solve problems that aren't properly stated. It's not fair to the students, yet, at least based on what's posted on PF, it seems really common.
 
  • #10
Yes there are some like that. More frequent is information in the question missed out by the student because he has not grasped its relevance and essentiality for any solution. More annoying is careless transcription of problems which we often intuit, but sometimes waste time on only to see the student come back with oh sorry, the question was...

However these are not the case here which is solvable and solved: the charges on the series capacitors are equal so the ratio of p.d.'s across them is given by V = Q/C .
 
  • #11
epenguin said:
However these are not the case here which is solvable and solved: the charges on the series capacitors are equal so the ratio of p.d.'s across them is given by V = Q/C .
OK but the annoying part is that the question wasn't about the series combination of C1 and C2. They could have modeled this as a single capacitor, but they chose the topologically confusing case of two series capacitors without providing any information about initial conditions.

If you let me choose the initial conditions, I can make the energy stored on C2 whatever you want it to be. If you doubt this consider the following thought experiment:
I take the network shown and add a voltage source connected across C2, let's say V2 volts. Then I allow the circuit to reach equilibrium (steady state) then disconnect the additional voltage source, let the circuit again settle to steady state (which takes 0 seconds, BTW), and ask what is the energy stored on C2?

This, in my mind at least, isn't a trivial issue. Experience analog EEs are hypersensitized to seeing capacitors in series, inductors in parallel, loops with only capacitors, nodes with only inductors, voltage steps on capacitors, current discontinuities in inductors... These are red flags that the analysis isn't conventional (i.e. singular, infinities, or depends an initial conditions, etc.).

Sorry, I expect better from EE professors.
 
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  • #12
DaveE said:
OK but the annoying part is that the question wasn't about the series combination of C1 and C2. They could have modeled this as a single capacitor, but they chose the topologically confusing case of two series capacitors without providing any information about initial conditions.

If you let me choose the initial conditions, I can make the energy stored on C2 whatever you want it to be. If you doubt this consider the following thought experiment:
I take the network shown and add a voltage source connected across C2, let's say V2 volts. Then I allow the circuit to reach equilibrium (steady state) then disconnect the additional voltage source, let the circuit again settle to steady state (which takes 0 seconds, BTW), and ask what is the energy stored on C2?

This, in my mind at least, isn't a trivial issue. Experience analog EEs are hypersensitized to seeing capacitors in series, inductors in parallel, loops with only capacitors, nodes with only inductors, voltage steps on capacitors, current discontinuities in inductors... These are red flags that the analysis isn't conventional (i.e. singular, infinities, or depends an initial conditions, etc.).

Sorry, I expect better from EE professors.
The students find the academic exercises hard enough and you want to talk about reality or EE practice?!

OK I see what you mean. In this sort of problem and this student level it is an unstated assumption that the parts of the circuit not attached to any source |-| have not been charged in the way you mention and are overall neutral (charge on one plate equal to that on the other but opposite sign).

This kind of question, very often with battery or DC source, figures quite prominently in elementary physics courses. Seems to be traditional and unquestioned; you might indeed from a professional EE point of view raise some questions about the value of this. You expect better from EE professors, but most people setting the problems we meet here are probably not EE profs.

To tell the truth my own interest is purely academic (though I do have some background in a related (more difficult) aspect of network theory). I have repeatedly come in on such questions- almost the only ones on the Elementary Physics board where I do - because of the unnecessarily heavy weather I often saw students making of such problems. For example right now there is an open problem which you can do in a minute in your head and that is unsolved after 52 posts!
https://www.physicsforums.com/threads/find-the-supply-voltage-of-a-ladder-circuit.996062/

Reasons for all this difficulty may be in the way it is taught, or certainly learnt. Some defects are:

unphysicality: e.g. It should be obvious physically that voltages and conductances in series will add up and similarly for charges and capacitances in parallel. But such things seem to be formulated as black letter pedantic algebraic exercises. As a result I doubt whether some of the learning from doing exercises is permanent;

inflexibility: e.g.inability to recognise the circuit if it is drawn in a different way. Demand that everything is series or parallel, if not they are lost for what to do;

failure to spot simplifying features such as symmetry.Coming back to your objections, I ask if is it not the case that even if you do charge an isolated part |-| and remove the source, the charge quite soon 'leaks' away? That even if attached to a source as in the problem the excess charge above what standard calculation says would leak away? So that to say the capacitor draws no current is an idealisation? Or is this different with more modern capacitors? I don't know - just asking.
 
Last edited:
  • #13
epenguin said:
Coming back to your objections, I ask if is it not the case that even if you do charge an isolated part |-| and remove the source, the charge quite soon 'leaks' away? That even if attached to a source as in the problem the excess charge above what standard calculation says would leak away? So that to say the capacitor draws no current is an idealisation? Or is this different with more modern capacitors? I don't know - just asking.
That is exactly why I stipulate that the circuit must be allowed to reach steady state when the voltage source is connected, by that time the "leakage" will be zero. You are correct though, the addition of a source across C2 will charge other capacitors too.
 

What is a capacitor?

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

How does a capacitor store energy?

When a capacitor is connected to a power source, one plate becomes positively charged and the other becomes negatively charged. This creates an electric field between the plates, which stores the electrical energy.

What is the unit of measurement for energy stored in a capacitor?

The unit of measurement for energy stored in a capacitor is the joule (J). However, it is more commonly measured in microjoules (µJ) or picofarads (pF).

Can a capacitor store an unlimited amount of energy?

No, a capacitor has a maximum energy storage capacity determined by its capacitance value and the voltage applied to it. Once this limit is reached, the capacitor will not be able to store any more energy.

How can the energy stored in a capacitor be released?

The energy stored in a capacitor can be released by connecting it to a circuit. As the capacitor discharges, the stored energy is released in the form of an electric current.

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