learning basic workings of capacitor

[PG1995]

Re: energy of charged capacitor

Hi

Q = It, where Q is charge, I is current, and t is time. In case of capacitor I is maximum, I0, at t0 (when the switch is just closed). The current will decrease gradually in exponential fashion. We can find the total charge, Q=It, which flows into the capacitor until the potential difference across it equalizes the potential difference of the battery.

Energy = VI x t. Therefore, we can write, Energy = VQ.

The battery supplied all the charge which went into the capacitor at constant potential, V.

But the formula for power or work done for capacitor is: 1/2 VQ. It's half the work done by battery. Where did the rest of the energy, which is half of the supplied energy, go?

This is what I think. As a capacitor gets charged, the battery finds it hard to push the charge into the capacitor because the capacitor starts resisting the inflow of the charge. Therefore, half of the energy gets lost as heat. This is analogous to the case when air is blown into a balloon. As the balloon gets inflated, it starts resisting the inflow of the air.

Do I make sense? Please let me know. Thanks.

Regards
PG

 

[Grossel]

But the formula for power or work done for capacitor is: 1/2 VQ. It's half the work done by battery. Where did the rest of the energy, which is half of the supplied energy, go?

The rest of the energy was lost in the battery's internal resistance. At least in theory. Then you'll have to add wire resistance, cap's internal resistance etc..

Just think of a resistor and a cap in serie. When charging that cap, the current will be the same both for the resistor and the cap. Therefore, the resistor will steal half of the energy from the voltage source.

 

[misterT]

The rest of the energy was lost in the battery's internal resistance. At least in theory. Then you'll have to add wire resistance, cap's internal resistance etc..

Just think of a resistor and a cap in serie. When charging that cap, the current will be the same both for the resistor and the cap. Therefore, the resistor will steal half of the energy from the voltage source.

That is a bit strange explanation, because the equations still hold for ideal capacitors.

PG:

Energy = VI x t. Therefore, we can write, Energy = VQ.

With constant current the voltage (at the capacitor) is a ramp from zero to the final value, so Energy = 1/2 VQ. If the supply voltage is constant, you must add this same energy to the battery. Otherwise the voltage at the battery will drop and you need to re-calculate.

When you calculated total Energy = VQ, you calculated a situation where the capacitor was charged from the battery with energy = 1/2 VQ and then the battery was charged back with the same energy = 1/2 VQ. So the total energy that was "moved around" is then VQ. (But only 1/2 VQ was added to the system)

 

[Cabwood]

If the resistance around the capacitor/battery circuit is negligible, energy lost as heat is also negligible.

The charging current will start at a maximum, and diminish to zero. The capacitor's voltage will start at zero and tend towards the battery's EMF. If you plot power (IV) against time, then total energy delivered to the capacitor is the area under the graph.

The graph will show that power starts at zero and ends at zero, peaking in the middle. A flat (constant) power graph might yield CV² as the total energy, but this is not flat. The total energy delivered to the capacitor is clearly not CV², but somewhat less - a half of that. And "the other half" never left the battery.

 

[Ratchit]

To the Ineffable All,

OK, let's do a constant voltage (E) energy analysis across a capacitor (C) and resistor (R) in series. R represents the sum total of all the resistance in the circuit.

The instantaneous current is the well known i = (E/R)*e^(-t/RC)

First lets calculate the energy stored in C . The voltage across C is the constant voltage minus the voltage across R. So that is E-Ri. The incremental energy stored in the capacitor is dW = (E-Ri)*i*dt . Integrating dW across a time interval T, we get W = ½E²C-½E²C*e^(-T/RC)+½E²C*e^(-2T/RC) as the energy stored in the capacitor. Letting T run to infinity gives ½E²C .

Now let's calculate the energy dissipated in R . dW = i²*R*dt , so integrating dW across T gives us W = ½E²C-½E²C*e^(-2T/RC) as the energy dissipated by the resistor, or ½E²C when T goes to infinity.

The same results are found when the RC circuit is driven by a constant current source.

In conclusion, half the energy from E will be lost across the resistor, no matter what its value is.

Ratch

 

[Cabwood]

I stand corrected, Ratchit.

The other half did heat something up.

 

[misterT]

To the Ineffable All,

OK, let's do a constant voltage (E) energy analysis across a capacitor (C) and resistor (R) in series. R represents the sum total of all the resistance in the circuit.

The instantaneous current is the well known i = (E/R)*e^(-t/RC)

First lets calculate the energy stored in C . The voltage across C is the constant voltage minus the voltage across R. So that is E-Ri. The incremental energy stored in the capacitor is dW = (E-Ri)*i*dt . Integrating dW across a time interval T, we get W = ½E²C-½E²C*e^(-T/RC)+½E²C*e^(-2T/RC) as the energy stored in the capacitor. Letting T run to infinity gives ½E²C .

Now let's calculate the energy dissipated in R . dW = i²*R*dt , so integrating dW across T gives us W = ½E²C-½E²C*e^(-2T/RC) as the energy dissipated by the resistor, or ½E²C when T goes to infinity.

The same results are found when the RC circuit is driven by a constant current source.

In conclusion, half the energy from E will be lost across the resistor, no matter what its value is.

Ratch

Empty capacitor and a constant voltage source gives you a potential energy of ½E²C. ½E²C is the amount of potential energy you have at your disposal. After you used it all, you've lost it all. You don't "waste the half". You just use it all.

 

[Ratchit]

misterT,

Empty capacitor and a constant voltage source gives you a potential energy of ½E²C. ½E²C is the amount of potential energy you have at your disposal. After you used it all, you've lost it all. You don't "waste the half". You just use it all.

The mathematics prove that half the energy taken from a constant voltage source will be "wasted" in energizing a capacitor to whatever voltage. This energy will be dissipated as heat in the inevitable resistance of the circuit no matter what its value is. You are on the wrong side of this discussion unless you can prove mathematically that it is wrong.

Ratch

 

[misterT]

This energy will be dissipated as heat in the inivitable resistance of the circuit no matter what its value is.

Ratch

The math is correct whatever the resistance is. And zero resistance does not dissipate any energy.

 

[Ratchit]

misterT,

The math is correct whatever the resistance is. And zero resistance does not dissipate any energy.

Now you are discussing infinities (current) and infinitesimals (resistance). Energizing a capacitor without any resistance will take a infinite amount of current. That is not what practical problems are about.

Ratch

 

[misterT]

misterT,



Now you are discussing infinities (current) and infinitesimals (resistance). Energizing a capacitor without any resistance will take a infinite amount of current. That is not what practical problems are about.

Ratch

Still.. that is what the math says. I know real circuits always have resistance. The problem here is in the constant voltage supply. Every action has a reaction.. where is the reaction in this example?

Imagine if I am floating in space with a steel ball (1 kg). I push the ball away from me at 1 m/s. The ball now has kinetic energy ½mv². But when I pushed the ball forward, I also pushed myself backwards and I also have kinetic energy ½mv².

Now.. imagine a situation that I pushed the ball forward, but I managed to remain still. Where did one of the ½mv² go? That is exactly the same situation when you use an ideal constant voltage source in your calculation. The constant voltage supply brings energy to the system and therefore you do not have a closed system.

 

[Ratchit]

misterT,

Still.. that is what the math says. I know real circuits always have resistance. The problem here is in the constant voltage supply. Every action has a reaction.. where is the reaction in this example?

What does the math say? What is the problem with a constant voltage supply? Every application of mechanical force has a reaction. There is no mechanical reaction for the example, only an effect.

Imagine if I am floating in space with a steel ball (1 kg). I push the ball away from me at 1 m/s. The ball now has kinetic energy ½mv². But when I pushed the ball forward, I also pushed myself backwards and I also have kinetic energy ½mv².

How do you figure that? You won't have the same velocity and mass as the ball.

Now.. imagine a situation that I pushed the ball forward, but I managed to remain still. Where did one of the ½mv² go? That is exactly the same situation when you use an ideal constant voltage source in your calculation. The constant voltage supply brings energy to the system and therefore you do not have a closed system.

I don't understand the question about where the ½mv² went. Why are you dragging in a mechanical analogy into this discussion? Who cares if the system is open or closed? The voltage source supplies the energy for both the resistance loss and the electric field of the capacitor.

Ratch

 

[misterT]

What does the math say? What is the problem with a constant voltage supply?

The constant voltage supply brings energy to the system.

How do you figure that? You won't have the same velocity and mass as the ball.

Different mass, different velocity, but the same kinetic energy. See my signature for a proof.

Who cares if the system is open or closed?

Every scientist I know. It is important to know if the system is open or closed. See my signature for a proof.


Ok. Can you show me a calculation for this: I have a charged capacitor C1 (100uF, 10V in it). I know that it has energy in it (½E²C) and I want to use that energy to heat up something. I decide to connect a resistor R1 and a capacitor C2 (100uF, 0V in it) in series with the charged capacitor. How much of the energy will get dissipated to heat? My intuition says "half".

"Why only half"? If I connect the resistor directly to C1, I will get all of the energy to heat up my stuff. So why does the capacitor C2 "waste" half of the energy. Well it does not waste it.. it only stores it.

Can you calculate the energy I will get if, after the first step, I connect the resistor R1 across both of the capacitors C1, C2? How much energy will I get out of them?

After the two calculations, could you add up how much total heat I got? My intuition says 3/4 E²C. How did I get more heat out of the system than was initially in the capacitor C1?

 

[Cabwood]

Since C1 and C2 have the same capacitance and current, whatever current flows when charging one from the other is lowering the voltage across one at the same rate that it is increasing the voltage in the other (dV/dt proportional to I). Current will stop flowing when the voltages are the same. If C1 started started with 10V, and C2 with 0V, C2 now has 5V, and C2 also has 5V. Since the voltage of C1 has halved, its stored energy is a quarter of its starting energy. C2 now has the same energy, a quarter. The remaining half still has to be lost in resistance around the loop.

C2 does not get half of the starting energy - it only gets a quarter, and only if it has the same capacitance as C1.

OK, Ratchit, I'm bracing for impact - let me have it.

 

[Ratchit]

misterT,

The constant voltage supply brings energy to the system.

As would a constant current supply or anything in-between. What is wrong with that?

Different mass, different velocity, but the same kinetic energy. See my signature for a proof.

That is not a proof, that is a cop-out. Momentum is a vector, and is conserved. Since you and the steel ball are traveling in opposite directions, the momentum of you and the steel ball are equal and opposite. Energy is a scalar quantity. There is no requirement that the kinetic energy distribution has to be equal, only that is has to add up to what the energy was beforehand.

Every scientist I know. It is important to know if the system is open or closed. See my signature for a proof.

That is not a proof, that is a cop-out. Closed and open systems are just a method of accounting for all the energy involved in an action. I believe I did that in my analysis of a capacitor energizing.

Ok. Can you show me a calculation for this: I have a charged capacitor C1 (100uF, 10V in it). I know that it has energy in it (½E²C) and I want to use that energy to heat up something. I decide to connect a resistor R1 and a capacitor C2 (100uF, 0V in it) in series with the charged capacitor. How much of the energy will get dissipated to heat? My intuition says "half".

Sure I can calculate it. Anyone can. Doing a simple loop with two 100 µf capacitors, one of which is energized to 10 volts, and 1 resistor of undefined value, we get a differential equation. Solving this for current, we get I = (10/R)*e^(-2000*t/R) . Notice that current becomes zero as time -> ∞ . To find the energy dissipated by the resistor, we integrate power over a time interval T. So power is I²R, and ∫I²R*dt from 0 to T is 0.0025-0.0025*e^(-4000*T/R), or 0.0025 joules when T -> ∞. The coulombs transferred from the energized cap to the unenergized cap is the integral of the current with respect to time over a interval of T. That gives ∫I*dt = 0.0005-0.0005*e^(-2000*t/R) = 0.0005 coulombs when T -> ∞ . Using the familiar V = Q/C we observe a voltage change of 0.0005/100E-6 = 5 volts. So both caps are at 5 volts after an infinite time.

So now lets add up and account for all the energy. The energy in the energized cap was at ½*100E-6*10² = .005 joules. After an infinite time, the energy in each cap is ½*100E-6*5² = .00125 joules. So adding everything up we get 0.00125+.00125+.0025 = 0.0050 joules, which is what we started out with. So half the energy is dissipated as heat. And the point is?

"Why only half"? If I connect the resistor directly to C1, I will get all of the energy to heat up my stuff. So why does the capacitor C2 "waste" half of the energy. Well it does not waste it.. it only stores it.

Of course, perfect caps never waste energy. Their storage mechanism is manifested as a voltage across their dielectric.

Can you calculate the energy I will get if, after the first step, I connect the resistor R1 across both of the capacitors C1, C2? How much energy will I get out of them?

Well, let's see. Two caps in parallel give a total capcitance of 200E-6. So you are discharging an effective 200 µf cap energized to 5 volts. That makes it ½*200E-6*5² = 0.0025 joules.

[After the two calculations, could you add up how much total heat I got? My intuition says 3/4 E²C. How did I get more heat out of the system than was initially in the capacitor C1?

Well, you got 0.0025 joules out of the first calculation and 0.0025 out of the second calculation. That equals 0.005 joules total, which is what was in the 100 µf cap in the first place.

Ratch

 

[Ratchit]

Cabwood,

I cannot find anything wrong with your functional description. You seem to have a good grasp of what occurs. I can fault you for you for your descriptive phrase "current flow", however. You should have described it as charge flow. Because current flow means literally "charge flow flow", it is redundant and ridiculous. Remember, charge flows, but current exists.

Ratch

 

[MrAl]

Hi,

Here is a graph of the energies during the process of charging a cap through a resistor powered by a voltage source.
The green trace is the total energy delivered by the source, the blue trace is the energy absorbed by the resistor, and the red trace is the energy stored in the cap.

It's plain to see that after a long time relative to the time constant (which is 1 here) we see the total energy is 1, and half the energy gets dissipated in the resistor and the other half stored in the capacitor. Thus, the power source put out 1 unit of energy while the capacitor stored only half of that.
The waveforms themselves are relative to the time constant in that if you decrease R the same waveforms occur just that it happens faster, and that does not change the energy balance.

This problem is part of what limits the processor speed in modern CPU's. The resistance causes heat and with each gate charge occuring faster and faster more total heat is generated in the same amount of area, so that area gets hotter and hotter. Eventually there's no heat sink that can dissipate all that heat with the required small area so it can not be clocked any faster.
If we could charge a capacitor and not lose any power, we could clock at much faster rates. It is true that this would take more current, but we would then only be left with the simple problem of increasing the power supply capability.

In theory a capacitor can be charged in zero time with an infinite current, but that doesnt exist in nature. In fact, many circuit analysis programs will generate an error message if a DC voltage source is connected directly across a capacitor unless it has some small default resistance value it uses in series with all caps when there is none drawn in the circuit schematic.

 

[Cabwood]

Now that's just not fair. I was only braced for a math-related correction or criticism, and I was not expecting a sucker-punch to my definition-groin parts. You are, of course, technically correct, but that's not the point. Well, it is, but my syntax-nads and pride have been needlessly injured. I hope you are ashamed of yourself, Ratchit.

I shall wait patiently. I will notice when you say "hotter temperature" or "another sun" or when you mix up momentum and inertia. And when that happens I shall be there, grinning and chuckling maniacally. And my riposte shall be unmerciful.

No, I won't do that. I'm having a laugh. You are right. I will however be most aggrieved should ever you fail to correct someone who says "voltage through". Now that I will be watching for.