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Consider a capacitor,

$$ C= \frac{Q}{V}$$

To derive the energy of this device,

$$ U = \int V \cdot dq$$

$$ U = \int \frac{Q}{C} \cdot dq$$ (from original capacitor equation)

$$ U = \frac{Q^2}{2C}$$

Plugging back in, $$ C = \frac{Q}{V}$$

$$ U = \frac{ QV}{2}$$

Now if a 'q' charge is pushed out the battery, then it must do work $$qV ,$$ so I was thinking this same energy would be given to the capacitor but only half is. Where did rest of the half go?

And, my next question is why is the potential not constant across capacitor? That is why can't we factor out the 'V' from the integral? Because according to the Kirchoff voltage law, the sum of voltage drops across a closed loop should be 0. Below I have attached a picture of a circuit with capacitor and battery, for it shouldn't the energy be

$$ V_{capacitor} + V_{battery}=0$$

$$ V_{capacitor} = - V_{battery}$$

usually we say that the potential of battery is constant, if the potential of battery is constant wouldn't this imply potential of capacitor is constant?

enter image description here

Ok, so suppose the actual voltage is changing in the capacitor as increase the charge on it, then what happens we reach the maximum charge on it and still have it plugged into the battery? Would the capacitor become defective?

Edit: A lot of people gave an answer that there is an aspect of inductance in this idealized circuit and also the release of electromagnetic radiation. I am now looking for a mathematical description of the inductance aspect of this radiation and also a proof of energy lost to radiation being .5 CV by application of Maxwell law's. I mean if it is indeed true then we must be able to bring out by Maxwell law's, right?

References:

Same problem is said here Why do I get the wrong answer when determining the charge in a capacitor using definition of voltage? but no explanation as to where exactly the half is coming from.

Edit:

What exactly is wrong in this derivation for k.v.l ( The Feynman Lectures on Physics, Volume II, Chapter 22: AC Circuits)

$$ \nabla \times E = \frac{ -\partial B}{\partial t} = 0$$ (maxwell faraday eqn)

Integrating over any loop in the region,

$$ \sum V_i = \int E \cdot dl = \int_{\partial S} \nabla \times E \cdot dS=0$$

i.e:

$$ \sum V_i = 0 $$

Which step is causing problems to have the seeming violation of kvl for the problem stated?

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    \$\begingroup\$ Try raising you capacitor voltage linearly from 0 volts to V volts and then find that what you take from the source will equal what the capacitor stores in terms of energy. \$\endgroup\$ – Andy aka Jul 21 '20 at 12:47
  • \$\begingroup\$ If I had capacitors at home I would try this but the only thing I have are textbooks \$\endgroup\$ – Buraian Jul 21 '20 at 12:48
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    \$\begingroup\$ You can do the math; a linearly rising voltage produces a constant current into the capacitor hence V.I is a linear ramp and the area under the curve is easily estimated as \$V^2\cdot I/2\$ and all is well. Applying a step voltage to a capacitor (as you have done in your question) requires infinite current and things don't work right (even the math). \$\endgroup\$ – Andy aka Jul 21 '20 at 12:51
  • \$\begingroup\$ Try doing the math, instead. \$\endgroup\$ – Hearth Jul 21 '20 at 12:51
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    \$\begingroup\$ No, it's not an impossible situation; read my answer. The series resistance of the wires and the power supply act as a heat-producing current limiter - that's where 50% of the energy disappears. \$\endgroup\$ – Andy aka Jul 21 '20 at 13:04
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The simple answer is that you cannot short circuit two ideal voltage sources without creating an over determined system of equations. So if you try to use network theory to describe the circuit you have drawn, it will always fail. As others have pointed out, to make this circuit calculatable, you need to add some parasitic component (parasitic resistance would be easiest to model and understand).

When we try to work with Kirchhoffs laws, there are two equations for the voltage over your capacitor (like you drew it).

$$ v_C(t) = V_{bat} \quad; (1) \\ v_C(t) = V_C(0) + \frac{1}{C} \int i(t) \mathrm{d}t \quad; (2) $$

(1) Follows from KVL because the battery is also a voltage source

(2) Because of the definition of voltage on a capacitance

Obviously, they can't both be true if there is any current flowing in the circuit. This is why any mathematical approach to describing your system without extra parasitics will fail.

There is an easy, non-mathematical explanation to why no source without parasitic resistance can exist in reality: Having a voltage source without internal resistance means that this source will deliver any current at its fixed voltage. This means that the source would be able to provide any or infinite power, which can't be true in a physical system.

Because of discussion in the comments:

Kirchhoffs laws are not some kind of mathematical super tool that apply to anything and everything you can think of. They laws can be understood as a special case of Maxwells equations for low frequencies. We believe that Maxwells equations are the best description of electromagnetic phenomena that we can work with (maybe some kind of quantum theory will supersede it one day). And even with Maxwells equations you couldn't find a consistent description of your ideal lumped circuit. You cannot use equations that describe reality and try to apply them to something made up.

What exactly is wrong in this derivation for k.v.l ( The Feynman Lectures on Physics, Volume II, Chapter 22: AC Circuits)

∇×E=−∂B∂t=0

In an earlier version, I tried to show some contradictions in Maxwells equations when applied to your circuit. This explanation was flawed because when transitioning to Maxwells equations to explain what's going on, we have to assume that there exists an inductance in the circuit. So already at this point, I was deviating from the actual lumped element model of your circuit, as Sredni Vashtar has pointed out.

This is because of Amperes law, one of the Maxwell equations: $$ \nabla \times \vec{H} = \vec{J} $$

This states that any current density is always linked to a magnetic field. Since inductance is a measure of how much flux a circuit produces per current, it can not be zero in reality. This also leads back to the explanation, that your lumped element model cannot be a description of a real thing.

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Jul 23 '20 at 20:46
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If you build this circuit in reality, there is some resistance. Half the energy provided by the battery is stored in the capacitor and half of it is turned into heat in the resistor.

You would think that you can reduce the resistance to decrease the wasted energy - but you cannot! If you have a total of 1\$\Omega\$ of resistance, and you change it to 0.2\$\Omega\$ - now the resistor wastes 5 times less energy at the same current, but the capacitor charges 5 times more quickly with 5 times the current. The current went up by 5 times, causing 25x the loss (\$P = I^2R\$), but the resistance went down, causing \$\frac15\$ the loss, and the time also went down, causing another factor of \$\frac15\$. The total lost energy is the same as before!

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    \$\begingroup\$ But there is no resistance in my circuit!! My derivation would give different results if there was a resistor in it \$\endgroup\$ – Buraian Jul 21 '20 at 13:14
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    \$\begingroup\$ @DDD4C4U One of your equations assumes that V = Vbattery and another one assumes that V goes up starting from 0. Those can't both happen at the same time! The only way they can happen is if there is something in between the capacitor and the battery. \$\endgroup\$ – user253751 Jul 21 '20 at 13:16
  • \$\begingroup\$ yes exactly that is the problem . According to kirchoff law it must be equal but it is obvious it shuld be different because depending on if you take it or not that value of integral is different \$\endgroup\$ – Buraian Jul 21 '20 at 13:19
  • \$\begingroup\$ but why? why is it like that? \$\endgroup\$ – Buraian Jul 21 '20 at 13:20
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    \$\begingroup\$ @DDD4C4U because you're trying to analyze a circuit that doesn't actually exist. You have found some type of mathematical singularity, but it's okay because this singularity doesn't exist in the real world. Instead of trying to compute the energy dissipation with 0 resistance, you could compute the limit as the resistance approaches 0. \$\endgroup\$ – user253751 Jul 21 '20 at 13:56
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This is called the two capacitor paradox. It is the reason charge pumps can not be 100% efficient. Although you have a battery instead of a capacitor, the problem is essentially the same. If you like, a battery can be considered approximately equivalent to a very large capacitor.

There are several ways to resolve the paradox. Here's one:

Kirchoff's voltage law (KVL) states:

The directed sum of the potential differences (voltages) around any closed loop is zero.

So let's try that:

schematic

simulate this circuit – Schematic created using CircuitLab

$$ 3\:\mathrm V + 0\:\mathrm V + 0\:\mathrm V + 0\:\mathrm V \ne 0\:\mathrm V $$

So by posing the question at all, KVL has been violated. So one resolution to the paradox: the initial conditions are invalid, and there's no point in continuing. The circuit as drawn is no more valid than "2+2=5" or other mathematical nonsense.

Perhaps the thing to realize is if you want to use network theory, the lines in a schematic are not wires. They are mathematical constraints which require that everything touching a line is at the same electric potential. As the circuit is drawn, the lines say the voltage across the battery and the capacitor must be equal. And then, you posit these voltages are not equal. Any further mathematical reasoning from this inconsistent set of constraints is bound to turn up contradictions.

One way to avoid violating KVL is to insert a resistor into the circuit:

schematic

simulate this circuit

Now the capacitor can start at 0V, because V1 can appear across R to satisfy KVL.

However, you must now additionally calculate energy lost in the resistor as the capacitor charges. You will find the "missing" energy has been lost as heat in the resistor.

The value of the resistor does not matter. A larger resistor will dissipate lower power for a longer time. A smaller resistor dissipates a higher power for a shorter time. Either way the same energy is lost.

Note that as the resistance approaches zero, the power in the resistor approaches infinity due to Joule heating: \$P = I^2 R\$. This is quite unlike a lot of circuits where as resistance approaches zero, energy lost to the resistance approaches zero. This is because in most circuits as resistance approaches zero, the voltage across the resistor approaches zero, but in this circuit it can not due to the voltage source and initial conditions of the capacitor.

This model with the resistor is a more accurate model of what happens in practice, as any battery, any capacitor, and any wire you might use to construct the circuit has some non-zero resistance.

If you could somehow build the circuit with zero resistance (superconductors, magic, or something) then you would still have to consider the inductance of the circuit. This inductance means the capacitor won't simply charge up, but instead the energy will oscillate between the capacitor and the inductor.

Probably not forever though, because eventually energy will be lost as electromagnetic radiation. To for the model to be accurate, the radiation resistance must be included.

In short, the paradox can only arise from an initial violation of the KVL. Incorporating resistance and inductance which must be present in any physically realizable circuit resolves the paradox.

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  • \$\begingroup\$ I worded my comment badly, but what I was asking is what part in the derivation of kirchoff voltage law failed so as to fail describing this case \$\endgroup\$ – Buraian Jul 22 '20 at 9:47
  • \$\begingroup\$ @user253751 why do you think I don't know what kvl is \$\endgroup\$ – Buraian Jul 22 '20 at 9:48
  • \$\begingroup\$ @DDD4C4U KVL assumes that rate of change of magnetic flux outside the lumped circuit element is zero. This condition is violated since your circuit does not model the wire inductance. That's why KVL fails. \$\endgroup\$ – sarthak Jul 22 '20 at 15:11
  • \$\begingroup\$ Great answer.... But if energy is lost in the LC oscillations as EM waves...why no EM waves are created in a normal LC oscillator circuit? \$\endgroup\$ – sarthak Jul 22 '20 at 15:14
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    \$\begingroup\$ @DDD4C4U You have misunderstood my comment...KVL says that \$\Delta V = 0\$ across a closed loop. This is only true if the electric field is conservative, which in turn is only true if there is no changing magnetic field inside the loop. This is not true for your circuit. \$\endgroup\$ – sarthak Jul 23 '20 at 15:40
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Comments I made before answering

You can do the math: a linearly rising voltage produces a constant current into the capacitor hence V⋅ I is also a linear ramp and the area under the curve is easily calculated as V²⋅ I ÷ 2 and all is well! Applying a step voltage to a capacitor (as you have done in your question) requires infinite current and things don't work right (even the math).

and

The series resistance of the wires and the power supply act as a heat-producing current limiter - that's where 50% of the energy disappears.

Hence, my original answer

Directly charging a capacitor from a voltage supply is inefficient: -

  • The energy consumed is C·V² but, the energy stored is only ½ C·V².
  • Consider a 1 μF capacitor charged to 1 volt and then connected to a discharged 1 μF capacitor. Charge (C·V) is conserved hence, the final voltage is 0.5 volts.

For energy there is loss: -

  • Initial energy: ½ × 1 μF × (1 volt)² = 500 nJ
  • Final energy: ½ × 2 μF × (0.5 volts)² = 250 nJ

Consider this: would you try and charge an inductor up from a constant current source (hint: what would be the terminal voltage induced?).

The same happens when objects collide; momentum is conserved but energy is lost.

Inductors are different; all the energy taken from a voltage supply is stored in the magnetic field. Unlike capacitors, inductors don’t cause a current surge. There’s no collision; current ramps up from zero amps in an orderly manner. Energy is preserved (except in the less-usual case of charging an inductor from a current source).

Mechanical analogy: if an ideal motor winds up a spring then, all the energy consumed is liberated when the motor is reused as a generator. However, if a spinning motor were “applied” to a flywheel then, the energy acquired by the flywheel is only 50% of the energy taken by the motor. There is a collision.

However, if a flywheel was progressively spun-up from rest then, energy transfer is 100%. Likewise, if a capacitor was charged by a ramping voltage, 100% transfer occurs.

Inductors are more useful with voltage supplies; energy stored can be released into a capacitor with perfect efficiency. A charged inductor connected to a discharged capacitor will ramp the voltage from zero and transfer 100% of the energy to the capacitor (as per a boost converter).

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Voltage Spike Jul 23 '20 at 15:38
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Just to add to the answer given by user25375 (half the energy is lost in a resistor in the circuit, no matter how small it is):

If you want to insist that there is no resistor in the circuit, then you have to ask what removed the kinetic energy that the battery gave to the electrons. There is excess kinetic energy during the charging process because the battery is at a higher voltage than the capacitor. Thus the electrons are accelerated, and have kinetic energy left over that did not get deposited on the capacitor.

This excess kinetic energy would normally be removed by whatever resistance is in the circuit. If you take away the possibility of a resistor, then you have taken away the mechanism that removes this kinetic energy from the electrons, yet you have assumed that the energy was lost anyway. That is where your 'missing' energy is.

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  • \$\begingroup\$ Electrons inside the superconductor cancel any electric field. \$\endgroup\$ – Helena Wells Jul 23 '20 at 9:56
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If we are modeling this as the movement of charges, the work done on an individual charge is qV, but that doesn't mean that the overall work done on all charges is QV.

The issue is that, as charges move from the negative terminal to a capacitor plate, the voltage between the terminal and the plate is reduced, meaning V is actually a function of time.

Since the voltage strictly decreases between those two points, the sum-total work done across all charges must be strictly < QV.


It appears we solved your paradox without introducing a phantom resistance. However, the above argument assumes that the charges aren't all released from the battery simultaneously; there must be some finite current. In the circuit model, this is represented by the resistance of the circuit! So we did introduce a "phantom resistance" after-all.

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You should be careful with the assumptions behind the equations you are using. Are we in a static, quasi-static or dynamic setting?

Because the charging process you propose would put your circuit into an electrodynamic setting: in this case if you insist in having perfect conductors you cannot neglect the associated self-inductance and corresponding L di/dt contribution that will create a radiating circuit (electrodynamics) where part of the energy will go in the EM field in the space around you.

Even if you manage to stay in quasistatic conditions, you will end up with an oscillating circuit where the energy will go back and forth between L and C.

If you want to slow down the charge so as to make the L di/dt contribution negligible in the quasistatic settings, you need to add a resistance, but in this case energy will be lost in heat in the resistor.

And if you are wondering what happens at equilibrium, when the conditions are static, well... there is no longer movement of charge. The cap will sit there storing whatever energy it has in its field at the end of the transient.

NOTE:
For a good introduction to electro-quasistatics and magneto-quasistatic conditions, have a look at Chapter 3 of Haus & Melcher's textbook on EM, "Electromagnetic Fields and Energy", freely available on the MIT OCW website.
Broadly speaking,
"Static": we neglect time derivatives in Maxwell's equations.
"Quasi-static": we neglect either the magnetic induction or the electric displacement current.
"(electro)Dynamic": the whole she-bang. We consider the effects of both dE/dt and dB/dt.

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  • \$\begingroup\$ are you saying capacitor would have inductance for the electrodynamic setting?? and can you tell what these all states really mean like static , quasi static and leectrodynamic \$\endgroup\$ – Buraian Jul 22 '20 at 0:09
  • \$\begingroup\$ the self-inductance I am referring to is that of the closed circuit formed by ideal voltage source-perfect conductors-ideal capacitor. When you put an ideal voltage source across a discharged capacitor, you are asking for a step in voltage that corresponds to an infinitely high di/dt. Even if we were able to supply such an infinite amount of power, this will require so fast a charge transfer that you would have to consider your system in an electrodynamics context, with full-fledged Maxwell equations (with both dB/dt and dE/dt contribution present). \$\endgroup\$ – Sredni Vashtar Jul 22 '20 at 0:16
  • \$\begingroup\$ What is this step in voltage? is it the abrupt voltage rise? I have heard of heaviside step functions , is this related to that? Why does fast charge require us to use maxwell description of the circuit? \$\endgroup\$ – Buraian Jul 22 '20 at 0:18
  • \$\begingroup\$ @DDD4C4U yes. When you connect the ideal voltage source for example by closing a switch, the capacitor voltage should change from 0V to whatever the source imposes. This will require a step in Vc(t), corresponding to an infinite dv/dt. If you look at the equation you will see that the capacitor voltage is the integral of the current and in order to have an abrupt step in v you would need an infinite amount of current. The current in your circuit would have to change from 0 (open circuit) to infinity (at t=0+) hence di/dt would be so big that any small L would become relevant. \$\endgroup\$ – Sredni Vashtar Jul 22 '20 at 0:40
  • \$\begingroup\$ The time rate of change of the fields could be so fast that you would have to consider the contribution of both dE/dt and dB/dt in Maxwell's equations - in this general electrodynamical case your circuit would not only oscillate but also radiate. Energy would be lost in what can be thought as a radiation resistance. \$\endgroup\$ – Sredni Vashtar Jul 22 '20 at 0:43
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Without any resistance at all, the proposed circuit will oscillate. No energy is lost as heat; therefore, as device A discharges, and charges device B, the rush of current has to be so great that it overshoots. Then B ends up with more voltage than A, and the situation reverses: B discharges and charges A, and so forth.

At the point where the devices have equal voltage, where the missing energy has gone is into the moving current, which has a drift velocity and mass.

This is no more difficult to understand than a ball oscillating on a spring or what have you. We don't even have to bring induction into the picture (though for correctness/realism, we must).

If induction didn't exist, then the analysis would have to boil down to considering the masses and inertia of the individual electrons. The voltage difference is associated with an electric field in which electrons accelerate, and carry kinetic energy like any other particles with mass. Without resistance, their energy doesn't dissipate. Once accelerated, they keep moving and the movement doesn't simply stop when the devices have equal voltage.

The phenomenon of induction ensures that electricity has a kind of apparent "kinetic energy" that is far in excess of the ordinary inertial kinetic energy of the drift current due to the electron mass alone. When we suddenly interrupt a circuit, the induction wants the current to keep going at the same value, and the effect is surprisingly powerful, out of proportion to the mass of the electrons and their drift velocity through the wires. This effect actually caused by the collapse of the induced magnetic field (which, by the way, stores energy).

Just like mass resists change in velocity (requiring acceleration), inductance resists change in current (giving rise to a slow change in current in response to a fast change in voltage, similar to acceleration). Effectively, inductance makes the slowly drifting electrons look as if they are a lot heavier than they are.

If we could build the circuit, such that there is no resistance (everything is superconductive), the inductance would, in the above sense, "amplify" the inertia of the current. It would slow down the discharge, as well as prolong the overshoot. The result is that we get oscillation that is a lot slower than what would be predicted by only the mass inertia of the super-conducted current.

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  • \$\begingroup\$ 'Mass resists change in velocity'.That is not true. Inertial mass resists change in velocity. Gravitational mass doesn't resist change in velocity. \$\endgroup\$ – Helena Wells Jul 22 '20 at 7:02
  • \$\begingroup\$ "the rush of current has to be so great that it overshoots" I see no evidence that this must be the case. \$\endgroup\$ – Elliot Alderson Jul 22 '20 at 12:22
  • \$\begingroup\$ @HelenaWells There is no point in mentioning the word "gravitational', if we have no intention of discussing gravitational effects; it only creates confusion. \$\endgroup\$ – Kaz Jul 23 '20 at 7:32
  • \$\begingroup\$ @ElliotAlderson It must be the case in the abstract model that the question is about, not in reality; we're not going to construct an apparatus that supplies empirical evidence. The question is, where does the energy go (in that model) between those two states of the system that contain the same charge distributed differently. Well, it cannot be the case that we moved to a static lower energy state, without losing energy. If that energy is still in the system, it must be that state of equal voltages isn't actually static. \$\endgroup\$ – Kaz Jul 23 '20 at 7:41
  • \$\begingroup\$ Kaz there are 2 types of masses. Inertial mass and gravitational mass. At low energies they are equivalent. At higher energies they are not equivalent e.g if the electron moves at 1/10 c then there is a difference so you have to specify. \$\endgroup\$ – Helena Wells Jul 23 '20 at 7:43

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