Understanding Energy Loss of 2 Joules in Circuits with Changing Capacitance

[Touuka]

Homework Statement

Capacitors A and B are identical. Capacitor A is charged so it stores 4J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now ____.

Relevant Equations

U = q^2/2C

I am given the answer is 2J and I know how to get there with U = q^2/2C. But what I don't understand is that why is the energy not conserved here? How can energy loses just because the capacitance is doubled?

 

[Delta2]

The energy is conserved always (at least in classical physics). You mean that the energy stored in the capacitors is not conserved, some is lost as heat or as electromagnetic radiation. This happens because there will be flow of charge (electron movement that is a current) till the two capacitors are at the same potential, and this flow of charge takes some energy away , initially as kinetic energy of the electrons and eventually as heat and EM radiation.

 

[Steve4Physics]

EDIT. Delta2 beat me to it.

The explanation is quite interesting. At the moment of connecting the charged capacitors, electrons have to jump across the gap between the ends of the connecting wires.

When the electrons jump, they radiate electromagnetic waves (hence lose some of the electrical potential energy).

This is an example of (the well-known classical physics) fact that accelerating charges radiate EM waves. If the capacitances and charges are large, you can actually see a spark. Even if a spark can't form (e.g. if the wire-ends are in an insulating medium such as oil), the EM waves are still generated and are quickly absorbed causing local heating.

 

[DaveE]

You can also analyze this with a lumped element circuit model. Suppose there are no resistive losses; this may be feasible if you make your circuit out of superconductors. As others have said, you can't move electrons without making a magnetic field. In the lumped element model this is an inductor connecting the two capacitors. Let's also assume that the losses to radiation from the fields is zero; this isn't realistic, but a good approximation for "very small".

When you solve this system you will get sinusoidal currents and voltages everywhere. there is no steady state capacitor voltage. The peak voltage (at zero loop current) is the original capacitor voltage, so the energy is the same at that time. The capacitors shift their energy back and forth as the circuit oscillates.

The peak current occurs when the two capacitors are equal at 1/2 of the initial voltage and is equal to $\frac {Vo}{Zc}$ (Vo = initial voltage), where $Zc=\sqrt{\frac{2L}{C}}$, the characteristic impedance. If you then calculate the energy stored in the inductor $E=\frac{LI^2}{2}$, you will find the rest of the energy you thought was missing.

This is exactly analogous to pulling a child back on a swing, letting them go, and asking where did the extra potential energy go when they reach the low point.

 

[DaveE]

In the case where you wait long enough that the radiated energy does make the circuit settle down to steady state, with 1/2 of the initial voltage on each capacitor and no current, you can model this as an RC circuit. This leads to a really useful rule of thumb for circuit designers, which is that when you charge a capacitor in a resistive circuit the energy lost in the process is the same as the energy you added to the capacitor. This is really useful when designing circuits with complex waveforms, like snubbers for SMPS switches. The answer is independent of waveform shape or resistor value.

edit: Oops! This isn't always correct. It's not independent of waveform. It is true for an RC circuit with a step voltage applied.

 

[kuruman]

This is analogous to a charging capacitor except that instead of a battery, we have a discharging capacitor. We assume some wire resistance $R$. If we let $q(t)$ be the instantaneous charge on the initially uncharged capacitor and $q_0$ the initial charge on the charged capacitor, then the Kirchhoff voltage equation is $$\frac{q_0-q}{C}-iR-\frac{q}{C}=0~~~~~\left(i=\frac{dq}{dt}\right).$$The solution of this differential equation is$$q(t)=\frac{1}{2}q_0e^{\frac{-2t}{RC}}.$$The energy dissipated in the resistor is $$E_{\text{loss}}=\int_0^\infty \left(\frac{dq}{dt}\right)^2R~dt=\left(\frac{q_0}{RC}\right)^2R\int_0^\infty e^{\frac{-4t}{RC}}dt=\frac{1}{4}\frac{q_0^2}{C}.$$Doing it by energy considerations, $$U_{\text{before}}=\frac{1}{2C}q_0^2~;~~U_{\text{after}}=\frac{1}{2C}\left(\frac{q_0}{2}\right)^2+\frac{1}{2C}\left(\frac{q_0}{2}\right)^2=\frac{1}{4}\frac{q_0^2}{C}.$$Therefore, $$U_{\text{before}}=U_{\text{after}}+E_{\text{loss}}.$$ Note that this result is independent of the assumed wire resistance $R$ so one can let it become arbitrarily small and the result would be the same. Of course there will be radiative losses too as others have observed, which are ignored in the simple model presented here. However, these will be under the $E_{\text{loss}}$ umbrella and will come at the expense of the Ohmic losses. In short (the pun is intentional), you can't beat Poynting's theorem.

 

[Oltunde]

This is analogous to a charging capacitor except that instead of a battery, we have a discharging capacitor. We assume some wire resistance $R$. If we let $q(t)$ be the instantaneous charge on the initially uncharged capacitor and $q_0$ the initial charge on the charged capacitor, then the Kirchhoff voltage equation is $$\frac{q_0-q}{C}-iR-\frac{q}{C}=0~~~~~\left(i=\frac{dq}{dt}\right).$$The solution of this differential equation is$$q(t)=\frac{1}{2}q_0e^{\frac{-2t}{RC}}.$$The energy dissipated in the resistor is $$E_{\text{loss}}=\int_0^\infty \left(\frac{dq}{dt}\right)^2R~dt=\left(\frac{q_0}{RC}\right)^2R\int_0^\infty e^{\frac{-4t}{RC}}dt=\frac{1}{4}\frac{q_0^2}{C}.$$Doing it by energy considerations, $$U_{\text{before}}=\frac{1}{2C}q_0^2~;~~U_{\text{after}}=\frac{1}{2C}\left(\frac{q_0}{2}\right)^2+\frac{1}{2C}\left(\frac{q_0}{2}\right)^2=\frac{1}{4}\frac{q_0^2}{C}.$$Therefore, $$U_{\text{before}}=U_{\text{after}}+E_{\text{loss}}.$$ Note that this result is independent of the assumed wire resistance $R$ so one can let it become arbitrarily small and the result would be the same. Of course there will be radiative losses too as others have observed, which are ignored in the simple model presented here. However, these will be under the $E_{\text{loss}}$ umbrella and will come at the expense of the Ohmic losses. In short (the pun is intentional), you can't beat Poynting's theorem.

Very good mathematical analysis, but you did not quite explain in simple terms, why energy is not conserved! So two capacitors acté like LC and radiate waves!! Unheard of, why don’t we used them instead of LC systems. Please just explain, how is different than a RC or LC system, where energy conserved!

 

[Oltunde]

EDIT. Delta2 beat me to it.

The explanation is quite interesting. At the moment of connecting the charged capacitors, electrons have to jump across the gap between the ends of the connecting wires.

When the electrons jump, they radiate electromagnetic waves (hence lose some of the electrical potential energy).

This is an example of (the well-known classical physics) fact that accelerating charges radiate EM waves. If the capacitances and charges are large, you can actually see a spark. Even if a spark can't form (e.g. if the wire-ends are in an insulating medium such as oil), the EM waves are still generated and are quickly absorbed causing local heating.

Why this does not happen in other circuits, such as LC? In LC it convert to waves!

 

[Oltunde]

Homework Statement: Capacitors A and B are identical. Capacitor A is charged so it stores 4J of energy and capacitor B is uncharged. The capacitors are then connected in parallel. The total stored energy in the capacitors is now ____.
Relevant Equations: U = q^2/2C

I am given the answer is 2J and I know how to get there with U = q^2/2C. But what I don't understand is that why is the energy not conserved here? How can energy loses just because the capacitance is doubled?

Excellent question. Did you find the answer to it. I did not find a good simple reply to youtpr question.

 

[kuruman]

Excellent question. Did you find the answer to it. I did not find a good simple reply to youtpr question.

Please note that this thread is 3 years old and the the OP has not been seen since December 15, 2020. You are unlikely to get an answer. If you have your own questions about this problem, I suggest that you start a new thread and post them there.

 

[berkeman]

Old thread is now tied off, to avoid attracting more necroposts.