2
\$\begingroup\$

This is a followup to my previous question: A capacitor stores half the energy when charged from a battery each and every time?

This is a real world circuit with losses. The losses are in the resistance of the battery, the switch, the capacitor, and the inductor.

The inductance value of the L1 inductor has been left out because it was not known or measured. It's not a value of 0 however as it has some inductance.

The capacitor starts out with an initial energy of 0 Joules as well as an initial voltage of 0 Volts.

The switch is closed only one time and left closed. The capacitor charges to the same voltage level as the battery of 12.33 V therefore it now has 76.014uJ of energy in it.


The question is:

Does the 50% loss of energy that normally occurs when charging a capacitor from a battery (without the inductor) still apply to this circuit, even with the L1 inductor in place?

In other words, the capacitor will charge up to the same voltage level as the battery of 12.33 V as confirmed with an actual physical circuit. Has only 50% of the energy been transferred into the capacitor than what came from the battery?

The actual physical circuit has actual inductance in the inductor, but it is unknown, but it's not 0.

Summarized and to the point:

In this circuit, how much energy is transferred from the battery and into the capacitor when everything comes to rest and all processes are completely done? 50%?, 99.8%?, how much energy was transferred from the battery to the capacitor?


schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • \$\begingroup\$ What is the inductance? \$\endgroup\$ – Voltage Spike Nov 2 '17 at 16:41
  • \$\begingroup\$ The inductance can be anything you would like it to be. I didn't know what the inductance was in actual practice, so I left it off because it's a more generalized question. \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 16:43
  • \$\begingroup\$ As I can use any value of inductance, I choose 0. So it'll make no difference whether the inductor is there or not. \$\endgroup\$ – Finbarr Nov 2 '17 at 16:46
  • \$\begingroup\$ Inductor is non-dissipative so yes final energy in capacitor would still be half of work done by battery. So 50% would still be dissipated \$\endgroup\$ – sarthak Nov 2 '17 at 16:46
  • \$\begingroup\$ @sarthank - By the inductor being non-dissipative, does that go for any inductance with any resistance or is that just for a value of 0 for everything including inductance and resistance? Trying to keep this as simple as possible. Should I add an inductance of anything to get a more accurate answer you think? \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 16:49
3
\$\begingroup\$

The LC tank, that you've drawn, will oscillate / ring, and as you've stated in your question, "This is a real world circuit with losses", it will come to a still. During the oscillations the same current will go back and forth several times through the same losses, aka resistances.

So having an inductor will drain more power than having no inductor. The total power lost (aka energy) will be greater with the inductor than without the inductor. I'm implying the same thing with different words.

If the LC tank is critical damped or overdamped then the current will only pass through the resistances once. If it's underdamped however.. Then you get oscillation / ringing.

Hmm, this is my gut feeling though, so I may be wrong. I don't feel that I have time to put up the math to prove myself wrong or right. It's not worth the effort.

When it has stopped oscillating it will reach a steady state and no more power will be lost. And then it will reach your strange 12.33 V.

And to reach the climax of your "question". The capacitance hasn't changed, the voltage hasn't changed, the equation for the energy stored in a capacitor hasn't changed. It doesn't matter how it was put in there. After the steady state, when everything has come to a still, the energy stored in the capacitor can be calculated, which you've done beforehand. So I'll take your word on it being "76.014uJ".


Your question is weird and I feel that it doesn't matter what I write, will the correct answer give you anything meaningful?

When I feel that I'm becoming misunderstood, I take a step back and ask myself
"Who is my audience?"
"Do they know what I know?"
"Do they need to know something else to grasp the information?"

In other words, The problem is 99% of the time with me. Perhaps.. you need to reword your question so you don't have to make 3 nearly identical questions within 24h.


Try #2 LC tank

At T=0 when the switch closes, the following things will happen:

The capacitor, together with the inductor, is like a loaded spring. But it will "decompress" and surge current and start oscillate. I'm explaining it weirdly, but it's my view of it. And according to the link, the energy stored in an LC tank which is what you got in that setup, is equal to \$\frac{1}{2}CV^2 + \frac{1}{2}LI^2\$. At T=0 then the current will be 0. If we know the voltage and the capacitance, which we do, then we can calculate it. In other words, the total energy in the LC tank will be what you had in your question. "76.014uJ". However, this energy will be moving around in the circuit. At some point in time the majority of it will be in the capacitor, some other point in time in the inductor.

Since the oscillations are going in both directions, then... at some point in time the capacitor will be fully charged in one direction, such as having 12.33V across it's terminals. Fully loaded. Then in some other point in time it's voltage will be entirely inverted and it will have the same amount of energy. Such as having -12.33V across its terminals. This is probably how you are seeing a value above 50%. The equation is \$\frac{1}{2}CV^2\$, if V is -12.33 or 12.33 then the energy stored will be the same, right? But if you calculate some weird energy difference then you might see 99.8% because it's swinging to the top of some cycle and then to the bottom of some other cycle. It's hard to explain but I can see that it's possible that you've come to terms that it's 99.8%. While in reality it's 49.9%.


After trying it out in a simulator I've come to terms that at T=0, both the AC and DC energy are put into the LC tank. After the AC part has decayed, then the DC part is left which is equivalent to \$\frac{1}{2}CV^2\$ J. In other words, at T=0, \$CV^2\$ is inside of the LC tank in the form of AC + DC, the AC will have an amplitude that is half of the DC, so the \$V_{PP}\$ of the AC is the same value as the DC level.

Solving it for DC first, you get what you had in your question, "76.014uJ", solving it for AC after a tiny amount of time has elapsed so everything is stored inside of the capacitor and none is inside of the inductor, you get "76.014uJ" as well. Then you put the AC component + the DC component together and you get an answer that is close to that of laptop2d, his has decayed a little bit which is why it's slightly less than 100%. I used superposition in order to solve this problem. I'm not a clever man. It took me 3 tries.

So the 50% part I spoke about earlier that was "swooshing" around, that's not a lie, but it was also a 50% DC component laying on top of everything which confused me and most likely you.

TLDR; At T=0 the system contains 2×76.014uJ≃152µJ. After the AC component is gone, the DC component, 76.014uJ is left inside of the capacitor.

\$\endgroup\$
  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Dave Tweed Nov 2 '17 at 19:53
2
\$\begingroup\$

Well I do not want to write complete differential equation this time. But I think simple energy conservation can give us the answer.
After all the transients have died, the capacitor is going to have a voltage \$V\$ across itself. So finally, the energy stored in the system is \$\frac{1}{2}CV^2\$. Because there is no current no energy is stored in the inductor. So final energy of the system is: $$\frac{1}{2}CV^2 + 0 = \frac{1}{2}CV^2$$ The charge across the capacitor is coming from the battery. That is the battery moves the total charge \$CV\$ from its negative terminal to its positive terminal which has a potential difference V. So work done by battery is: $$W_B = CV*V = CV^2$$ So, once again half of it will be lost in the steady state with no energy in the inductor.

\$\endgroup\$
  • \$\begingroup\$ I have accepted your answer. In your opinion, what do you think of laptop2d's answer to this question? Is it right, wrong, or just poorly worded? Thank you very much for this answer it was succinct and to the point! \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 18:21
0
\$\begingroup\$

In this case, the capacitor never really 'charges' to a DC value. As soon as the switch is thrown, the circuit forms an LC circuit with a step input. The LC circuit then resonates (depending on the value of L and C). At this point you can't really say "how much energy is transferred" without looking at the circuit over time.

You can see below how the energy transfer from the supply (light blue trace) and the capacitor (red trace happens).

If you integrate the energy over 1.2mS after the switch was thrown. 970uJ was sourced by the supply, 218uJ ended up in the capacitor and 750uJ and 218uJ ended up in the inductor. (with a value of 0.1uH for the inductor)

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Thank you for the answer, it is much appreciated. From what you have written there it looks like nearly 100% of the energy has actually been transferred but in the form of oscillations. Am I interpreting that correctly? So when the capacitor ends up with a final voltage of 12.33 V then this isn't the actual amount of energy that's been transferred from the battery then but only appears to be? \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 17:10
  • \$\begingroup\$ The capacitor actually ends up with 0V after 1.2ms (with an inductance of 0.1uH) So how much energy is 'in the capacitor' is very much time dependent. You can think of it the capacitor and the inductor like a sloshing tank of water, and the switch being opened like giving the tank a kick. The water sloshes around and then dies down and stops. Same thing here, except the voltage would be like the height of the water. Asking about the height of the water (or energy at that point in the tank) doesn't really get you anywhere. \$\endgroup\$ – Voltage Spike Nov 2 '17 at 17:15
  • \$\begingroup\$ The problem is, now your talking about AC, and not DC so you have to use equations that also incorporate the amount of time that the circuit has been oscillating to look at energy transfer. Looking at the values instantaneously doesn't tell you much \$\endgroup\$ – Voltage Spike Nov 2 '17 at 17:17
  • \$\begingroup\$ That makes sense. But in general almost all the energy has been transferred but in the form of an AC type of energy sloshing back and forth as in a tank circuit, is that one way of looking at it then? \$\endgroup\$ – Marc Striebeck Nov 2 '17 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.