# Energy Lost

Discussion in 'General Electronics Chat' started by vks_foe, Nov 12, 2010.

1. ### vks_foe Thread Starter Active Member

Jan 11, 2009
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Suppose we have an ideal capacitor(C), connected to an ideal voltage source (V) through an ideal switch. Let capacitor be discharged initially and the switch is open. On closing the switch, what will be the energy lost by the source ??
Will it be equal to energy stored in cap i.e. 1/2(CV^2) or will there be an equal amount of loss as EM radiation leading to total energy lost by source as CV^2 ??? (based on max. power transfer theorem)

2. ### marshallf3 Well-Known Member

Jul 26, 2010
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If this were a totally theoretical problem your only loss would be through the resistance of the wiring.

Jan 18, 2008
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4. ### mik3 Senior Member

Feb 4, 2008
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Everything is ideal!

5. ### vks_foe Thread Starter Active Member

Jan 11, 2009
55
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Thanx to all!!
But, this paper says that there is some radiation loss in case of instantaneous charging and discharging of cap.. In this paper, there were 2 capacitors, so we could directly apply charge conservation. Unfortunately, we have a voltage source here .

But still, here it is instantaneous charging. So there must be some radiation loss. Isn't it ???

6. ### Wendy Moderator

Mar 24, 2008
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Yes, there could be a spike of RF, but it would be incredibly small. You are into the realm of fourier calculus, which is the math of describing how waveforms are made from many different frequencies.

There is another aspect to consider, while everything you mentioned was ideal, how about the inductance of the wiring? Not ideal, but neither is your problem, you are picking and choosing what to ignore.

The inductance of the wiring will affect the frequency output much more than anything else.

7. ### jpanhalt AAC Fanatic!

Jan 18, 2008
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I wasn't suggesting that you turn that paper in for your homework. I realize it is not the same question. It is a study of a situation involving perfect capacitors in which energy seems to be lost. That two-capacitor question is a classic. There are lots of papers and theories on that subject. I gave the link to help get you thinking of the subject in a different way. Have you searched Google?

8. ### vks_foe Thread Starter Active Member

Jan 11, 2009
55
5
@Bill : you mentioned that there will be a small RF spike. I have a doubt. Suppose the connecting wires had some resistance R (no inductance for now, to make life easy). So, the energy dissipated in the resistance while charging will be 1/2CV^2. This is independent of the actual value of resistance. So if we keep reducing this resistance, energy lost is constant. So, will that small RF spike radiate the same energy i.e. 1/2 CV^2 ??

@jpanhalt: I'm sorry if you didn't like my question to be posted here. I thought discussion in this forum is better than google.

9. ### Wendy Moderator

Mar 24, 2008
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No, it is not. You are fundamentally describing a DC scenario. After 5 time constants that capacitor is charged. After 1 time constants the rate of change is extremely low.

The rate of change is what is going to generate RF. When I say low, think times 10-15 power of what the DC is. You may create a small "pop" in the RF spectrum, but that is it.

Here is a RC curve I made for future graphical layout. The 1st TC (time constant) is 68% of the curve. The second TC is around 90%.

RF is not that easy.

You said ideal as part of the initial conditions. You are changing the problem mid stream, which doesn't help a bit.

OK, you have resistance in the wire. Their will be some heat generated during the charging process, but once the capacitor is charged, that is it. The energy is stored, and there is no rate of change.

One other ideal you might want to get rid of, capacitor leakage. Some caps are less ideal than others (namely electrolytic). They leak their charge away slowly due to internal resistance, their electrolyte is not a perfect insulator.

When thinking of these problems consider rate of change, it is where the energy is.

10. ### vks_foe Thread Starter Active Member

Jan 11, 2009
55
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Do you mean to say that energy dissipated is 10-15 times 1/2 CV^2??

I agree and this is where i have doubt. The "some heat generated" is independent of the resistance of the circuit or the charging rate or time and is equal to 1/2 CV^2. So what will happen if this resistance goes to zero. I am interested in finding the total energy lost during charging process. As, I'm of afraid of the complex mathematical equation of wave theory, a qualitative solution will be easier for me to understand.

11. ### Wendy Moderator

Mar 24, 2008
20,772
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Capacitive reactance (and inductive reactance) while expressed in ohms, does not generate heat. It is a function of the phase shift for AC (which is not what you described), the square root of a -1. If you haven't had this in your course you will.

The fact that reactance is used extensively to increase the efficiency of many devices, such as fluorescent bulbs. The basic fluorescent bulb, like an LED, does not limit current, so a coil is used for its reactance.

ALL heat generated within a circuit is the direct result of resistance, real resistance, period.

I meant to say $10^{-15}$. When I said low power, I meant really really low power.

Last edited: Nov 13, 2010
12. ### russ_hensel Distinguished Member

Jan 11, 2009
820
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Try including a series resistor with an ideal voltage source, then look at the behavior as r approaches 0. Beware of current approaching infinity, which is correct but has some problems as real wires tend to vaporize under such conditions.

13. ### BillB3857 Senior Member

Feb 28, 2009
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Bill.
Your statement of "After 5 time constants......." is true for "all practical purposes", but in an IDEAL situation, I was taught that a capacitor will never fully charge. It will charge to a point that an increase can on longer be measured, but it will continue to charge. Ideal scenarios always cause problems since the can't really exist.

14. ### Wendy Moderator

Mar 24, 2008
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Yep, and this was discussed to death in the 0.999... = 1 thread. How does this help the OPs question though?

Calculus would state, approaches the power supply voltage as a limit. A tech would say, 5 time constants is close enough.

15. ### BillB3857 Senior Member

Feb 28, 2009
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I wasn't trying to criticize your post, Bill. Only trying to point out that any time IDEAL circuits are discussed, confusion reigns. There is no ideal circuit The OP's original question is invalid since ideal components don't exist. Of course, theoretical questions must be pure in as much as real world and theory cannot be mixed, as you pointed out in an earlier post when you said that the OP simply choosing which parameter to ignore. Otherwise, results are also invalid. As an example: does a perfect wire have resistance? NO. Does a perfect wire have inductance? In my opinion, no. Reasoning is, length would have an effect on inductance just a it would have on resistance. Perfect wires would have zero length, zero ohms and zero inductance and zero distributed capacitance to other components.