# Interesting argument between friends

 P: 2,251 Alan, no matter what you want to assume, when the transistors are ON, their resistance is at least as much as a wire of short length. when Q2 discharges the capacitor, if there is no other resistance in the loop, that energy will be dissipated in whatever resistance that current sees. in the wires, even those inside the capacitor, the soldered nodes, and in any other conductive element in the loop (i.e. Q2). specifically, getting back to the original question, a computer consuming 1 kW is rather unlikely, but a computer left on that continually draws W watts will dissipate nearly the very same amount of heat and heat the room just as well as a space heater that also consumes the same W watts. only energy that somehow crosses the boundary between what is inside the room and what is outside (sound leaking out, light going out a glass window), only that energy will not heat the room. a battery charger (drawing W watts) will heat the room less if energy is going into and stored in the battery. an electric motor (drawing W watts) with a shaft going through the wall to the outside will heat the room less than a space heater consuming the same W watts.
 P: 3,904 I know your argument, I am just confused about the current need to charge and discharge the cap. I just take everything ideal and look at one part. Assuming there is no ohmic loss, is there power deliver into the box?
 P: 140 If there are no resistances anywhere in that circuit. The circuit will not consume any energy. The charge stored in the capacitor will "return" to the flow when the capacitor is discharges and then charged again. We we were talking about positive charges moving, they would first move onto the positive part of the capacitor (charging it). Then they would move around loop 2 and end up on on the bottom (discharging the capacitor). When the capacitor is charged again, these positive charges would flow from the ground node and through the voltage source. So the circuit would be a perfect, forever circuit (if your voltage source was ideal also).
 P: 3,904 I don't see how, the current that charge the cap is from the battery with voltage, the current discharge the cap is by Q2, nothing to do with the battery. Far as the battery concern, it only see the charging cycle time after time. I need to be convinced that there is no power transfer from the battery into the box using an idea case. Then worry about the real resistance that cause power loss and heat.
 P: 140 Actually it's not so easy at all. You would have infinite currents somewhere and its generally stupid to think about it this way. There WOULD be resistances everywhere, and because of this, the energy is dissipated in these resistances. As Russ has stated several times, almost all energy in a circuit will be dissipated as HEAT in that same circuit. A little bit of energy will perhaps be radiated as radiowaves, but again, these will turn into heat when they hit the walls of the house. So for all practical purposes, the energy consumed by a computer will be dissipated in that computer as heat because of resistances in that circuit.
 P: 140 As soon as you add a resistance in the wires all these problems dissapear, and all power equations will be satisfied. The power in supplied by the voltage source, will be dissipated by the resistances. By wanting to have no resistances you present an impossible scenario with infinite current. Infinite current means the electrons move infinitely fast. They will bang into stuff when they reach the end of the wire and be dissipated as heat. But as soon as you add just 1 Ohm of resistance in the wires everything becomes apparent = The power is dissipated in the resistances.
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P: 22,313
 Quote by yungman I know your argument, I am just confused about the current need to charge and discharge the cap. I just take everything ideal and look at one part. Assuming there is no ohmic loss, is there power deliver into the box?
Setting aside the issue of infinite current, if the capacitor is ideal and there is no resistance elsewhere, there us no power either used or dissipated.

But how does that help us?
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P: 22,313
 Quote by yungman I don't see how, the current that charge the cap is from the battery with voltage, the current discharge the cap is by Q2, nothing to do with the battery. Far as the battery concern, it only see the charging cycle time after time. I need to be convinced that there is no power transfer from the battery into the box using an idea case. Then worry about the real resistance that cause power loss and heat.
If you write a conservation of energy statement to calculate that, it reduces to 1=0 or gives a divide by zero error, so that's why assuming no resistance doesn't clarify things:

I^2R=VI

Set R=0...
 P: 1,042 With non-zero R value in the cap charge/discharge network, the power lost is the same regardless of value, as long as the RC time constant is short compared to the period of switching. An Rdson in the FET of 0.10 ohm, 0.010 ohm, or 0.001 ohm results in the same power dissipation. If the Rdson is actually 0 ohm, the current is still limited, as it cannot be infinite due to the inherent inductance present. Since the energy stored in the cap is not returned to the battery it is either dissipated or radiated. In the R=0 case it is radiated. The parasitic inductance value of the FET, herein "L", forms an LC network with the capacitor. When the cap is fully charged w/ the FET off there is energy W = C*V^2/2. Then the FET turns on. The cap discharges into L, the FET inductance, which results in L storing energy. When the cap energy is depleted, V=0, but I continues through L charging C negatively. This is a resonant LC network. Oscillation takes place & energy is radiated. The small value of L results in a high resonant frequency. If the energy is not dissipated as heat, as in the nonzero R case, then it is radiated due to LC resonance. It does not recharge the battery because there looks to be no path for that to happen. A real world FET will have nonzero Rdson, so I don't think we should dwell on radiated power. The energy in the cap is dissipated in Rdson on a per cycle basis. Anything unclear can be explained if needed. Claude
 Sci Advisor PF Gold P: 3,745 i agree with russ thought experiment: place that computer (tv, stereo, whatever) inside a black box in center of the room. Whatever energy goes in (1kw of electrical) must come out into room light will impinge on walls of black box and warm them sound will echo around inside till it is converted to heat warming the air(recall Newton's error in calculating speed of sound) RF will be absorbed by the computer's internal shielding and warm it the heater FEELS warmer because it radiates heat at higher temperature than the others but recall temperature is not heat. my 2cents old jim
 P: 5 This seems to be the apex of the argument. While I agree that the "black box" concept is true, recall that our goal is to HEAT the room. To do this we must add thermal energy to the room faster than the surroundings can take it away. While at the end of the day all the energy will go to thermal energy (as everything does in the universe), a heater turns "nearly" all of it into thermal energy, while with something like a computer or a television the energy gets "sloshed around" via other means and does not contribute to heating the room when the heat is needed
 P: 235 But again that goes back to the difference in heat and temperature. I don't think anyone would argue that a heater is more efficient at changing the temperature in a room than a computer, but that doesn't mean that the two aren't producing the same amount of heat. The biggest difference there is a heater is designed to transfer thermal energy to the room very rapidly. In a computer the primary source of heat transfer is heat sinks. Heat sinks are designed to have a higher thermal mass, meaning they transfer thermal energy slowly (a high thermal mass can store a lot of thermal energy). This means they aren't going to quickly change the temperature of a room, even though the same amount of heat is involved in the two scenarios because the heat sink stores the thermal energy and releases it over a longer period of time.
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P: 22,313
 Quote by JustNobody This seems to be the apex of the argument. While I agree that the "black box" concept is true, recall that our goal is to HEAT the room. To do this we must add thermal energy to the room faster than the surroundings can take it away. While at the end of the day all the energy will go to thermal energy (as everything does in the universe), a heater turns "nearly" all of it into thermal energy, while with something like a computer or a television the energy gets "sloshed around" via other means and does not contribute to heating the room when the heat is needed
While strictly speaking you are correct, in practical terms the difference is negligible since the only portion that 'sloshes around' for more than a few seconds, is the heat absorbed by the mass of the computer. That's a small amount and it only causes a few minutes of lag for that small amount of heat it holds.
 Sci Advisor Thanks PF Gold P: 12,256 I have just laboured my way through this thread, reading as much as I could bring myself to and scanning through the rest of it. It amazes me that otherwise sensible people seem to be arguing about the principle of Conservation Of Energy in the specific case of someone's front room! How can it not apply here, same as in the nucleus and everywhere else? Unless the energy leaves the room / data centre in the form of sound (I don't think so) or EM radiation (a few Watts at the most) it must end up by heating up the air, walls or people in there. Why are you wasting time trying to show that energy in a circuit magically disappears because Mosfets and Capacitors are involved (are the REALLY that ideal?). It's crazy; isn't this supposed to be a Physics and not a Magic forum? Where heating is concerned, there are a lot of details involved in where the heat ends up in the room and how our body actually becomes aware of it. Cold feet in an otherwise warm room is not comfortable, for instance. But that is an issue that heating engineers and serious environmental engineers know all about. A watt hour is a watt hour and it's not how you produce it that counts - it's how you use it. End of rant.
 P: 3 If we are talking about purely heat, then yes, I guess they would give off the same amount of heat. But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors. I guarantee a heater gives off more heat per second than a computer.
P: 2,536
 Quote by LawRooney If we are talking about purely heat, then yes, I guess they would give off the same amount of heat. But if some guy is trying to argue that using a 1000W computer will keep you as warm as a 1000W heater will, I'd say that guy definitely needs to switch majors. I guarantee a heater gives off more heat per second than a computer.
A large portion of this thread has had discussion that disagrees with you.
Maybe I could suggest that many of the responders here could then head on over to the general math forum and start a new topic on $0.\dot{9} \neq 1$. Hurkyl would just love that.