What is the voltage at points A, B and C?
What did adding R2 do to the potential energy level at point B?
What is the voltage at points A, B and C?
What did adding R2 do to the potential energy level at point B?
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Um.. it did nothing! the potential energy is still 10v /100 ohms.
Wrong answer.
Global warming?
Tony/Steve
You are wrong!
Sparks is correct as Tony/Steve has failed to define a point of reference from which to measure. Now you may say the -ve is the place but without difinition this is an arbitrary judgement, and leaves the question subject to misinterpretation.
P.d. or Potential Difference is the difference between two places.
Therefore you are all free to use any point of reference you find convenient.
See below for Steve/Tony’s reply and clarification.
Sorry that is the current in amps – at least it is what I was taught at school.
V/R = I in amps.
U = R x I, the voltage (potential energy) decreases due to the passage of the current (charge) through the resistance, the charges loose energy in the resistance… what’s the point of the question?
Wrong answer
Ok, the potential energy at each point of the circuit doesn’t change by the addition of any resistance, because it’s defined by the source… but what is the point of the question?
Assuming DC, VA = 0 V, VB = 5V, VC = 10 V
You beat me to it.
Share your cookie with me?
Exactly
Actually, it doesn’t make any sense to ask “What is the voltage at point A.”, unless you establish a reference point as “zero volts”. Or, in other words, It only makes sense to ask what is the voltage difference between two points.
It is almost impossible to have an intelligent conversation about anything around here
Glad you pointed this out Albert, I was going to say the same thing.
I’m really not trying to be a smarty pants. I do this kind of stuff for a living, so it’s the first thing that came to my mind. No reference was stated, so it appeared obvious to me that the question was intended to determine if the viewer understood that a reference was necessary in this sort of measurement.
Albert, Your post was right on.
I started training as a hardcore EE in the late 1960s. So being old school I worry about details like where is ground. Ohm, Thevenin, Kirchoff, Nodal analysis, Norton analysis were are bread and butter. All done with slide rules !
Please stop. You are driving me nuts with this endless inanity, folks.
There is no such thing as an absolute voltage. People measure voltage by putting the black lead on negative and the red lead on positive. You read the voltage off the nice little LED on your meter.
If people want to be anal retentive and talk in circles about stupid shit endlessly, do it somewhere else.
Good to hear it, EW. Wouldn’t want to be misunderstood here. I’m trying to learn something about climate science and the way it’s being manipulated by politics. I must say, this has all been very enlightening, and I’m very thankful to SG for putting this all together.
Sorry, SG. I’ll be quiet now.
I never liked this + convention on DC power sources. Not the flow or anything like that. I just feel it’s lacking in information. I took EE course for a few years and not once was it mentioned that one side is used as reference. Many people were confused why it’s 10V on one side, but 0V on the other when the potentials are identical. That’s why I’m not that big a fan of this example. If you use the opposite as convention, then the voltage actually dropped instead of increased.
It doesn’t make the slightest bit of difference what you use as a reference. The voltage at B is increased by 5V because of R2.
The source is 10V. It may be regarded as +10V one side and 0V on the other, or 0V on one side and -10V on the other, or +5V one side and -5V on the other – they are all functionally the same.
If the source is +10 to 0v, B = +5V
If the source is 0 to -10V, B = -5V
If the source is +5 to -5V, B = 0V
If I’m measuring the voltage across R1, then the addition of R2 into the circuit reduces the measured value.
Voltage is not an absolute value, it does not exist at a point, it exists only as potential difference between two points. You answers are with reference to point A (i.e. the negative battery terminal).
However, to answer the question fully, Steve asked about potential energy. The only potential energy described in the system above is the chemical energy stored in the battery… so, I would say that adding R2 to the circuit makes the battery last twice as long before running flat. I think that’s the key to all of this.
Adding R2 raises the potential at B by 5 volts. This is the most basic circuit problem.
It sounds to me like the answer should be a voltage answer, but I know the right answer is global warming. If there is a problem, the answer always is global warming.
You sir, have great potential.
The answer is simple.. By adding R2 at point B the resistance was doubled. The resistance at point A is 100 ohms.. point B and C have a combined resistance of 200 ohms.
The idea of inserting another resistance in the circuit is tricky, because if we begin with a circuit without resistances the voltage at each point would depend on the geometry of the circuit.. but given a point X with a given, finite, initial voltage (current) the addition of another resistance at points of higher voltage will not affect the voltage at X, because the addition of more resistances only affect the current, at each point.
There is nothing tricky about it.
It could be, if you think in terms of a superconductor circuit, with little resistances of normal material in between! 🙂
It’s not tricky at all.. the circuit is in series therefor it RTotal = R1+ R2+ etc..
Thanks.
The strictly technical answer for the circuit as shown is “The potential energy is zero” in both cases because without any means to do work (or delivery any current) the joules/coulomb is zero.
For those who do not understand, in order to measure a voltage, one requires two points. Usually one is defined as “0V” so the other is the potential relative to this point.
However, with no connection to “0V”, which is the implied second point of connecting a voltmeter, there is only one connection. As one connection does not permit any flow of current, there is no potential of delivering any energy to the voltmeter so there is no “joules/coulomb” which is the definition of potential in this case.
Wrong answer
Resistors do not store energy.
That is equivalent to saying that dams don’t store energy. Why are you pursuing this ridiculous line?
Voltage A = 0 Volts
Voltage B = 5 volts
Voltage C = 10 volts
Presuming a closed system. As current flows through the resistor heat is produced. More resistance = more heat. The potential energy in the system decreases. If this were powered by a battery the potential chemical energy would be converted into heat energy at the resistors. If the voltage source is an outlet, it’s no longer a closed system.
Not gonna happen with constant EMF. Loss to heat is I^2 * R (I-squared R) so greater R is compensated twice by lower I.
Think of it this way, crank the R right up to infinity (open circuit), how much heat now?
I once explained this to the non technical missus.
If you had a water tank in the loft, with a pipe and a tap(faucet)
The higher the tank, the more the pressure, thats voltage.
The more that flows through the pipe, the amount, thats current
The tap(faucet) can be closed, opened a little or a lot, thats resistance
In my example, the resistance in the system is added together in a single place, the tap.
My guess is that the people that Steve is disputing are using a similar idea. They are not interested in potential difference, just the current.
The point is that greenhouse gases increase the surface temperature, analogous to how a resistor increases the voltage at point B.
That’s why I said the definition is tricky, it’s not the resistor that produces the voltage, it’s the source, the addition of resistors does not necessarily change the voltage at a given point, only if they’re placed at points of lower voltage.
In the case of GHGs the source of heat is the Sun, but it acts not only on Earth’s surface but also on the atmosphere, therefore the analogy is not completely valid I think.
It’d be as if the source of voltage in the electric circuit was also altering the resistances, to some extent.
Adding the resistor increased the voltage at point B by 5 volts. Nothing tricky about it.
Ok, there are clear definitions to calculate these circuits, I agree.
Placing a resistor at a given point does necessarily change the voltage at that point (after the resistor).
If the voltage is constant it does, because the new resistor alters the current of the entire circuit.
Increases the resistance, the voltage is decreased by the resistance.
OK. Point taken. But does replacing one resistor with another resistor increase or decrease voltage? Is it a more efficient or less efficient conductor? co2 has a lower heat retaining unit level than virtually any measurable gas it may be replacing (at xxppm minutia)
Well, the first question was what ‘what is the voltage’…
I’m a bit hazy on what he meant by ‘potential energy’ in the second question, though. The total power dissapated in the resistors in the second circuit is half a watt, distributed evenly between the two resistors. That’s power, though, and the question was regarding potential energy.
A-0V
B-5V
C-10V
taking (-) as reference
The Potential Energy is 0 in both cases
Any point left of R1 is equal to C and any point right of R2 is equal to A. Even an R3 right of R2 will not make any point left of R1 increase in value.
OK. The resistor is a solid [blanket] and traps heat, but CO2 is a gas, and when it is heated it expands and rises.
In the water cycle the water evaporates [Heat of Vaporization]; rises and gives up that heat[Heat of Condensation] to the upper atmosphere to radiate to space.
Back to my boiling water; the burner evaporates the water; the water vapor rises through the water still in the pot and expands into your kitchen. The water is still at 100C, and the burner is not made hotter by back radiation. The GHGs are coolants.
The hotter the burner the faster the water boils; The hotter the sun the faster the oceans-et-al evaporate. Wind helps this by a factor of 3, I believe.
When you sweat, the sweat evaporating, cools your body. Faster when in the wind.
There is no reason to obfuscate basic physics with unnecessary complexity.
Can I put everyone out of their misery yet?
Geez, how come no one mentioned Ohm’s Law and Kirchhoff’s Law? is that forgotten?
Since circuit 1 is a simple series circuit with one resistive elment, all applied voltage (10V) is dropped across that element. Measuring where indicated (A) yields 0V as both meter leads would be on the same side of the element, giving no difference in potential. To read the potential voltage, one would have to measure the voltage drop ACROSS the element.
In circuit 2, since both elements are in series the total applied voltage is dropped across both elements. Here Kirchhoff’s Law would apply. Since both elements are equal in value, the applied voltage is dropped equally across both elements. Hence, point C is 10V and point B is 5V.
Basic electronics doctrine.
The voltage at points A, B and C are as follows..
Point A = 0v
Point B = 20v
Point C = 10v
Is that one of those magical voltage-amplifying resistors?
lmao 😉
No, that odd looking value Sparks posted for point B is the answer as calculated by a climate modeling calculator.
Climate model calculators contain pixie dust and unicorn farts to generate magical values that match theories exactly.
(OK, /sarc, maybe….)
Very good!!
Oh, and to answer the question: It increased it 5V
Point B (if I read what you are doing correctly, is the same point as point C in the first schematic. Point C in the first configuration is 0V. If that is now point B in the second schematic, then potential voltage is increased to 5V as only half the applied voltage is dropped by R1.
A=0, B=5, C=10. Adding R2 raised the potential (voltage) at B (formerly A) from 0 to 5V
It did not raise anything, adding R2 doubled the resistance of the circuit and divided the voltage by 2 at point B.
That is completely incorrect. The voltage at that point was raised by 5 volts because of R2. This is the most basic electronics topic. What are you talking about?
I understand the analogy you’re going for, look at it this way, keeping it straightforward.. The resistance of R1 and R2 lowers the voltage at point B, it’s a step down in voltage because the circuit is configured in series. If you add another 100 ohm resister [R3] what will the voltage at point B be? Higher or lower?
Damn, I forgot the feedback… All bets are off.
You also didn’t factor in the aerosol effects of the smoke in the wires.
I used to use analogies a lot, but people attack the analogy then think they won the argument. Much better to stick to the subject and provide an explanation if asked. Otherwise you are flogging a dead horse.
I’m an ME and a pumping/piping system is analogous to an electrical system. Basically HP=flow*head/3960. If I double the head/resistance for a fixed HP, flow has to halve. If I want the same flow, HP has to double. And there’s the pressure proportional to square of flow issue as well. Energy is the ability to do work. Power is energy delivered over time. Most of the public, including the professions, don’t understand the difference. So if the voltage source in the example is infinite, potential energy never changes, but the power/rate of energy release does. Waiting to see what steve-tony says.
Went back to earlier comments. GHGs don’t increase surface temp directly. They heat the atmosphere which through conduction/convection/radiation heat the surface/sea/land. It’s just that air is a poor heat transfer medium compared to the water vapor cycle.
…at this point, what difference does it make?
Look, I don’t KNOW what the voltage at point B is, alright?
Well nobody knows what do you believe is the voltage is what matters.
🙂
It is elemenary until you allow CO2 into the room with the resistors, then the battery voltage rises due to positive forcing (+ve feedback).
Of course TOBS has to be accounted for and subtracted from the initial figures.
Steven/Tony
On the serious note, to be able to measure a potential difference (p.d.) you’ll have to define a point of reference. Without this there is no difference. With out this reference I am free to choose any point of reference, and oddly my meter alway reads zero for A, B, and C.
And I am correct.
Assume negative is ground.
From the question the only thing you can be certain of is in adding a second resistor of the same value the current is halved from 0.1 amps to 0.05 amps and the voltage across each resistor is 5 volts. If the negative terminal of the battery is the reference point then the voltage at A is zero volts B is positive 5 volts and C is positive 10 volts but if the reference point is the positive terminal then the voltage at A is minus 10 volts B is minus 5 volts and C is zero volts. With one resistor the power consumed is 1 watt and with two resistors the power consumed is 0.5 watts.
Here’s another quiz question.
If I point my infra red thermometer at the sun why does it measures 300°C? .
It usually is, unless the V supply is bipolar- like some of these answers. 😉
IF the above circuit was to model diurnal earth energy balance (not saying it is) some added complexity is necessary, say, for instance, a photoresistor or photodiode for differentiating daytime Insolation from nighttime Radiative heat-loss. :0
The first Circuit has no voltage as point A is at the reference.
(Incidentally the current through the whole circuit is V/R and gives 0.1Amps (100mA)
At point B we have 5v (a simple equal ratio voltage divider)
The long-winded way is –
Current through both resistors is equal and from
I = V/R
we get 10/200
This is 50mA (0.05Amps)
From V=IxR we get
50mA x 100R = 5V
As each resistor is equal 5volts is across each of them.
Sanity check –
The voltage across the resistor network can not be greater that the supplied voltage.
Each resistor has 5volts a cross it giviving 10volts in total, therefore OK.
At point C is 10v as this is connected to the +ve side of the power source.
All voltages are wrt -ve terminal of the power source.
1. When you ask “what is the voltage” (at each point) you mean relative to what? (You need to show a gnd connection or something, presumably to one end or the other of the 10v supply.)
2. I’ve always used the term “potential energy” in contrast to “kinetic energy,” as in, for example, a mass held up by a support, against the force of gravity. Do you mean voltage?
3. Point “B” isn’t shown in the first diagram, so I’m not sure what you mean by the effect on it of adding R2. If you mean what is the voltage there compared to point A in the first diagram, then the voltage at B is 5v higher.
Ha, I see tom beat me to it. (I started typing a while ago, and then the phone started ringing…)
I think it is all pretty clear.
Answer: Who cares??
This example has almost nothing to do with the previous example of “Water behind a Dam” as “Potential Energy”… all the critics and Science Geeks out there just need to chill out and relax… breathe in the simplicity of Water behind a Dam…
adding R2 increased the potential energy at point B from 0 volts to 5 volts. Since the flowing current is 0.05 A in the lower image, the potential energy went from zero to 0.25 W.
Sorry. energy is Watts, not volts. I mis spoke. 😦
Energy is Watt*sec=J (Joule)= Nm (Newtonmeter)
There is no voltage “increase” going on in this circuit. Vs is 10volts. The sum of all voltage drops in a DC circuit will be the source voltage for the circuit and that is the basis of Thevenin’s Theorem.
In the first example all 10 volts is dropped across R1 which is drawing 100 milliamperes of current and the power dissipated by that resistor is 1 watt. The voltage at Point A relative to the negative terminal of Vs is 0v as they are connected to each other so all you are measuring is a single point relative to itself. The power supply needs to be capable of 10vdc at 100 mA (or 1 Watt.)
In the second example you have a series circuit with two 100Ω resistors and is the most basic form of a voltage divider. The circuit is half the total power of the 1st example, draws 50 milliamperes of total current and uses 500 milliwatts of power. 5 volts is dropped across each resistor. Each resistor dissipates 250 milliwatts of power. (All of the current in a series circuit flows through each resistor.) The voltage relative to 0 (the minus sign on Vs) at Point B is 5v and is electrically measuring the voltage drop across R2. The voltage between Points B and C is 5v (measuring across R1.)
Point C is 10v and it doesn’t matter if the resistors are there or not as they are no longer part of the circuit as point C is electrically equivalent to the “+” side of Vs. The power supply needs to be capable of 10vdc at 50 mA or 1/2 Watt.
Enough of this bullshit already. These responses are pathetic, and make me feel like I am wasting my time here.
away with that. we come here because we think you are onto something. 🙂
just steer clear of these anaogies imho
If we can’t accept the basic principles of science, we can’t move forwards.
So elucidate those principles for us obvious novices. You had a point, we’ve been offering our thoughts, what, exactly, was your point?
The point is that impeding the flow of energy, increases the potential energy upstream.
The same reason why greenhouse gases increase the temperature at the surface of the Earth.
Tony, with respect, your DC circuit with resistors has nothing in common with any optical process/system design.
Just an FYI. I have gradate degrees in both electrical and optical engineering, several US patents in both fields and 30 plus years of experience modelling, designing, and testing electro-optical systems (some currently orbiting the Earth and sending back pictures). Not an argument from authority, just a word to the wise, your circuit analogy looks foolish.
It just does not work that way, no matter how much you want to believe it.
We will just have to disagree, one of us is the bigger dummy, time will tell who, it always does,
Cheers, Kevin.
Anything which impedes energy flow raises the potential energy upstream, whether it is a resistor limiting the flow of electrons, or an insulator restricting the flow of heat.
Ok, Tony, whatever you say.
So how exactly is IR energy flowing away from the surface of the Earth (at very nearly the speed of light in a vacuum) “impeded” again ? Do the “back-radiation” photons hold up tee tiny little speed limit signs ? Speed Limit: 1/10 c ? Are there speeding tickets involved ?
I love the rest of your blog, maybe just allow for the possibility that the “science is not settled” and move on ?
All respect, Kevin.
If a photon is absorbed and the energy reradiated back towards the surface, the amount of time required for that energy to escape into space goes up by many orders of magnitude.
In an electronic circuit a resistor does not just “impede” the flow of current, it consumes power and dissipates it as heat. In your first example R1 is going to radiate 1 watt of heat into its environment and nothing is being “held back” as any resistance in an electrical circuit is part of the load and it consumes power.
The potential energy is not raised it is simply used at a lower rate. The higher the resistance the less current is used and current is the number of electrons passing a particular point. You cannot raise the value of the “upstream” supply over its no load voltage without applying additional energy into the system.
I disagree. Impeding the flow doesn’t increase the potential energy, just slows its rate of discharge.
Increasing the voltage is by definition increasing the potential energy.
The voltage at C does not change so the potential energy of the system does not change. All R2 does is spread it around.
Stop
In physics we speak of the “electric potential”, which often is the voltage to infinity, or to the chassis. (To calculate the potential energy multiply with the charge to be transported).
Thus voltage is the “potential difference” between two points. That is why every voltmeter needs to have 2 connections, and why it imprecise language to talk of “the voltage at point A”.
And it is so important to confuse the issue and waste time with pointless semantic arguments.
You should have added a ground symbol at the negative terminal, and the question should have been either:
“What ist the voltage between Points A and G?”
or
“What is the potential at point A?”
I resent sloppy physics as much as I resent sloppy programming, sloppy statistics or sloppy climatology.
“If a photon is absorbed and the energy reradiated back towards the surface, the amount of time required for that energy to escape into space goes up by many orders of magnitude.”
Yes, I agree, this results in a time delay, not a velocity change. Thermal insulators work by slowing the velocity of thermal energy flowing through them. A time delay is not the same as a slower velocity. The “GHE” merely delays the flow of energy through the system. This total delay is on the order of tens (perhaps hundreds) of milliseconds (distance to TOA times speed of light times “several” bounces).
Given that the “fundamental frequency” of the system is 1/24 hrs = a period of ~ 86 million milliseconds this delay does not “trap energy”. Further, once you consider the thermal capacities of the elements involved (oceans vs “GHG’s”) this delay only changes the response time of the gases in the atmosphere. As “GHG’s” increase all the other gases warm up slightly faster after sunrise and cool down slightly slower after sunset. This effect is so small we probably cannot afford to measure it and the historical temperature databases do not contain the necessary data (even with extensive waterboarding applied).
Cheers, Kevin.
The net effect of the delay is exactly the same. It causes the potential energy upstream to increase,
But how does the voltage increase? Upstream of the 1st resistor it stays the same. After the 1st resistor the voltage drops per V=IR. When the 2nd resistor is inserted the current drops and the original voltage drop is split in two. Not that I’m disputing that climate change is FUBAR, just don’t like your analogy. See my comments about heat capacity and sensible vs latent heat transfer.
http://www.writerbeat.com/articles/3713-CO2-Feedback-Loop
The potential energy at point B increases, just as the potential energy in the lower atmosphere increases due to greenhouse gases impeding the outwards flow of energy.
The hydraulic analogy has been done, but I’ll try again.
There is a water tank with 30 feet of water/voltage/potential energy in it. An inlet control valve maintains that 30 feet/voltage/potential energy. There is one valve/resistor in the discharge line 50%/500 Ohms open. Water/current will flow through the valve/resistor based on its Cv/resistance causing a pressure/voltage drop according to the relevant equation. If I install another identical valve/resistor the flow/current will decrease, the level/potential energy/voltage in the tank will not change, the pressure/voltage drop that occurred across one valve will now occur across the two valves. Yes, the pressure/voltage after valve/resistor 1 & before valve/resistor 2 has increased, so what, the system’s energy has not changed.
I would think the global energy balance, the gozintaz and gozoutaz, and CO2’s role in all of that would be more informative. CO2 absorbs/retains heat in the atmosphere, leaving less to emit from the atmosphere. So where does it go? A CO2 sensibly heated atmosphere can’t sensibly heat the ocean nearly as fast as latent heat/evaporation cools those oceans. And the newly discovered heat at 2,000 meters didn’t come from the surface, not without help.
Tony, Ok, let’s talk “microprocessor” speak for a moment, I’m sure you are familiar with the concept of a “pipe-lined processor”. In such a system a computation (say multiplying two numbers) is accomplished in multiple sequential steps. During the first clock cycle part of number A is multiplied by part of number B. In the next clock cycle the rest of number A is multiplied by the rest of number B while the first result is stored in a register waiting for the final step. In the third clock cycle the two results are combined for a final answer.
In such a pipe-lined system you get an output (answer) on every clock cycle, but the delay from the time you input the numbers to the result is three clock cycles. If you carefully plan your steps is is possible to multiply many numbers in a continuous process at the clock rate.
This is a very high level description for the general audience. And we will likely quibble a bit about my description. But a pipe-lined processor can spit out results faster than a regular processor (given that all you want to do is lots of sequential multiplying, a common computer operation).
Now think just a bit about the “GHE”, it is really similar to a pipe-lined processor, photons are bouncing back and forth at the clock rate (~distance to TOA / speed of light), but they are being “spit out” at TOA at the same rate they are being input into the atmosphere at the surface. No impedance present.
Just another way of looking at the problem.
Cheers, Kevin.
The analogy is incorrect.
The greenhouse effect changes the latency of energy, not the flux – which has to be in balance with incoming energy.
A deeply pipelined processor also has a higher latency.
Yes, the greenhouse effect changes the latency of energy (AKA: the delay time, how long it takes for the “comes-outta” to show up after the “goes-inta” is present), EXACTLY, I think you are starting to get it.
Cheers, Kevin.
As with any insulator, the impedance of heat flow forces the temperature upstream to rise in order to maintain a constant flux. I don’t think you are starting to get it yet.
Kevin: Whenever I read your comments, I am reminded of the saying that, to a man who only has a hammer, everything looks like a nail. Your optical experience is completely misleading you with regard to radiative heat transfer. The issues have nothing to do with the physical speed of radiation, as you keep insisting.
Get an engineering heat transfer textbook. You will not find lots of discussion about resistance (impedance) to radiative transfer, as Tony is talking about. You will not find any analysis involving physical speed or time delays. That is simply not a useful way of looking at the problem.
Tony, whatever you say. But your own data analysis shows that there is no correlation between increasing CO2 and temperature…..
The impedance of air at the surface of the Earth is ~377 ohms in electrical speak and ~1.0 (no units) in optical speak. Regardless of the “GHE”, “back-radiation” reaching the surface has no effect on the impedance, it is still 377 ohms (sparky talk) or 1.0 (optical talk).
Your are spinning, you admit the “GHE” simply increases latency, then you also claim that it changes the impedance. Optical systems are more challenging than electrical systems, that’s why I stopped designing digital circuits 20 years ago, after a while they all look the same.
Enjoy your understanding of things.
Cheers, Kevin.
That is because the CO2 absorption spectra is nearly all absorbed already, so adding more CO2 has very little additional effect.
Curt;
“Kevin: Whenever I read your comments, I am reminded of the saying that, to a man who only has a hammer, everything looks like a nail. Your optical experience is completely misleading you with regard to radiative heat transfer. The issues have nothing to do with the physical speed of radiation, as you keep insisting.
Get an engineering heat transfer textbook. You will not find lots of discussion about resistance (impedance) to radiative transfer, as Tony is talking about. You will not find any analysis involving physical speed or time delays. That is simply not a useful way of looking at the problem.”
Curt, with respect, I have designed precision temperature controlled systems with temperature control of a surface within 0.1 degree C (using heat transfer textbooks).
I have many hammers; electrical circuit design, thermal system design and optical system design. among them.
I have included “radiative insulation” when it helps the system design. I designed a precision (+/- 0.1 degree C (3 sigma)) thermal surface which included “radiative barriers” (i.e. mirrors) to control the heat loss via radiation. The system in question was at about 350 degrees F (+/-0.25 degrees F). A bit more complicated than the usual; “is this heat sink big enough so my transistors won’t overheat” ?
I have also designed thermal control systems for laser diodes (+/- 0.05 degrees C), you would be surprised to learn that modulating the current in a laser diode at less than a few hundred kilohertz results in a temperature modulation of the laser frequency. The thermal capacity of the optical resonant cavity is so small that you can modulate the length of the cavity and change the optical wavelength, yes that’s correct, a temperature dependent effect at a hundred kilohertz frequency. I did not believe it at first either, but that was the unmistakeable result.
If you wish to still consider “thermal effects” as a “dc circuit’ go ahead.
I think the person with a “single hammer” in this case is yourself, you insist that everything is covered by a “heat transfer textbook”, OK, if you insist. Seems that your hammer is the “heat transfer” textbook.
But you may want to investigate the “temporal response” of an optical integrating sphere, it turns out the the “speed” at which energy flows through a system does indeed affect the outcome.
Thanks for your input, Kevin.
Kevin: Every thermodynamics curriculum I have ever seen starts with steady-state (“DC”) analysis, not because we see a lot of steady state systems in the real world, but (1) because they are easy to do, and (2) if you don’t get the steady state case right, it is not even worth starting on the dynamic case.
You really ought to publish your explanation of using optical methods for heat transfer analysis, explaining why it is better than what generations of scientists and engineers have been taught. It would be truly novel!
Simply Astounding. Here is the correct analysis.
Circuit A; Total Voltage available, 10v DC, Total Current available: not specified Total load, 100r
All DC circuits measured from ground or negative terminal. Also in this case since there are no reactive components, and the resistors are for all practical purposes considered as “pure” resistances, this analysis applies also to AC as well.
1a. The current flowing in a series circuit is the SAME for all points with respect to ground.
1b. From ohms law 10/100=.1a or 100 milliamps. 1 milliamp being equal to a thousandth part of an ampere. Power consumed is 10x.1=1watt
2. Point A is to considered as a direct short to ground and henceforth no voltage will be measured.
Circuit B: Total voltage available 10v. Total Current specified: not specified
Resistors in series add. so R1 + R2 = 200r Therefore accordingly 10/200 = 50 milliamps or .05 of an amp. Since R1 and R2 are identical then the voltage across each of them is the same.
Therefore point B is +5v and point C is considered to be the same as putting your meter directly across the power source and will read as +10v. The power consumed of course is one half that of circuit A or one half watt.
In my localized part of the universe 5v + 5v = 10v. The purpose of R1 and R2 is to divide potential but at the expense of current. This increase of potential energy is complete nonsense and the idea of “potential energy upstream” is even worse. These two circuits as a matter of fact are perfect examples of potential loss and not gain since some of the power in each circuit is lost through any heat generated in each resistor.
Apparently the owner of this blog believes he can get something for nothing.
By the way SG there is a situation where you can easily reduce the potential of the power source. Do you know what that is?
Unbelievable nonsense.
R2 increases the voltage at point B by 5 volts. Where do you people come from?
By George, I think I’ve got it! For the current/water flow to remain constant (I don’t recall that conditional.) after the second resistor/valve are installed, the potential energy will have to increase, a 15 volt battery in place of the 10, 60 feet of water head instead of 30. But the analogy still isn’t good because the global atmospheric system doesn’t work that way. Nobody turns up the solar energy.
Solar energy enters the global system at a relatively fixed rate. Energy of a certain wavelength excites CO2 molecule electrons which subtract the work function and emit energy in a wave length that agitates (not excites) water vapor molecules (like the microwave) which heats the atmosphere in general which transfers that heat to the surface & oceans which release more CO2 and loopy-de-loop. With more CO2, more energy is absorbed leaving less to leave the system and more to stay behind heating up the atmosphere. No net change in potential energy, just relocated.
If air were any good at transferring heat to water nobody in Phoenix would have to heat their swimming pools, they’d all rise to 105 F and have to be cooled. The latent heat of evaporation keeps that from happening. The oceans could handle the missing hiatus heat by increasing their evaporation/cloud formation .001%.
IPCC AR5 admits in TS.6 they have a poor understanding of the magnitude of the CO2 loop forcing especially over land, and high uncertainty about the water vapor/precipitation cycle, and a whole list of other rather critical unknown unknowns, like the ice sheets/caps, oceans below 2,000 meters, etc.
Whatever mechanism raised global temperatures during the first half of the twentieth century, the hiatus makes it abundantly clear it wasn’t CO2, not then, not now, not tomorrow. Coincidence equals cause plus an anti-coal agenda stirred together with egos, money, politics, and voila, the beast arises.
http://www.writerbeat.com/articles/3713-CO2-Feedback-Loop
You must be kidding right? Unbelievable Nonsense? That’s all you have? So 5 + 5 = 11 or something (lol). R2 doesn’t increase anything it simply D-I-V-I-D-E-S. THE 5 VOLTS YOU ARE READING ON YOUR METER IS PART OF WHAT YOU ALREADY HAVE. And you call this physics! Refute Properly. Everything in my previous post is demonstrably accurate. If You insist on this line of delusional thinking I will be forced to hand you your head on a scientific silver platter.
AND it won’t take long either. Also by the way you never answered my last question in my previous post. Why ? Don’t know the answer? Those who have been schooled in pseudoscience have an excuse.
The voltage at point B increases by 5 volts. Take your stupid straw man arguments elsewhere.
Well, yes, but the voltage at C stays at 10.
That is not in dispute and irrelevant
That’s because before point b you have no potential at all. Silly man.
That is exactly the point.
You have no potential because both of your meter probes are connected to the same spot. One to the -ve the other to just before b, which is a wire connecting to the same place. No hope of reading any potential there. Is there? Its a direct short. What on earth are you going on about. Do you live in this universe or an alternate one. You think resistors don’t heat up.
The answer is “violence”. As Chuck Norris will tell you, violence solves everything.
Aha. Clever analogy for the greenhouse effect. Never studied electronics much, so I had to think it over before I got it.
Anyways, it’s nice you’re taking on the greenhouse effect skeptics. Hate them so much.
The potential energy of the system is 10 volts – with one resistor, with two resistors, with a hundred resistors, in series, in parallel, painted fluorescent pink. As current flows and the resistors heat, the potential energy, i.e. voltage, can only decrease, entropy. Hardly irrelevant.
Complete nonsense,
The potential energy at B increases by 5 volts. This is the most basic circuit problem.
Only relative to ground. One 500 Ohm resistor drops 10 volts at 20 ma, with the downstream side at 0 volts relative to ground. Two 500 Ohm resistors will each drop 5 volts, at 10 ma, 5 volts between, 0 volts relative to ground. W/ four 500 Ohm resistors each will drop 2.5 volts, at 5 ma, 7.5, 5.5, 2.5 volts between in three places, 0 volts relative to ground. But the only potential energy that matters is the undiminished 10 volts, until the switch is closed, anyway. Can’t leave out the current. As I pointed out earlier, the flow decreases with increased resistance.
So, interesting back & forth, but what does this have to do w/ CO2, climate change, etc.?
Nonsense, It doesn’t make any difference what the reference point is. The voltage at point B increases by 5 volts.
The total energy of the system remains the same.
Enough straw man arguments.
There’s no straw about. Yes, the voltage at B increase from zero to 5 volts with the 2nd resistor, but so what? The system’s potential energy, 10 volts, does not change.
You are completely missing the entire point of the exercise.
Energy is not, and never will be, measured in Volts.
I might point out that energy is not equivalent to temperature either.
Yes I accept that circuits can be analogous to many other systems, but energy is energy. If your analogy wonks around with energy then there’s a problem.
Absolutely correct.
Energy is measured in Volts*Amps*seconds = VAs = J = Ws = Nm.
Seems some people do not understand the difference between “potential energy” and “electric potential”.
The first is measured in J, the second in V.
And that point is?
+1
“The point is that greenhouse gases increase the surface temperature, analogous to how a resistor increases the voltage at point B.”
Went back to the top and scrolled down. GHGs in general or CO2 specific? I read that no 3 (water vapor is 1) GHG methane has been dismissed as lacking the needed exciting frequencies.
CO2 absorbs solar energy of specific frequency, electrons jump orbits, falling back subtract work function, emit the remaining energy, heat up water vapor molecules, heat the atmosphere, heat the ocean, release more CO2 and ’round and ’round it goes. That’s a little more involved than one or two resistors plus I don’t see any analogous form of “resistance” in that loop.
“Climate Change in 12 Minutes” does a good job on this loopy loop.
So did I win?