TinMans reserch and experiments into free energy devices.

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poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #100 on: 2015-05-28, 13:53:28 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: TinMan on 2015-05-28, 07:20:09 TK If both the average voltage on the CSR and across the globe are negative,how do we get more power in the positive direction. So if we use our peak voltage X's our average current,and we get our average current by using ohm's law to calculate current,we would have I=V/R. So that would be -40mV/1 Or we take our peak current,and X's that by average voltage. So (about)700mV x 1ohm = 700mA. So now it is 700mA X's minus 200mV -->or do i have this wrong,and we have to use RMS voltages or something?. How did we get on this kick of using average voltage and/or current for computing Pout? The only time you can use average voltage or current with switched circuits is when computing Pin, AND when the VDC supply is steady. As follows: Pin(avg) = Vdc (bat or psu) x I(avg). However, IF your load is a pure resistance and you have a good true rms meter (or scope), you can use the rms voltage across the resistor and compute Pout via the following: Pout(avg) = Vrms²/R. Just make sure you know what the "hot" value of R is when you are computing the power. If the load is not purely resistive, then you must use the instantaneous method. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #101 on: 2015-05-28, 14:08:23 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: TinMan on 2015-05-27, 18:03:45 I So im at a loss as to why the voltage and current traces seem so constant. Already explained. The time constant of your L1I coil is large relative to the time constant of L1O. Quote Anyway,the blue trace is my current-over a 1 ohm CSR,and my yellow trace is across the globe. Great. But you can not use the average current or voltage to compute the Pout. It is quite possible for the average current to come out to 0 Amps, as you saw already. How then could you say you have 0 Watts going to the bulb? You can't, right? But that is what your computation would tell you if you used average current. If the load is purely resistive, and you know the "hot" value of the resistor, you can compute the output power either using the rms current or rms voltage as follows: Pout(avg) = Vrms²/R, or Irms²R. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

TinselKoala	Re: TinMans reserch and experiments into free energy devices. « Reply #102 on: 2015-05-28, 14:22:52 »
Group: Guest	Here I think there's some problem with the definition and the use of the term "average". If you are computing the power in, say, a waveform that consists of in-phase (resistive load) square pulses of current and voltage, and you are concerned about comparing the power in the "on" portion of the waveform with the power in the "off" portion, then the "average" across the "on" portion is the same as the peak value for that portion. You can consider the voltage and current to be constant at the value of the top of the "on" pulse and these values will also be the "average" values for that portion of the waveform since the pulse is flat on top during the "on" portion of the period. So to find the energy in the ON portion of a squared-off pulse you multiply the "average" or constant peak value of the current, times the "average" or constant peak value of the voltage _for that ON portion only_ and then multiply that result by the duration of the "on" portion. Ditto for the "off" portion. But this is a different use of the term "average" than what is shown on the "average" measurements on TinMan's scope, since I'm talking about portions of one period during which current and voltage are constant, and the scope is computing "averages" based on the entire screenfull of data: some number of full and partial periods.

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #103 on: 2015-05-28, 14:32:17 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	I agree TK, there may be some confusion. I also agree with your statements. It is fine to call the power over a portion of the cycle "the average", as long as one understands what that is referring to. I could be wrong, but it seems Brad is using measured average values from his scope and DMM's to compute Pout. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

TinselKoala	Re: TinMans reserch and experiments into free energy devices. « Reply #104 on: 2015-05-28, 16:48:44 »
Group: Guest	I just posted a video demonstration of what I was talking about with respect to the measurements being taken from the screen data only rather than some larger portion of data in the scope. http://www.youtube.com/watch?v=Q_fOJw8rePg

partzman	Re: TinMans reserch and experiments into free energy devices. « Reply #105 on: 2015-05-28, 16:53:49 »
Group: Experimentalist Hero Member Posts: 1808	All, I am somewhat hesitant to jump in here but I would like to offer several thoughts on the results TinMan is seeing on the output and the measurements involved. What I would recommend TM is to measure the current thru L1 so the current is measured during the off phase of the mosfets. What i believe you would find is that the current continues to flow thru the primary during the off phase because the low voltage across D2 is "clamping" the primary collapsing field. The current will decrease on a slope determined by the D2's voltage drop, the resistance of the primary, and the current in the loaded secondary. This is the source of the negative current/voltage you are witnessing across the lamp loaded secondary IMO. If you have removed D2, you have already found that the output negative voltage and current will diminish rather quickly after the IRF840's come out of avalanche at ~500v. They should reach avalanche unless there is an appreciable amount of capacitance in the drain circuit on the primary in which case the output will still diminish rather quickly and a half resonant sine will be seen on the drain during this period. EDIT: If you do not experience a decrease in the output negative voltage and current with D2 removed, then you do have an anomaly Regarding average or rms measurements, there is a time for each. If we are trying to determine if a particular DC powered circuit driving a resistive load is OU or not, we will measure the average DC current drawn from the supply in most cases. This is not always the case but sufficient for this discussion. For the load, we chose to make a true rms measurement and compare the two for the resulting COP. The problem here is that we are comparing apples to oranges in the fact that rms does not represent the true power in regards to equivalent heating as the average or mean measurements do. To really have an accurate energy or power measurement, one must use the average or mean on both input and output measurements to really be meaningful. TK has already pointed out how to take the approximate average or mean energy in joules of the output waveform and this IMO is the preferred method. If one's scope has vertical cursor capability, then use these cursors to take the average or mean measurement of energy in each portion of the negative and positive occurrences for the complete cycle, convert to power and sum. There are exceptions of course but in this circuit that will suffice. If one has a math channel, it is possible that the math function will integrate the average of the instantaneous values or samples over time and this should yield a rather accurate result. I can say that the TEK 3000 series scopes operate in this manner but now sure about any other brands or series. regards, paartzman

TinMan	Re: TinMans reserch and experiments into free energy devices. « Reply #106 on: 2015-05-28, 17:21:58 »
Group: Elite Experimentalist Hero Member Posts: 4728	Quote from: poynt99 on 2015-05-28, 14:32:17 I agree TK, there may be some confusion. I also agree with your statements. It is fine to call the power over a portion of the cycle "the average", as long as one understands what that is referring to. I could be wrong, but it seems Brad is using measured average values from his scope and DMM's to compute Pout. No I only used average voltage,and instantaneous current to measure power disipated through the globe test i posted not so long back,as some assumed that more power was being disipated through the globe that was brightest.. As it was a clean square wave that was 0V to peak V on each cycle,and never went below the 0 volt line,then i assumed this was the correct way to do it-->same as P/in we went through some time back Darren. There were no inductors,just a resistive load,and power being delivered via the pulsed mosfets. The way i have been obtaining average power for both the top half and bottom half of the wave forms is simply to rectify the output with two diodes on each leg of the secondary(cathode and annode together on either leg),and send the power from the off cycle to a cap with a resistor across it,and the same for the on time via a different cap of the same type and value with the same value resistor across it. Now,if i have a higher voltage across the cap/resistor that is collecting the off time power than i do across the cap/resistor that is collecting the on time power,then i must have more power output from the secondary during the off time than i do during the on time-correct?. I have also used average current X's supply voltage when the wave is square,and not below the 0 volt line. my DMM was within .8mA,so im guessing i was preaty close to being correct in this case. When it comes to AC type waves that have offset's,that is when i need help. PW said something about using the averages,and others have said to use RMS values-->so who knows. I think i will just stick to my rectifying,and cap/resistor combo's to get P/out --------------------------- Never let your schooling get in the way of your education.

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #107 on: 2015-05-29, 13:02:32 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: partzman on 2015-05-28, 16:53:49 All, I am somewhat hesitant to jump in here but I would like to offer several thoughts on the results TinMan is seeing on the output and the measurements involved. The problem here is that we are comparing apples to oranges in the fact that rms does not represent the true power in regards to equivalent heating as the average or mean measurements do. Perhaps you should have pondered this statement a little more before making it. Using rms voltage or current in the Pout computation with a pure resistance is perfectly valid (and this IS what I have been saying). Why the reference to apples and oranges? A valid power measurement is a valid power measurement. It is perfectly valid to use an average input current times the DC supply voltage to obtain Pin, and it is perfectly valid to use the rms output voltage and hot resistance to obtain Pout. Both results (Pin and Pout) are average power. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

TinMan	Re: TinMans reserch and experiments into free energy devices. « Reply #108 on: 2015-05-29, 13:18:59 »
Group: Elite Experimentalist Hero Member Posts: 4728	Quote from: poynt99 on 2015-05-29, 13:02:32 Perhaps you should have pondered this statement a little more before making it. Using rms voltage or current in the Pout computation with a pure resistance is perfectly valid (and this IS what I have been saying). Why the reference to apples and oranges? A valid power measurement is a valid power measurement. It is perfectly valid to use an average input current times the DC supply voltage to obtain Pin, and it is perfectly valid to use the rms output voltage and hot resistance to obtain Pout. Both results (Pin and Pout) are average power. Quote Using rms voltage or current in the Pout computation with a pure resistance is perfectly valid (and this IS what I have been saying). Well this is all good when we want the total power out,but what when we want to measure only the top or bottom half of the wave form?. Is this not when you would use the average voltage of say the bottom half of the wave,and use your average current from the bottom half of the wave form to get your average power from the bottom half of the wave form?. --------------------------- Never let your schooling get in the way of your education.

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #109 on: 2015-05-29, 13:57:02 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Yes. But that method with diodes isn't going to work too good if you're dealing with output voltages of 0.7V or less. Brad, Remove D2 (flyback across PRI) and post a scope shot of the current and voltage across the load again. Is there any difference with it In vs. OUT? --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

TinselKoala	Re: TinMans reserch and experiments into free energy devices. « Reply #110 on: 2015-05-29, 14:34:23 »
Group: Guest	Quote from: poynt99 on 2015-05-29, 13:57:02 Yes. But that method with diodes isn't going to work too good if you're dealing with output voltages of 0.7V or less. Brad, Remove D2 (flyback across PRI) and post a scope shot of the current and voltage across the load again. Is there any difference with it In vs. OUT? I installed a switch on the 1n5408 diode to make it easy to do this comparison. On my build, the only difference I can tell is in the "bottom right corner", that is, the trailing edge of the pulses. The negative offset during the OFF time isn't really affected much. But I don't have quite the magnitude of that negative part that TinMan does have. See the shots below. First one is with diode disconnected, second with diode in circuit. (ETA: The shots were taken in the "Average" acquisition mode with 8 averages. If I use "Normal" mode there is a big noisy ringing on the TE of the Iout trace (but not the Vout trace) instead of the small rings that show in these shots. ) ------------------------ TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeOut.JPG (73.63 kB, 800x480 - viewed 794 times.) TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeIn.JPG (73.6 kB, 800x480 - viewed 736 times.)

TinMan	Re: TinMans reserch and experiments into free energy devices. « Reply #111 on: 2015-05-29, 14:39:49 »
Group: Elite Experimentalist Hero Member Posts: 4728	Quote from: poynt99 on 2015-05-29, 13:57:02 Yes. But that method with diodes isn't going to work too good if you're dealing with output voltages of 0.7V or less. Brad, Remove D2 (flyback across PRI) and post a scope shot of the current and voltage across the load again. Is there any difference with it In vs. OUT? i will post a scope shot when i put it all back together Removing D2 decreases the P/in by a third-the amp draw go's from 430mA down to 320mA @ 7 volts.The P/out go's up. The forward side(above the 0 volt line)rises in voltage,and the reverse side(below the 0 volt line also increases. D2 seems to be a big power killer in both the P/in and P/out --------------------------- Never let your schooling get in the way of your education.

partzman	Re: TinMans reserch and experiments into free energy devices. « Reply #112 on: 2015-05-29, 14:52:03 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: poynt99 on 2015-05-29, 13:02:32 Perhaps you should have pondered this statement a little more before making it. Using rms voltage or current in the Pout computation with a pure resistance is perfectly valid (and this IS what I have been saying). Why the reference to apples and oranges? A valid power measurement is a valid power measurement. It is perfectly valid to use an average input current times the DC supply voltage to obtain Pin, and it is perfectly valid to use the rms output voltage and hot resistance to obtain Pout. Both results (Pin and Pout) are average power. Poynt99, I wholeheartedly agree that rms is a valid measurement and compares equally with the mean measurement as long as the signal being measured is a square wave. If however, we are measuring the rms of any wave shape other than square, the result will not be comparable or equal to the mean. For example, we have a circuit that produces a sine wave output of 10v rms across a 10 ohm resistive load for an output of 10w rms. The averaged mean input power from the DC supply also measures 10w so we're thinking we reached an efficiency of 100% with our circuit. Remembering however that the average or mean of a sine wave is 90% of the rms, our output voltage is really 9v mean. This calculates to 8.1w mean which means our true efficiency is 81%. In order to reach 100& efficiency with this circuit when comparing the mean input power to rms output power, the rms output voltage would have to be 1.11 times higher or 11.1 volts to be 100% efficient. This is the apples to oranges I was referring to. With complex wave shapes this difference can become quite large and therefore IMO, the only true method of accurate measurement is the average or mean. Regards, partzman

TinselKoala	Re: TinMans reserch and experiments into free energy devices. « Reply #113 on: 2015-05-29, 15:02:01 »
Group: Guest	Wowsers. I've been using a duty cycle of 5 percent HI. When I change to 16 percent HI, the picture changes drastically and my mosfets, both diodes, and load resistors get quite hot. (using 12 v in from battery) ------------------------ TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeIn16Hi.JPG (78.55 kB, 800x480 - viewed 742 times.) TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeOut16Hi.JPG (82.01 kB, 800x480 - viewed 718 times.)

TinselKoala	Re: TinMans reserch and experiments into free energy devices. « Reply #114 on: 2015-05-29, 15:58:29 »
Group: Guest	I dropped back to 10 percent Hi to cut down on the heating. Here's where I'm at now. ------------------------ TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeIn10Hi.JPG (78.84 kB, 800x480 - viewed 703 times.) TinMans reserch and experiments into free energy devices. TKsTinManWPotCoreDiodeOut10Hi.JPG (80.63 kB, 800x480 - viewed 530 times.)

TinMan	Re: TinMans reserch and experiments into free energy devices. « Reply #115 on: 2015-05-29, 16:09:57 »
Group: Elite Experimentalist Hero Member Posts: 4728	Quote from: partzman on 2015-05-29, 14:52:03 Poynt99, I wholeheartedly agree that rms is a valid measurement and compares equally with the mean measurement as long as the signal being measured is a square wave. If however, we are measuring the rms of any wave shape other than square, the result will not be comparable or equal to the mean. For example, we have a circuit that produces a sine wave output of 10v rms across a 10 ohm resistive load for an output of 10w rms. The averaged mean input power from the DC supply also measures 10w so we're thinking we reached an efficiency of 100% with our circuit. Remembering however that the average or mean of a sine wave is 90% of the rms, our output voltage is really 9v mean. This calculates to 8.1w mean which means our true efficiency is 81%. In order to reach 100& efficiency with this circuit when comparing the mean input power to rms output power, the rms output voltage would have to be 1.11 times higher or 11.1 volts to be 100% efficient. This is the apples to oranges I was referring to. With complex wave shapes this difference can become quite large and therefore IMO, the only true method of accurate measurement is the average or mean. Regards, partzman I would agree when measureing either the top half or bottom half of the wave form,but when measureing as a whole,i would think that you would have to use the RMS value,as i have managed(as can be seen in some of my scope shots) to achieve a 0volt average both on current and voltage-->so dose that mean we have 0 power?-->something was providing the energy to light the globe. --------------------------- Never let your schooling get in the way of your education.

Aking.21	Re: TinMans reserch and experiments into free energy devices. « Reply #116 on: 2015-05-29, 16:52:34 »
Sr. Member Posts: 420	Looks like squeaky bum time. (To quote the former manager of Man United) --------------------------- Electrostatic induction: Put a 1KW charge on 1 plate of a capacitor. What does the environment do to the 2nd plate?

Aking.21	Re: TinMans reserch and experiments into free energy devices. « Reply #117 on: 2015-05-29, 17:14:14 »
Sr. Member Posts: 420	On a more serious note: this is Tesla's version of Tinman's trafo. http://www.teslauniverse.com/nikola-tesla/patents/us-patent-433702-electrical-transformer-or-induction-device Tesla states that the main object is to secure a 90 degree phase difference between primary and secondary. Tee Hee: Is there no limit to Tesla's genius? --------------------------- Electrostatic induction: Put a 1KW charge on 1 plate of a capacitor. What does the environment do to the 2nd plate?

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #118 on: 2015-05-29, 19:21:00 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: TinselKoala on 2015-05-29, 15:58:29 I dropped back to 10 percent Hi to cut down on the heating. Here's where I'm at now. Good stuff TK, looks to me like you have a very close match to Brad's results. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #119 on: 2015-05-29, 19:34:49 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: partzman on 2015-05-29, 14:52:03 Poynt99, I wholeheartedly agree that rms is a valid measurement and compares equally with the mean measurement as long as the signal being measured is a square wave. If however, we are measuring the rms of any wave shape other than square, the result will not be comparable or equal to the mean. That is incorrect. As long as the wave form is repetitive (does not change significantly from cycle to cycle), the rms value of ANY wave form (or shape) in conjunction with the resistance value can be used to obtain Pout. Quote For example, we have a circuit that produces a sine wave output of 10v rms across a 10 ohm resistive load for an output of 10w rms. The averaged mean input power from the DC supply also measures 10w so we're thinking we reached an efficiency of 100% with our circuit. I can assure you that if the Pin and Pout are properly measured using the methods I've stated, you will not be able to obtain power measurements of equal power for input and output. If you do, then you might have something. What you are stating above is either dreamed up or the result of poor measurement methodology. Quote With complex wave shapes this difference can become quite large and therefore IMO, the only true method of accurate measurement is the average or mean. You seem a knowledgeable person, however something seems amiss. I invite you to show a real example or even a simulation to support these notions of yours. --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

partzman	Re: TinMans reserch and experiments into free energy devices. « Reply #120 on: 2015-06-01, 15:50:20 »
Group: Experimentalist Hero Member Posts: 1808	Quote from: poynt99 on 2015-05-29, 19:34:49 That is incorrect. As long as the wave form is repetitive (does not change significantly from cycle to cycle), the rms value of ANY wave form (or shape) in conjunction with the resistance value can be used to obtain Pout. I can assure you that if the Pin and Pout are properly measured using the methods I've stated, you will not be able to obtain power measurements of equal power for input and output. If you do, then you might have something. What you are stating above is either dreamed up or the result of poor measurement methodology. You seem a knowledgeable person, however something seems amiss. I invite you to show a real example or even a simulation to support these notions of yours. Poynt99, I have to admit the error in my mistaken understanding for years that the area under a waveform would create the equivalent to DC heating. I now see that the rms is equivalent and I apologize to any I have misled. The positive side of this for me is that I have actually developed OU circuits thru these past years and didn't realize it. However, it is like A.King said "just another squeaky bum"! partzman

TinMan	Re: TinMans reserch and experiments into free energy devices. « Reply #121 on: 2015-06-01, 16:10:31 »
Group: Elite Experimentalist Hero Member Posts: 4728	Quote from: partzman on 2015-06-01, 15:50:20 Poynt99, I have to admit the error in my mistaken understanding for years that the area under a waveform would create the equivalent to DC heating. I now see that the rms is equivalent and I apologize to any I have misled. The positive side of this for me is that I have actually developed OU circuits thru these past years and didn't realize it. However, it is like A.King said "just another squeaky bum"! partzman Please share as i (and many others here)have over the years. But i would probably remove the OU from the equation,and just call it something like-very efficient circuit. --------------------------- Never let your schooling get in the way of your education.

Aking.21	Re: TinMans reserch and experiments into free energy devices. « Reply #122 on: 2015-06-01, 17:44:07 »
Sr. Member Posts: 420	Tinman: On that note, how is your very efficient toroid circuit? Any further results to share? --------------------------- Electrostatic induction: Put a 1KW charge on 1 plate of a capacitor. What does the environment do to the 2nd plate?

poynt99	Re: TinMans reserch and experiments into free energy devices. « Reply #123 on: 2015-06-01, 19:59:11 »
Group: Administrator Hero Member Posts: 3217 It's not as complicated as it may seem...	Quote from: partzman on 2015-06-01, 15:50:20 The positive side of this for me is that I have actually developed OU circuits thru these past years and didn't realize it. Care to show some proof? --------------------------- "Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa

Farmhand	Re: TinMans reserch and experiments into free energy devices. « Reply #124 on: 2015-06-02, 01:35:47 »
Group: Guest	Quote from: Aking.21 on 2015-05-29, 17:14:14 On a more serious note: this is Tesla's version of Tinman's trafo. http://www.teslauniverse.com/nikola-tesla/patents/us-patent-433702-electrical-transformer-or-induction-device Tesla states that the main object is to secure a 90 degree phase difference between primary and secondary. Tee Hee: Is there no limit to Tesla's genius? Aking, that is a "constant current transformer" nothing special, the device would provide the same output current even if short circuited. That means no matter what the load the maximum current output is limited to a known maximum. It is not OU or even highly efficient. It is a transformer designed for a specific purpose. To provide a constant current to all loads. What makes you think it is special or OU ? Did you read the patent then notice the 90 degrees phase difference and wet your pants ? The 90 degree phase difference means the input and output are limited, that's all. Where Tesla says all loads he errs a bit because with a current limit of 1 ampere, if one was to connect a 3 watt bulb to the output the current would be less than if one applied a 100 watt bulb of the same voltage rating, however if the current was limited to say 1 amp then the 100 watt bulb could only get a maximum of 1 amp of current through it. Even with a maximum of 1 amp of current not every load we can imagine will receive 1 amp of current. And actually we can achieve a constant current "effect' by using a too high of a frequency for a given transformer, the frequency induced impedance limits the output and input and this can show us the effects of a reduction in current when a load is applied and also the acceleration under load effect neither of which are OU and both are very inefficient. Tesla's constant current transformer would be more efficient than the previous examples for limiting the output. Basically it is a transformer that can be shorted without overloading the source. ..

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