The Back emf misnomer

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TinMan

Re: The Back emf misnomer « Reply #100 on: 2018-08-22, 03:57:12 »

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Quote from: Smudge on 2018-08-21, 20:19:06

Just got my PC working again after moving house and I note the disagreement between two individuals here. I see them making opposing statements that are completely opposite yet they are both correct from their different perspectives or viewpoints. I'll elaborate on that in a later post but I will say that it took me a long time to work out why you don't get a 180 degree phase shift across a transformer, and it all comes down to perspective. We say that there is a voltage drop across a resistor but we don't talk of a voltage drop across an inductor. We are taught that V=-L*di/dt (note the minus sign) but also that V=omega*L*i (note the absence of a minus sign) and that second V is the AC voltage drop across the reactance. So voltage drop is a sort of back emf and that also applies to resistors. Apply a voltage to a resistor and clearly the polarity of the voltage across the resistor is the same as the that of the supply as seen by an external viewer. But now take an internal view going around the circuit and what is the polarity as seen by a moving electron? In the battery the polarity is in the same direction as the current, in the resistor it is in the opposite direction. So we could say that the resistor creates that opposing polarity, but we don't, we call it voltage drop.
Smudge

The phase relationship of the seconary to that of the primary will depend on what configuration your transformer is.
Take a toroid transformer for instance,where primary and secondary are wound around the core in the same direction together. The phase relationship will be in phase.
If we take a MOT for example,the phase relationship between primary to secondary will be 180* out-but that is to be expected,as the end of the primary is inducing the start of the secondary,and so the primaries output is inverted to that of the seconday. Simply swap the scope probes over on either,and now we have an in phase relationship.

But the question here is--is the polarity of the CEMF across the coil the same as that of the applied EMF

And-is the value of the CEMF at it's greatest just after T=0, or when the current flow has reached its maximum value?.

Simple really.

Brad

---------------------------

Never let your schooling get in the way of your education.

TinMan	Re: The Back emf misnomer « Reply #101 on: 2018-08-22, 10:57:26 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Room3327 on 2018-08-21, 21:40:00 I thank everyone who posted on this maybe one day we can all be on the same page with this topic if we keep at it. Yes.lets stick with it ,and work it all out. I have been watching many video's on youtube,and read many electronics web pages on the subject,and all i can say is-WTF. How can all these !so called! ex-spurts have it so wrong. Anyway,im sure Smudge will join in here. In the mean time,here is a video i found,that is actually correct,and shows the polarity of the CEMF to that of the applied EMF from T=0 As you can see,the polarity of the induced CEMF across the coil,and the applied EMF from the source is the same polarity. You will also note that the instant the switch is closed,(T=0),is when the current flow is at it's minimum,and the CEMF value is at it's maximum. https://www.youtube.com/watch?v=4LuwwIFZD_A Brad --------------------------- Never let your schooling get in the way of your education.

TinMan	Re: The Back emf misnomer « Reply #102 on: 2018-08-22, 11:13:04 »
Group: Elite Experimentalist Hero Member Posts: 4727	Here is a video showing BEMF. Once again,you will see that the BEMF voltage polarity across the motor,is the same polarity to that of the source. https://www.youtube.com/watch?v=qwtiX3pFCIE&t=26s Brad --------------------------- Never let your schooling get in the way of your education.

Smudge

Re: The Back emf misnomer « Reply #103 on: 2018-08-22, 19:27:57 »

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Quote from: TinMan on 2018-08-22, 10:57:26

As you can see,the polarity of the induced CEMF across the coil,and the applied EMF from the source is the same polarity.

Of course it is as measured from the common ground. But a physicist who is interested in the forces on an electron will look t it differently. He will look at the polarity, not as measured relative to ground, but as seen relative to the velocity direction of the moving electron. Then the battery polarity is in the opposite direction to the battery's electron velocity while the CEMF polarity is in the same direction as the coil's electron velocity. He looks at CEMF as being the reverse polarity to the forward EMF, one is driving the electrons and the other is retarding the electrons. And that is where the minus sign comes in for V=-L*di/dt and for V=-N*dPhi/dt. You are correct from your perspective, and so is the physicist from his perspective. It's a pity we are not taught that V=-R*i for resistors, instead we we are given V=R*i and told to consider it as a voltage drop. But clearly the voltage across the resistor is retarding the electron flow.
Smudge

gyula

Re: The Back emf misnomer « Reply #104 on: 2018-08-22, 20:39:34 »

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Hi Smudge,

Okay, I understand what you wrote and would like to make a notice as an addition.

I took the attached picture from the video TinMan linked to ( https://www.youtube.com/watch?v=qwtiX3pFCIE&t=26s ) and indicated the polarity of the V_Back virtual voltage source (the CEMF source) inside the motor and indicated the polarity of the 100V battery driving the motor.

So it shows that the two positives of them are common and their negatives are common, this is the only way to have the two current values calculated at start-up (10 A) and then at the full and unloaded rotor speed (1.5 A).
So the polarity of the CEMF and that of the input voltage is the same. This has been one of the points TinMan (and I) wanted to clarify.

Thanks, Gyula

PS Edited for more clarity.

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The Back emf misnomer

CEMF polarity.jpg (58.92 kB, 896x527 - viewed 528 times.)

« Last Edit: 2018-08-22, 22:53:29 by gyula »

muDped

Re: The Back emf misnomer « Reply #105 on: 2018-08-22, 20:49:43 »

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Additionally, the resistor is dissipating power as it
'resists' electron flow. Inductors and Capacitors
dissipate very little power as they go about their 'magic.'
L and C both have a small resistive characteristic
which is a bit of a nuisance.

---------------------------

For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

TinMan	Re: The Back emf misnomer « Reply #106 on: 2018-08-23, 01:14:15 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Smudge on 2018-08-22, 19:27:57 Of course it is as measured from the common ground. But a physicist who is interested in the forces on an electron will look t it differently. He will look at the polarity, not as measured relative to ground, but as seen relative to the velocity direction of the moving electron. Then the battery polarity is in the opposite direction to the battery's electron velocity while the CEMF polarity is in the same direction as the coil's electron velocity. He looks at CEMF as being the reverse polarity to the forward EMF, one is driving the electrons and the other is retarding the electrons. And that is where the minus sign comes in for V=-Ldi/dt and for V=-NdPhi/dt. You are correct from your perspective, and so is the physicist from his perspective. It's a pity we are not taught that V=-Ri for resistors, instead we we are given V=Ri and told to consider it as a voltage drop. But clearly the voltage across the resistor is retarding the electron flow. Smudge Smudge We are looking at the polarity of the CEMF to that of the applied EMF,not the flow of electrons. But,even looking at the flow of electrons,we can clearly see ,although the direction of flow is opposite,the polarity across the source,and the CEMF across the coil is the same. This is hard to show with an inductor,but very easy to show with say a PM DC motor,where we can increase the value beyond that of the applied EMF. I believe some here are mixing up BEMF with inductive flyback,in which case the voltage across the inductor dose indeed invert. Brad --------------------------- Never let your schooling get in the way of your education.

muDped	Re: The Back emf misnomer « Reply #107 on: 2018-08-23, 03:37:18 »
Group: Tinkerer Hero Member Posts: 3055	The polarity reversal of the Induced Voltage is essential to the operation of a Buck Converter. The Induced Voltage while the Inductor is charging reverses as the Inductor goes into discharge. Charge and Discharge Current move in the same direction. While a Capacitor doesn't reverse polarity while charging or discharging, the direction of the current does reverse. Charge Current and Discharge Current move in opposite directions. --------------------------- For there is nothing hidden that will not be disclosed, and nothing concealed that will not be known or brought out into the open.

Smudge	Re: The Back emf misnomer « Reply #108 on: 2018-08-23, 09:22:03 »
Group: Professor Hero Member Posts: 1953	Quote from: TinMan on 2018-08-23, 01:14:15 Smudge We are looking at the polarity of the CEMF to that of the applied EMF,not the flow of electrons. But,even looking at the flow of electrons,we can clearly see ,although the direction of flow is opposite,the polarity across the source,and the CEMF across the coil is the same. Let's look at a dual carriageway road where there is a strong wind blowing parallel to the road. Clearly the wind is in the same direction at both carriageways. The polarity of the wind is the same. But that is from the perspective of a stationary observer looking down on the road. If you are a driver, on one carriageway you have a headwind slowing you down and on the other carriageway you have a tail wind speeding you up. So what perspective is the most important? You say we are not looking at the flow of electrons, yet the so called "experts" that you think have got it wrong are like the car drivers, they are interested in what is slowing the electrons down or what is speeding them up. You are like the guy looking down onto the road, you see the wind/voltage polarity from that perspective. Both you and the "experts" are right, so why continue arguing about it? Smudge

TinMan	Re: The Back emf misnomer « Reply #109 on: 2018-08-23, 14:30:17 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Smudge on 2018-08-23, 09:22:03 Let's look at a dual carriageway road where there is a strong wind blowing parallel to the road. Clearly the wind is in the same direction at both carriageways. The polarity of the wind is the same. But that is from the perspective of a stationary observer looking down on the road. If you are a driver, on one carriageway you have a headwind slowing you down and on the other carriageway you have a tail wind speeding you up. So what perspective is the most important? You say we are not looking at the flow of electrons, yet the so called "experts" that you think have got it wrong are like the car drivers, they are interested in what is slowing the electrons down or what is speeding them up. You are like the guy looking down onto the road, you see the wind/voltage polarity from that perspective. Both you and the "experts" are right, so why continue arguing about it? Smudge We were just trying to be clear as to the polarity of the CEMF,thats all. Now-next Which is correct ? 1-The CEMF is at it's greatest value at T=0 (moment the voltage is dropped across the coil) 2- The CEMF is NOT at it's maximum when the switch is thrown. it's at it's minimum, Brad --------------------------- Never let your schooling get in the way of your education.

Itsu

Re: The Back emf misnomer « Reply #110 on: 2018-08-23, 15:19:21 »

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Quote from: forest on 2018-08-21, 17:29:43

You only need very fast scope 4-channel and a coil to check all this and end this topic.

i tried to, but its not that easy, or be more specific on how.

see below circuit, 12V SLA, switch (mr hand) and a 5.9mH coil.

The screenshot shows the:

battery voltage (blue)
coil voltage (yellow)
coil current (green).

I don't think the yellow (coil voltage) is the Cemf, its just the voltage across he coil when switch is active.
It follows the sagging battery voltage (blue) trace.

Green is the current following the normal inductor current build up curve.

Itsu

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The Back emf misnomer

cemf circuit.png (5.62 kB, 690x442 - viewed 426 times.)

The Back emf misnomer

cemf.png (28.41 kB, 640x600 - viewed 410 times.)

gyula	Re: The Back emf misnomer « Reply #111 on: 2018-08-23, 15:48:03 »
Group: Tech Wizard Hero Member Posts: 1242	Hi Itsu, Would you mind measuring the DC resistance of the coil? Just to learn about coil time constant. (of course the resistance could be calculated from the measured current) Thanks,, Gyula

Itsu	Re: The Back emf misnomer « Reply #112 on: 2018-08-23, 16:02:58 »
Group: Elite Experimentalist Hero Member Posts: 4235	Gyula, the coil measures 0.6 Ohm. Itsu

TinMan	Re: The Back emf misnomer « Reply #113 on: 2018-08-23, 16:18:17 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Itsu on 2018-08-23, 16:02:58 Gyula, the coil measures 0.6 Ohm. Itsu I calculate that at the point where you have 1 amp of current flowing,the CEMF will be 600mV lower than that of the source EMF Looking at your scope shot,at the point of 1 amp of current flow,the source EMF is at 12.6v. This means that your CEMF has a value of 12v. Brad --------------------------- Never let your schooling get in the way of your education.

gyula	Re: The Back emf misnomer « Reply #114 on: 2018-08-23, 16:55:31 »
Group: Tech Wizard Hero Member Posts: 1242	Thanks Itsu. The peak current is about 14 A and the battery voltage drops to about 9 V from the unloaded 12.6 V. The voltage drop across the internal battery resistance is 12.6-9 = 3.6 V so the battery resistance is about 3.5/14 = 0.25 Ohm. This should be added to the coil DC resistance: 0.6 + 0.25 = 0.85 Ohm so time constant, L/R is 0.0059/0.85 = 0.0069 second i.e. 6.9 ms. I indicated the switch-on moment in the scope shot as t=0 and drew the 1xL/R (6.9ms) and the 5xL/R (34.5ms) vertical lines to cut the exponential curve of the coil current. I agree with Brad CEMF calculation at the i = 1A coil current moment. I indicated the 1A coil current in a horizontal pink line and projected it up to cut the battery voltage. The measured and calculated values are well within the ballpark. Gyula ------------------------ The Back emf misnomer cemf 1.png (34.87 kB, 640x600 - viewed 416 times.)

Smudge	Re: The Back emf misnomer « Reply #115 on: 2018-08-23, 17:05:02 »
Group: Professor Hero Member Posts: 1953	Quote from: TinMan on 2018-08-23, 14:30:17 We were just trying to be clear as to the polarity of the CEMF,thats all. Now-next Which is correct ? 1-The CEMF is at it's greatest value at T=0 (moment the voltage is dropped across the coil) 2- The CEMF is NOT at it's maximum when the switch is thrown. it's at it's minimum, Brad 1 is correct. The moment the voltage is applied to the coil at switch closing the current (and hence the flux created by the current) starts to rise and the value of that flux rise (dPhi/dt) is at its greatest. The dPhi/dt always creates a CEMF of a value such that the sum (you must always sum the voltages as you go round a circuit) of the forward EMF from the battery and the CEMF from the coil equates to the voltage drop across the series resistance wherever that might be (could be in the battery, the coil or as an added resistor or all three in series). So at T=0 the initial current is 0, there is no voltage drop across the resistance and the CEMF equates in value to the supplied EMF. Thereafter the current rises in value while its rate of change (di/dt) and hence also the flux dPhi/dt and the CEMF fall, leading to the well known exponential waveforms. As T tends to infinity the current plateaus at a fixed value determined by the resistance, dPhi/dt tends to zero hence CEMF also goes to zero. Smudge

Itsu	Re: The Back emf misnomer « Reply #116 on: 2018-08-23, 17:21:00 »
Group: Elite Experimentalist Hero Member Posts: 4235	OK Guys, i have zoomed in on the switch close point and increased the vertical amplitude / div. value for yellow/blue. I also changed the current probe to 1A/div. and moved it out of the way of the yellow trace botom line. The blue cursors are the unloaded battery voltage versus the coil voltage at 1A time (480mV delta). Itsu ------------------------ The Back emf misnomer cemf1.png (26.74 kB, 640x600 - viewed 394 times.)

Smudge	Re: The Back emf misnomer « Reply #117 on: 2018-08-23, 19:03:35 »
Group: Professor Hero Member Posts: 1953	I have added the battery resistance and the coil resistance to Itsu's circuit. I also show the emf polarity arrows. V₁ is battery voltage, V₂ and V₃ are the voltage drops aross the two resistors and V₄ is the CEMF. As it's a closed circuit the voltages around the circuit must sum to zero. Thus V1-V2-V3-V4=0 (note the minus signs!). At any point in time the CEMF V₄ is easily calculated from Itsu's measurements where he scopes V3+V4 and also the current. At any point in time the magnitude of the CEMF V₄ is equal to Ldi/dt and also equal to NdPhi/dt (Phi is flux, not flux-linkage) but it is normal practice to show the counter polarity by including a minus sign. Smudge ------------------------ The Back emf misnomer cemf circuit.jpg (59.14 kB, 519x350 - viewed 318 times.)

TinMan	Re: The Back emf misnomer « Reply #118 on: 2018-08-24, 00:17:04 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Smudge on 2018-08-23, 17:05:02 1 is correct. The moment the voltage is applied to the coil at switch closing the current (and hence the flux created by the current) starts to rise and the value of that flux rise (dPhi/dt) is at its greatest. The dPhi/dt always creates a CEMF of a value such that the sum (you must always sum the voltages as you go round a circuit) of the forward EMF from the battery and the CEMF from the coil equates to the voltage drop across the series resistance wherever that might be (could be in the battery, the coil or as an added resistor or all three in series). So at T=0 the initial current is 0, there is no voltage drop across the resistance and the CEMF equates in value to the supplied EMF. Thereafter the current rises in value while its rate of change (di/dt) and hence also the flux dPhi/dt and the CEMF fall, leading to the well known exponential waveforms. As T tends to infinity the current plateaus at a fixed value determined by the resistance, dPhi/dt tends to zero hence CEMF also goes to zero. Smudge Thank you Smudge. « Last Edit: 2018-08-25, 13:33:37 by TinMan » --------------------------- Never let your schooling get in the way of your education.

forest

Re: The Back emf misnomer « Reply #119 on: 2018-08-24, 10:11:08 »

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Ok guys I cannot write much because I have here only Polish language and it continously correct je while typing. What I thought is the coil with pulse square wave input current probe between input and the coil and the measure of voltage across first turn of the coil. Third and fourth channels used to measure voltage across second turn od he same coil and a shunt resistor and measure current induced. In second turn there should be the current from the source minus the induced current. Maybe we can lewej something from this setup

Room3327

Re: The Back emf misnomer « Reply #120 on: 2018-08-25, 17:58:52 »

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Smudge, Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong. I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel with the source, it is actually in series with the source. Is this correct or am I still wrong? I really would like to come to a consensus on this matter for everyone to understand this stuff.

I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC. One reduces and one adds to the AC signal. I think? Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down.

Respectfully
Room

---------------------------

"Whatever our resources of primary energy may be in the future, we must, to be rational, obtain it without consumption of any material" Nicola Tesla

"When bad men combine, the good must associate; else they will fall one by one, an unpitied sacrifice in a contemptible struggle." Edmund Burke

TinMan	Re: The Back emf misnomer « Reply #121 on: 2018-08-26, 04:35:40 »
Group: Elite Experimentalist Hero Member Posts: 4727	Quote from: Room3327 on 2018-08-25, 17:58:52 Smudge, Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong. I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel with the source, it is actually in series with the source. Is this correct or am I still wrong? I really would like to come to a consensus on this matter for everyone to understand this stuff. I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC. One reduces and one adds to the AC signal. I think? Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down. Respectfully Room How close do you wish to look Room?. First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF. So,the CEMF has the same polarity across the inductor as that of the applied EMF, + to +,and - to -. Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0 Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect. At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor. The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged. This has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time. So now,at this point in time,our inductor is sending energy back to the source. This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer. But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time. Brad --------------------------- Never let your schooling get in the way of your education.

Smudge	Re: The Back emf misnomer « Reply #122 on: 2018-08-26, 16:02:35 »
Group: Professor Hero Member Posts: 1953	Quote from: Room3327 on 2018-08-25, 17:58:52 Smudge, Thank you for your input on this, so just to be clear on this, TinMan and I were both right and we were both wrong. I was wrong on the minimum and maximum CEMF/BEMF, had it backwards in my old memory and TinMan was wrong about the CEMF/BEMF being in parallel with the source, it is actually in series with the source. Is this correct or am I still wrong? I really would like to come to a consensus on this matter for everyone to understand this stuff. The CEMF/BEMF is both in series and in parallel so you are both right and wrong! Now that might sound daft but as always it depends on how you look at it and what you see. We usually draw circuits with the source on the left and the load on the right both standing up from the common ground along the bottom and connected across the top. Then you see them in parallel. But you could also draw it with both the source and the load across the top in series, with the LH and and RH end of this series both connected down to ground. Then you see them in series. From the parallel viewpoint the polarity arrows both point in the same direction (say upwards) and that is Tinman's view. From the series viewpoint the polarity arrows point in opposite directions. For solving a closed circuit where you wish to determine the current produced by the various EMF's you must follow the series viewpoint around the circuit and sum the various EMFs using those polarity arrows. Quote I am still not feeling right on the naming convention CEMF/BEMF because it happens on both current rise and current fall with AC and DC. One reduces and one adds to the AC signal. I think? Should they both be called the same name, I believe it is rather confusing? I don't know the answers, but again we should nail this stuff down. Personally I consider counter EMF and back EMF to mean the same thing, but Tinman has used references to suggest that they apply to different circumstances. I have suggested that simple resistance produces a "counter" or "back" EMF proportional to current in order to illustrate the offending series/parallel viewpoints. However the accepted wisdom is that these EMFs are proportional to the rate-of change of magnetic flux. (As an aside Mr Lenz doesn't help when he talks about induced current and not induced voltage.) Perhaps it would be better if we used a common term like MIEMF (magnetically induced emf) for all these situations, then we would not be wasting energy on arguing which term is correct. I don't think you can say that one reduces and one adds to the AC signal without specifying the circuit you are referring to. It is common in most AC circuits for both addition and reduction to happen at different parts of the cycle. Smudge

Room3327	Re: The Back emf misnomer « Reply #123 on: 2018-08-26, 18:53:13 »
Group: Experimentalist Hero Member Posts: 571	Quote from: TinMan on 2018-08-26, 04:35:40 How close do you wish to look Room?. First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF. So,the CEMF has the same polarity across the inductor as that of the applied EMF, + to +,and - to -. Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0 Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect. At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor. The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged. This has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time. So now,at this point in time,our inductor is sending energy back to the source. This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer. But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time. Brad Brad Thank you for your reply, yes I have a lot of years experience but that does not make me an expert on everything, I have never claimed to be an expert on anything except Islam. You rebooted this thread to get to the bottom of CEMF/BEMF and I was trying to help, I also would like to know the bottom line on this stuff, it's OK I can be wrong, but I bet many have a better handle on this now then they did, I'm not here for me, I am just trying to help. I meant no disrespect to you and you are an excellent experimenter. Room --------------------------- "Whatever our resources of primary energy may be in the future, we must, to be rational, obtain it without consumption of any material" Nicola Tesla "When bad men combine, the good must associate; else they will fall one by one, an unpitied sacrifice in a contemptible struggle." Edmund Burke

Smudge	Re: The Back emf misnomer « Reply #124 on: 2018-08-26, 19:08:03 »
Group: Professor Hero Member Posts: 1953	Quote from: TinMan on 2018-08-26, 04:35:40 How close do you wish to look Room?. First i must say the value of the EMF and CEMF are measured in volts,and volts shows us a polarity,where we have 0,ground,negative as a reference point to measure the value of the EMF. So,the CEMF has the same polarity across the inductor as that of the applied EMF, + to +,and - to -. From your viewpoint yes, but not from the viewpoint of electrons travelling around the circuit. It is from that other viewpoint that the words "back" and "counter" were used to describe the polarity. Quote Now,in relation to Smudges comment,and most accepted ,is this-->So at T=0 the initial current is 0 Depending on how close you wish to look,and how accurate you wish to be,the stated in blue is actually incorrect. It is correct for a pure inductor and the rate of rise from zero is given by di/dt=V/L Quote At T=0,there is an instantaneous rise in current which come before or at the rise of voltage across the inductor. The value of this current,and it's time duration depends on the configuration of the inductor it self. The value of this current at T=0 can be greater than half that of the steady state current through the coil after the 5th time constant. This instant rise in current is the self capacitance of the inductor being charged. Or any other shunt capacitance that might be there. It comes from the rate of rise of voltage and is given by i=C*dV/dt. Quote This has a side effect,that being the CEMF across the coil rises to a greater value than that of the source EMF. Just after the capacitance has been charged,the resultant CEMF that now has a higher value than that of the applied EMF makes the inductor now act as a negative resistance,where the inductor becomes the source for a short time. So now,at this point in time,our inductor is sending energy back to the source. In any LC circuit energy can slosh back and forth Quote This is the area i am researching ATM in another thread,where there is a tight capacitive coupling between the primary and secondary of a transformer. But like i said,it depends on who you wish to believe--those with many years of experience,or a hack(me) that messes around on the bench in his spare time. It's hacks like you that will find the odd behaviour that is often hidden behind the theories. Good luck with that. Smudge