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Registered Member #54278
Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367
Turning the page.....
DerAlbi wrote ...
1) VL = 0
I AM RESPONDING IN A SLOPPY MANNER NOW, IN ALL CAPS BECAUSE I HAVE ONLY A FEW SECONDS FOR THIS RIGHT NOW (PLEASE EXCUSE THE FORMAT) I WILL CLEAN LATER.
WHY should i use that? That is only true if the current is constant. Its called a steady state. And its actually expressed by VL = L*I/dt. -> if I is steady I/dt == 0. But its simply not true. The full truth is VL = L*I/dt. If all textbooks are wrong about Kirchhoff, i would like to know any textbook that sais that the voltage over a coil is unconditionally zero....
FARADAY'S LAW IS CONSIDERING THE INTEGRAL OF E•dl THROUGH THE INDUCTOR WHERE IT ASSUMED E=0
BTW: for the capacitor use VC = q/C, NOT Vc=I/C*dt ...I think you meant to write Vc=I*dt/C above
Sry, what is the difference between Vc=I/C*dt and Vc = I*dt/C ? 1/2*3 = 1.5. 1*3/2 = 1.5 And NO, even some americans tend to read it wrong, there is not bracket implied by the notation... if i want a bracket there, then i would have written it. (I assume you read it the **** american way like Vc=I/(C*dt). There is no reason for that - multiplication and division do have the same priority (Both over + and - and below the "power-of"-operator) *****WHAT A WASTE OF TIME*****
IF YOU DON'T WRITE THE CAPACITANCE EXPRESSION AS VC=Q/C HOW CAN YOU SET UP THE DIFFERENTIAL EQUATION??, THIS SORT OF QUESTION, BTW, IT IS BEGINNING TO MAKE IT LOOK LIKE MAYBE YOU HAVE NOT SOLVED THE APPROPRIATE DIFFERENTIAL EQUATIONS.
VC = q/C
...and q is the integral of the current over time, right? so... q = I*dt. you get Vc=I*dt/C ....or I/C*dt for that matter
I REALLY DON'T THINK ANY MATHEMATICIAN WOULD EVER EXPRESS DIFFERENTIALS IN SUCH A MANNER (ONLY THE SINGLE VARIABLE AS A DIFFERENTIAL IN AN EQUATION)
Now set up the equation using 1) and 2). What do you get? (post it--we can start here)
Sry to deny your request, i wont participate in such BS. 1) is wrong. 2) is true. Unfortunately mixing false and true adds up to false in this case.
HUMOR ME FOR NOW--DON'T SHOOT ME DOWN--YOU MAY LEARN SOMETHING THAT YOU DON'T EXPECT. I THINK IT WOULD BE GREAT FOR ALL TO WORK TOGETHER CONSTRUCTIVELY. YOUR ATTITUDE IS ONLY GOING TO HAVE A NEGATIVE EFFECT AND SHORTCHANGE YOU---AT TIMES YOU HAVE REALLY UNDERESTIMATED THE USERS HERE IF YOU NEED ME TO SET UP AND SOLVE ANY RLC-TYPE DIFFERENTIAL HERE I WILL BE GLAD TO.
Are you sure in the video the sum of the voltages was set to zero??
a few seconds before Endscreen.
IT'S THE SUM OF EACH COMPONENT'S INTEGRAL E•DL = -L DI/DT... LOOK AT ABOUT THE TIMESTAMP 6:05 ALSO LOOK AT 5:00
And pleeeeas that guy is kind of a dude who should not teach..... In his last example he charges the coil to steady state current so that there is no voltage drop across the inductor and uses that VL=0 as condition for non steady state analysis. The only thing that is true in that case is that dI/dt equals 0 which implies VL=0, but writing only Zero and neglecting the true mathmatical formula is... i can not tell you that in words. i actuallly can understand your confusion if you watch stuff like that.
DUDE...YOU'RE TALKING ABOUT THE FAMOUS PHYSICIST. WALTER LEWIN, OF MIT! YOU GLANCE ONCE AT A 15 MIN. YOUTUBE LECTURE OF A SUBJECT YOU MAY BE UNSURE OF AND DECIDE. quote: ""THAT IS THE KIND OF DUDE THAT SHOULD NOT TEACH""...CHILL MAN!
Again and forever: The voltage across a coil is determined by its inductance and the rate of current change only. If the current change happens to be zero - fine, coilvoltage is then zero, BUT that is in no way the general case and should NEVER be asumed in a non-steady state.
PLEASE USE SPELL-CHECK, WITH ALL THE RED ON THE SCREEN I CAN'T TELL MY BAD SPELLING FROM YOURS
Registered Member #135
Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735
I'm going to poke my head in just for a laugh here because I know it won't make a difference, but I thought I would have some fun with it.
So let's look at this guy for a moment
I grabbed this from Wikipedia. V1 + V2 + V3 + V4 = 0. And this is STEADY STATE, no instantaneous currents, and no switches thrown at t = 0+, and no phantom induced voltages from the aether or orgone fields. This is Kirchhoff's voltage law, and we know what to expect, the sum of the voltages is zero.
Let's also look at Ohm's Law for a second here, V = IR. So we have a "VOLTAGE DROP" across a resistance with a given current, just like in the example above. This we also know and have verified time and time again.
Both together also prove the sums of the currents in the loop equal zero.
Now let's also look at V = L di/dt. What does this mean? This means the Voltage across the inductor changes as the inductance times the "RATE CHANGE" of current over the "RATE CHANGE" of time. Meaning this current changes as the time changes, it is NOT time independent.
At some very large time t, the magnetic field stops changing as the current has stopped changing, as in a solenoid coil with a battery attached to form an electromagnet. We have all done this with a coil and a nail as children.
At this time, where t is very large, we can integrate, moving L outside of the integral, which gives us V = L (integral) di/dt, the INDEFINITE integral then gives us V = LI, much resembling V = IR our friend Ohm's Law, now how about that!
Now if we took out R2 and replaced it with L, Kirchhoff's summation loop would look something like this
V1 + V2 + V3 + V4 = 0
WOW! IT DIDN"T CHANGE! HOW SO! THERE MUST BE A TRICK!
No, because we replaced R2 with an inductor and TIME t = infinity.
RI (1) + LI (2) + RI (3) + V4 ( Battery) = 0
AND the sum of the currents around the loop also equal zero.
Kirchhoff's Law is restored.
Now go ahead and shoot holes through me, I already have my degree so it doesn't matter.
Registered Member #2906
Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727
At this time, where t is very large, we can integrate, moving L outside of the integral, which gives us V = L (integral) di/dt, the INDEFINITE integral then gives us V = LI, much resembling V = IR our friend Ohm's Law, now how about that!
Just a tiny but important hole to poke in L*I does not add up to Voltage if you consider the units. So that must be wrong
Lets consider Significations PDF (just one little part as an example)
Ok. Lets asume there is no electrical field between the inductor-leads...
Now: lets change the measurement setup. We create a constantly changing B-Field with a 1H-Inductor (superconducting) and 1V aplied. That will make a dI/dt of 1A/s rising to infinity. Within that B-Field we put a perfeclty coupled 1H Inductor and measure the voltage at the leads. But not with a resistive measurement. We just observe the electric field between the leads (there should be none according to the PDF) At the leads we connect a small capacitor so store the (non existing) electric field for later measurement.
So after we disconnect the capacitor after some time, what voltage will it have?
We all know it: The capacitor will be charged upto 1V, because all we have built is a 1:1 transformer. Applying 1V at the primary side will result in 1V at the secondary side. Since the voltage at the capacitor is constant (=1V) there is no current flowing. However: if there is an electric field between the capacitor-plates it will be also between the capacitor leads. And heeeey the capacitor leads are actually connected to the inductor leads. so somehow there is an electric field between the inductor-leads...
Registered Member #54278
Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367
Hazmatt_(The Underdog) wrote ...
I'm going to poke my head in just for a laugh here because I know it won't make a difference, but I thought I would have some fun with it.
At this time, where t is very large, we can integrate, moving L outside of the integral, which gives us V = L (integral) di/dt, the INDEFINITE integral then gives us V = LI, much resembling V = IR our friend Ohm's Law, now how about that!
Now go ahead and shoot holes through me, I already have my degree so it doesn't matter.
I think any reply here will make a difference--NO SHOOTING--I, for one, would like to really get all this straight, if possible, and welcome any opinion. -----------------------------------------
-------------- OK, Actually, you could take L from under the integral sign ANYTIME--it is constant. It appears you are 'solving' just V =L di/dt. If you consider, the addition of a voltage source and a resistance, say, V4 and R2 from the Wiki... (let's call them V and R) and apply Faraday's law, going around the loop and encountering the circuit elements: L, then, R, then V, in that order, the proper application of Faraday's law---which is :
the closed loop integral of E•dl = -L di/dt; yields the differential equation:
0 + IR - V = -L di/dt.
This is useful in TWO ways:
FIRST: rewriting this solution in the form: V - Ldi/dt = iR is very illustrative...this form shows how the time-dependent, Ldi/dt, after a long time period, vanishes, Leaving us with the steady state equation:
V = iR; Ohm' law, unchained.
************* + DerAlbi ***************** (YES, I agree units must check--In my option they can often be more important than magnitudes)
SECOND: We can see how the IMPROPER application of Faraday's law gives the RIGHT answer---how most books do it.
They wrongly assume Faraday's law is:
1) "The closed loop integral of E•dl = 0" (This is the special case of Faraday, called Kirchhoff that is valid when there is no changing flux) 2) the component of the INTEGRAL of E•dl through the inductor is L di/dt
Reworking the same problem above yields the same answer...but remember the right way and the wrong way...
RIGHT WAY: ---------------------------------------------
closed integral E•dl = -L di/dt inductor component INT E•dl = 0
WRONG WAY: --------------------------------------------
closed integral E•dl = 0 inductor component INT E•dl = L di/dt
I prefer getting the right answer the VALID way--NOT out of a misapplication due to misunderstanding that just happens to give the right answer in this case.
@ DerAlbi: Speaking of units revealing much information: You considered a 1H inductor (L = 1H) with 1A applied through it, and mention it 'rising to infinity' Aren't the units of a Henry (Volt-Sec/Amp)? I like to re-arrange the 'wording' of the units to a much more intuitive: One Volt per (Amp/Sec). In other words, your 1H inductor will see a rise of in potential of 1 Volt if di/dt = 1 Amp/s. I may be missing the complete picture here, but if di/di = 1 A/s and L = 1H, with a constant 1V applied, shouldn't L reach a steady state potential rise of 1V in 1s?? What does your 'infinity' refer to? I may be looking at something wrong as I feel I need to re-read your post. . . ...Was there a capacitor in the PDF circuit? Anyway, the E field of the capacitor is not IN the inductor.
Registered Member #2906
Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727
Speaking of units revealing much information: You considered a 1H inductor (L = 1H) with 1A applied through it
Noohh 1V applied!
Aren't the units of a Henry (Volt-Sec/Amp)? I like to re-arrange the 'wording' of the units to a much more intuitive: One Volt per (Amp/Sec). In other words, your 1H inductor will see a rise of in potential of 1 Volt if di/dt = 1 Amp/s.
Its not "a rise of" it is 1V (absolute). I think thats a difference.
I may be missing the complete picture here, but if di/di = 1 A/s and L = 1H, with a constant 1V applied, shouldn't L reach a steady state potential rise of 1V in 1s??
No. Only current rises. Voltage is simply there as soon as the coil is exposed to a changing B-Field.
What does your 'infinity' refer to?
The infinitiy only refers to the fact that if you apply a constant 1V on a superconducting 1H-Coil then the current will increase to infinity in infinite time with the rate of 1A/sec.
...Was there a capacitor in the PDF circuit? Anyway, the E field of the capacitor is not IN the inductor.
No there wasnt. But thats an interesting modification! The electric field is not only inside the capacitor! The potential difference is the same everywhere outside the capacitor. If the voltage between the plates is 1V, then the voltage between the leads is also 1V. Since those leads are connected to the coils leads, also the coils leads will have a potential difference of 1V. Therefore there will be a electric field at the coil even if there is no current flowing (cap voltage is constant 1V). That kind of contradicts the PDF in my opinion.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
I've been watching this thread, and while there seems to be some consensus regarding the 'steady state' (DC) stuff, if you apply Ohm's law to Der Albi's 'superconducting inductor', I=V/R gives infinite current, once the 'steady state' is reached, as R=0.
I don't have a degree, but I've known Ohm's law since I was a kid.
EDIT: Assuming the current source can supply infinite current.
Registered Member #54278
Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367
So, isn't it time to at least form a REAL circuit (even is the values are a bit unusual) then we can actually solve an RLC circuit: Let's suppose the capacitor, C, is the only energy *source*, similar to a (dry) coilgun circuit with Vo=1v.
So, from above we let: L=1H v(0) = 1v Now, if someone will kindly suggest any old R and C please? C=? R=?
The paragraph DerAlbi quotes, is obviously wrong. It says there, that the voltage measured by the voltmeter is not the potential drop across the inductor but instead a measure of the time rate of change of magnetic flux in the voltmeter circuit. If that would be the case, then would placing the voltmeter outside the field cause a zero reading? It would not. The electric field would be in the inductor, even if there is no voltmeter present. Dr. Slacks betatron example shows, that there are fields to e.g. accelerate electrons without any coils or voltmeters picking the field up. That is a consequence of Maxwells equations and Faradays law, which is part of them.
Signification wrote:
However, I have found it just as easy in most "everyday" circuit problems like we are discussing to use Faraday as opposed to Kirchhoff, even where Kirchhoff holds true. Since I learned the proper use and "respect the way", I tended to not use an "invalid" analysis, even though this established 'incorrect' method does give the proper answer--at the expense of thinking wrongly.
I tend to think of Kirchhoff to be a consequence of Faradays law. Exactly valid, if there are no changing fluxes and often a very good approximation, if there are. There are cases, where Kirchhoff is quite wrong, but there are also cases, where Faradays law has problems, e.g. in the quantum domain. So you say, Kirchhoff is invalid, I will do this correctly using Faradays law. Then I say, Faraday is incorrect, you must use quantumelectrodynamics and I won't consider using an "invalid" analysis. We could go on debating this forever. At one point, one has to say, we are accurate enough. Argueing in terms of final validity doesn't lead anywhere.
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All physics textbooks wrong in the setup and derivation of the RLC series circuit equation?
IF YOU NEED ME TO SET UP AND SOLVE ANY RLC-TYPE DIFFERENTIAL HERE I WILL BE GLAD TO.
L*I does not add up to Voltage if you consider the units. So that must be wrong 
