Really Neat Science Page
Really Neat Science Page
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http://www.williamsonlabs.com/480_rlcl.htm
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BTW I actually performed the experiment at the bottom of the page in my high school science lab around '61 or '62. We used 60 Hz mains current.
The inductor core was a bundle of iron wires. We used a series LC arrangement with the circuit tuned so that maximum current flowed with the core sitting on the table (about 1/2 way out of the coil).
http://www.williamsonlabs.com/480_rlcl.htm
*
BTW I actually performed the experiment at the bottom of the page in my high school science lab around '61 or '62. We used 60 Hz mains current.
The inductor core was a bundle of iron wires. We used a series LC arrangement with the circuit tuned so that maximum current flowed with the core sitting on the table (about 1/2 way out of the coil).
Engineering is the art of making what you want from what you can get at a profit.
What exactly is the derivative of an applied durrent? Current/1? Say for instance you have a current of 10 watts, what do you do to find the derivative of the current to find the counter EMF?This expanding Field induces a current (Counter E.M.F.) possessing an opposite polarity from that of the applied voltage. The amount of this counter E.M.F. is directly related to the derivative of the applied current,
Or is that just a fancy way of saying 'potential'?
Current is measured in lumens. Care to try again?EricF wrote:What exactly is the derivative of an applied durrent? Current/1? Say for instance you have a current of 10 watts, what do you do to find the derivative of the current to find the counter EMF?This expanding Field induces a current (Counter E.M.F.) possessing an opposite polarity from that of the applied voltage. The amount of this counter E.M.F. is directly related to the derivative of the applied current,
Or is that just a fancy way of saying 'potential'?
Engineering is the art of making what you want from what you can get at a profit.
d'oh. Had current and power mixed up. My Forrest Mimms book has current measured in amperes.MSimon wrote:Current is measured in lumens. Care to try again?EricF wrote:What exactly is the derivative of an applied durrent? Current/1? Say for instance you have a current of 10 watts, what do you do to find the derivative of the current to find the counter EMF?This expanding Field induces a current (Counter E.M.F.) possessing an opposite polarity from that of the applied voltage. The amount of this counter E.M.F. is directly related to the derivative of the applied current,
Or is that just a fancy way of saying 'potential'?
Still don't understand how the derivative fits in I understand the slope of a tangent line, but I don't know how that turns out in real world applications.
As is my won't  I was pulling your leg.
Forrest Mims is an excellent  "isn't this interesting and fun" introduction.
He is worthless for understanding. For that you need a text. Go to a used book store and find something. And get your math up to the point where you understand derivatives and integrals. You don't need to be good enough to calculate them (under most circumstances) but you have to know what they mean.
dI/dt = amps per second. And amps are coulombs per second. So a rate of change of current is coulombs/sec^2.
A current increasing at the rate of 1 amp per second would (if it started at zero) be a current of 1 amp at t= 1 sec. 2 amps at t= 2sec. etc.
Where it gets interesting is:
L dI/dt = V
and
C dV/dt = I
What does all that say? If the current is changing through an inductor (assume a zero resistance inductance) there will be a voltage across it. If it has a resistance the equation is:
L dI/dt + IR = V
If there is a changing voltage across a capacitance there will be a current through it.
for capacitance this is a good one to remember:
CV = It
Now when things are varying in odd ways the math gets some harder. But for most usual situations it has been worked out for you and you are left with algebra or trigonometry to get the results (to close enough).
Forrest Mims is an excellent  "isn't this interesting and fun" introduction.
He is worthless for understanding. For that you need a text. Go to a used book store and find something. And get your math up to the point where you understand derivatives and integrals. You don't need to be good enough to calculate them (under most circumstances) but you have to know what they mean.
dI/dt = amps per second. And amps are coulombs per second. So a rate of change of current is coulombs/sec^2.
A current increasing at the rate of 1 amp per second would (if it started at zero) be a current of 1 amp at t= 1 sec. 2 amps at t= 2sec. etc.
Where it gets interesting is:
L dI/dt = V
and
C dV/dt = I
What does all that say? If the current is changing through an inductor (assume a zero resistance inductance) there will be a voltage across it. If it has a resistance the equation is:
L dI/dt + IR = V
If there is a changing voltage across a capacitance there will be a current through it.
for capacitance this is a good one to remember:
CV = It
Now when things are varying in odd ways the math gets some harder. But for most usual situations it has been worked out for you and you are left with algebra or trigonometry to get the results (to close enough).
Engineering is the art of making what you want from what you can get at a profit.
To a rough approximation yes. It really means as delta x gets infinitesimally small. i.e. what is the instantaneous rate of change.EricF wrote:Waiiiit a minute. Does the d in dt or di stand for delta. As in, the triangle they use to denote 'change in'? :?
So a study of limit theorems in calculus if you want to get deeper into how the math is derived would help. But it is not absolutely essential.
Engineering is the art of making what you want from what you can get at a profit.
This means that as long as the field lines are expanding or contracting, they will induce an electric force in a nearby conductor (like a magnetic muscle ?) but if they remain static there is no effect on the neighboring conductor?If one were to place two Conductors side by side; one passing a steady Electric Current; the other wire is unaffected. Although there are Magnetic Flux lines "cutting" the second Conductor, No E.M.F. is generated since there is No Relative Motion.
However, if the Electric Current is made to vary in Magnitude and/or Polarity, then there would be an Induction of an E.M.F into the second Conductor, i.e., this variation/change in Magnetic Flux has the same Effect as Relative Motion between the Magnetic source and the Conductor.

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Hi EricEricF wrote:I'll have to ponder this for a while. I'll head out to the bookstore tommorrow. Is there anything specifically from Trig (I'm up to graphing the trig functions) that I need to focus on to understand derivatives and integrals?
thanks
I highly recommend "The Calculus Tutoring Book" bu Carol Ash
Excellent but low key. Like a good friend explaining calculus while writing and drawing on a sheet of paper.
Optimist: Glass is half full
Pessimist: Glass is half empty
Engineer: Someone made that glass twice
as large as it needs to be.
Pessimist: Glass is half empty
Engineer: Someone made that glass twice
as large as it needs to be.
Yes. But not exactly. Steady currents can develop a force with another conductor with a steady current (the motor principle). What they will not do is induce a current in a conductor (closed circuit) with no current.EricF wrote:This means that as long as the field lines are expanding or contracting, they will induce an electric force in a nearby conductor (like a magnetic muscle ?) but if they remain static there is no effect on the neighboring conductor?If one were to place two Conductors side by side; one passing a steady Electric Current; the other wire is unaffected. Although there are Magnetic Flux lines "cutting" the second Conductor, No E.M.F. is generated since there is No Relative Motion.
However, if the Electric Current is made to vary in Magnitude and/or Polarity, then there would be an Induction of an E.M.F into the second Conductor, i.e., this variation/change in Magnetic Flux has the same Effect as Relative Motion between the Magnetic source and the Conductor.
And it all works by superposition. In a conductor with a steady current an induced current will be additive (with the proper sign).
Engineering is the art of making what you want from what you can get at a profit.
An inductor opposes a change in current. If a constant current (no matter how large) is running through an ideal inductor, there is no "resistance" (actually reactance/impedance) and there is no voltage across the inductor. If the current changes slowly, there is a small "resistance" and small voltage across the inductor. When the current changes faster the "resistance" and hence voltage across the inductor gets larger.
As stated above, for an inductor the key equation is v = (L) x (di/dt). di/dt being the rate of change rate of change of current*. The faster current changes, and the steeper the tangent and the greater is (di/dt), the greater the voltage across the inductor.
To highlight the concept, imagine a lightning strike on a transmission line with a big inductor at the end of it. The strike causes a voltage and current surge to run down the line  but the current in the inductor cannot change instantaneously. When the surge comes upon the inductor, the momentum of the moving wave of charge causes a build up on the line side (by analogy like a wave against a pier) causing the voltage to rise on that side. The voltage on the other side of the inductor hasn't changed, so the voltage is contained across the inductor.
Edit:
* Simon's comment below. Note, treat the (di/dt) term as a single algebraic variable that you substitute a single value for  that is the rate of change of current. [I hope thats a reasonable way to put it.]
As stated above, for an inductor the key equation is v = (L) x (di/dt). di/dt being the rate of change rate of change of current*. The faster current changes, and the steeper the tangent and the greater is (di/dt), the greater the voltage across the inductor.
To highlight the concept, imagine a lightning strike on a transmission line with a big inductor at the end of it. The strike causes a voltage and current surge to run down the line  but the current in the inductor cannot change instantaneously. When the surge comes upon the inductor, the momentum of the moving wave of charge causes a build up on the line side (by analogy like a wave against a pier) causing the voltage to rise on that side. The voltage on the other side of the inductor hasn't changed, so the voltage is contained across the inductor.
Edit:
* Simon's comment below. Note, treat the (di/dt) term as a single algebraic variable that you substitute a single value for  that is the rate of change of current. [I hope thats a reasonable way to put it.]
Last edited by BenTC on Sun Mar 14, 2010 11:51 am, edited 1 time in total.
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