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LAN_403

Signification

Wed Jun 03 2015, 11:05PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

I was trying to write some proofs last night that involved the old rules:
Current through an inductor lags the voltage across it by 90 degrees--vise-versea for capacitors.

For capacitors, I started with
1) Q=CV and
2) i = dQ/dt
3) Finally: i(t) = C*dV/dt

For inductors, (Let's call magnetic flux "F"), I started with
1) F is proportional to i
2) F = L*i (L is the prop. constant)
3) V = dF/dt; let "V" = EMF (Faraday's law)
4) V= -L*di/dt (applying Lenz's law)

We know that current lags voltage in an inductor by 90 degrees. So, suppose, keeping things simple:
i(t) = sin(t)
then V = -L * d/dt [sin(t)]
Which gives voltage across the inductor as:
V = -L*cos(t)
Since L cannot be negative AND sin LEADS cos, the conclusion seem to be inductor current LEADS voltage!! BUT it's actually the other way around!

Can anyone give a proper and rigorous mathematical proof? (I think it has something to do with the "-" sign called for by Lenz's law---BUT, it can't be ignored---this is how all things work, which the mathematics must reveal!!

DerAlbi

Wed Jun 03 2015, 11:45PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

University teacher incomming

Reformulate your equations and decide if you want to use sin(t) or cos(t).
If you decide for sin(t) then transform every cos(t) into sin(t+90Â°) to clarify your problem.
This tells you exactly who leads or lags.
Additionally remember that -L*cos(..) is the same as L * -cos(...).
And -sin(t) = sin(t +/- 180Â°). (same for -cos(t)=cos(t +/-180Â°)

That would lead to V = L * sin(t - 90Â° + 180Â°)
(the 90 is the conversion from cos to sin and the 180 is the negative sign)

That gives you a V=L*sin(t + 90Â°) from a current that was sin(t).
See, the Voltage is ahead of the current (+90Â°) thats what you wanted.
(or: the current is lags the voltage by 90Â°. doesnt matter how you formulate it)

And btw: flux is the capital greek letter "phi", not F.

Edit: Stupid Albi. I Said....

every cos(t) into sin(t+90Â°)

and then i did

sin(t - 90Â° + 180Â°)

Of course thats wrong.

It directly contradicts the statement above. sry. The correct transformation would be sin(t+90Â° - 180Â°) and then the phases are mixed up and we start wondering again. Down below i tell you maybe why...

Signification

Thu Jun 04 2015, 11:12AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

@DerAlbi:
I don't quite get it! In my message I derive:
1) V = -L*cos(t) {I'm expecting V= +L*cos(t)}
Then you state a few --useful-- identities, and write:
2) V = L * sin(t -90 +180)
Which is already IDENTICAL to: V = +L*cos(t)
You can't just PUT a "+" where you don't want a "-".

BTW: I know flux is represented by upper case "phi", but how do you type it?

mister_rf

Thu Jun 04 2015, 11:38AM

Registered Member #4465 Joined: Wed Apr 18 2012, 08:37AM
Location: Bucharest, Romania
Posts: 145

List of Alt Codes for entering Greek Letters

Alt 224 Î± Alpha
Alt 225 ÃŸ Beta
Alt 226 Î“ Gamma
Alt 235 Î´ Delta
Alt 238 Îµ Epsilon
Alt 233 Î˜ Theta
Alt 227 Ï€ Pi
Alt 230 Âµ Mu
Alt 228 Î£ Uppercase Sigma
Alt 229 Ïƒ Lowercase sigma
Alt 231 Ï„ Tau
Alt 232 Î¦ Uppercase Phi
Alt 237 Ï† Lowercase Phi
Alt 234 Î© Omega

DerAlbi

Thu Jun 04 2015, 01:03PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

4) V= -L*di/dt

maybe that minus here is wrong. it depends on what you want to express.

generally speeking: I = 1/L * integal( V ) dt
There is no negative sign involved. rewriting it yields V = L * I/dt.

Edit: i am sorry for the above statement. I made a mistake. Plase see the above Edit.

Uspring

Thu Jun 04 2015, 03:16PM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Signification wrote:

3) V = dF/dt; let "V" = EMF (Faraday's law)
4) V= -L*di/dt (applying Lenz's law)

3) is correct, but you mustn't apply Lenz's law here. It's just Faraday's law. Lenz's law would apply if there is some external source creating the flux change. It would create a voltage in a loop according to Faraday's law. If you would short circuit this loop, there would be a current opposite to that of the current generating the field.

It makes a difference, if you apply a voltage to a loop (your case) or if you short circuit a loop that contains flux from another source, even though the voltages may be the same. In this case, the current is reversed, just as Lenz's law requires.

Signification

Mon Jun 08 2015, 10:18AM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

Thanks, guys
This time, in attempt to make everything more easily understandable, I will, somewhat rigorously, derive the instantaneous voltage across a capacitor and the instantaneous current through an inductor:

CAPACITOR
-----------
q prop to v
q=Cv; C>=0
i(t)=dq/dt
i(t)=C dv/dt

INDUCTOR
----------
Î¦ prop to i; where Î¦ is magnetic flux
Î¦=Li; L>=0
v(t)=d/dt (Î¦)
v(t)=Ldi/dt

(moving from differential to integral operations) INTEG represents 'the integral of':

INTEG{dv} from v(t) to v(o) = 1/C INTEG{i(t)dt}
(v)t) - v(0) = 1/C INTEG{i(t)dt}; letting v(0)=Vo,
thus:
=======================
v(t) = 1/C INTEG{i(t)dt} + Vo
=======================

INTEG{di}; from i(t) to i(0) = 1/L INTGE{v(t) dt}
i(t) - i(0) = 1/L INTEG{v(t)dt}; letting i(0)=Io
thus:
=====================
i(t) = 1/L INTEG{v(t)dt} + Io
=====================

@mister rf: Thanks for the extended ASCII printables--I didn't realize the 'alt' trick worked post-DOS.
@Albi: You now see Î¦ (forget to look up the "prop to" code)

@Uspring: You say a lot in few words! So I guess Lenz only kicks in when external actions are acting.

So I wonder now---when solving the second order 'textbook' differential equation for the RLC series circuit (with no external energy source--just a non-zero capacitor Vo), is Lenz's law UN-applicable here?

DerAlbi

Mon Jun 08 2015, 12:01PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

i would have been satisfied if you had written "phi"

The textbook euqations are in short:
Cap: dV = I*dt / C and I = C * dV/dt
Ind: dI = V*dt / L and V = L * dI/dt
Res: R = V/I
As you mentioned in your post, which is correct.
The "*dt"-noatation is the integral and "dX/dt"-notation is the deriviation. (thats a little bit easier to read)

The question is now: what do you want with the Lenz Law? The RLC-Circuit is completely defined by the above equations.. so what is gained by Lenz?

Lenz is more a description of why an inductor behaves the way it does including inductor coupling and stuff.
Lenz says very basically that the cause is compensated by its effects.
The slow rising current of an Inductor is a cause of this law, but allready covered in "dI = V*dt/L"

Signification

Mon Jun 08 2015, 12:26PM

Registered Member #54278 Joined: Sat Jan 17 2015, 04:42AM
Location: Amite, La.
Posts: 367

DerAlbi wrote ...

The question is now: what do you want with the Lenz Law? The RLC-Circuit is completely defined by the above equations.. so what is gained by Lenz?

I have this gut feeling that you want to know "what I want with Lenz's law WRT coilguns"

OAN: I think I saw where there is an ASCII print for the integral sign.

DerAlbi

Mon Jun 08 2015, 02:58PM

Registered Member #2906 Joined: Sun Jun 06 2010, 02:20AM
Location: Dresden, Germany
Posts: 727

No, no, i really wanted to know what you want tot do with an RLC-Circuit where every element is allready described.. you would add another equation.. to do.... what?

But since you statet the "WRT coilgun" keyword, i would like to know what "WRT" stands for... is the "R"... maybe "R"eluctance ?

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