Analysing equations of inductance and capacitance

Thread Starter

kiroma

Joined Apr 30, 2014
40
Hello.
Ic=C*dv/dt comes from Q=CV.
V(Voltage across inductor)=L*di/dt comes from what?

I found that integral of (v*dt)=L*integral of (di)
Replacing v by R*I, and replacing I by dq/dt we have that
Integral of (R*dq/dt*dt)=L*integral of (di)
Integral of (R*dq)=L*integral of (di)
OBS.: limits are from 0 and 0 to q and i, respectively.
R*Q=L*I
L=RQ/I

So that means inductance is proportional to the resistance of the inductor?
I think I did a big mistake and a maybe a big confusion but, at least, I tried to find.
The question remains the same.
 

studiot

Joined Nov 9, 2007
4,998
Replacing v by R*I, and replacing I by dq/dt we have that
Integral of (R*dq/dt*dt)=L*integral of (di)
This doesn't work.

The ideal inductor has zero resistance so your second equation above, is nonsense. Where did you get the idea that v = ir in an inductive circuit?
 

MrAl

Joined Jun 17, 2014
7,757
Hello.
Ic=C*dv/dt comes from Q=CV.
V(Voltage across inductor)=L*di/dt comes from what?

I found that integral of (v*dt)=L*integral of (di)
Replacing v by R*I, and replacing I by dq/dt we have that
Integral of (R*dq/dt*dt)=L*integral of (di)
Integral of (R*dq)=L*integral of (di)
OBS.: limits are from 0 and 0 to q and i, respectively.
R*Q=L*I
L=RQ/I

So that means inductance is proportional to the resistance of the inductor?
I think I did a big mistake and a maybe a big confusion but, at least, I tried to find.
The question remains the same.

Hi,

If i understand you right i think you want to go back to Faraday which would lead to:
N*dPhi/dt=L*di/dt

for N the number of turns, Phi the magnetic flux, L the inductance, i the current.

or another interesting way to look at it is:
dPhi/di=L/N
 

Thread Starter

kiroma

Joined Apr 30, 2014
40
This doesn't work.

The ideal inductor has zero resistance so your second equation above, is nonsense. Where did you get the idea that v = ir in an inductive circuit?
Haha
That's why I thought I was doing a big mistake
Answering your question, sincerely, I don't know, haha.

Thank you MrAl.
 

Thread Starter

kiroma

Joined Apr 30, 2014
40
Hi,

If i understand you right i think you want to go back to Faraday which would lead to:
N*dPhi/dt=L*di/dt

for N the number of turns, Phi the magnetic flux, L the inductance, i the current.

or another interesting way to look at it is:
dPhi/di=L/N
So it means that N*dPhi/dt=Vl=L*di/dt?
Did Faraday found this?
 

MrAl

Joined Jun 17, 2014
7,757
Hi,

Faraday found:
E=N*dPhi/dt
or the signed form:
E=-N*dPhi/dt (i think attributed to Lenz)

L is the constant of proportionality that relates the flux to the current:
L=Phi/I

just as capacitance C is the constant of proportionality that relates the charge to the voltage:
C=Q/V

L has a physical origin just as C has a physical origin, both dependent on the geometry.

Another view is through the stored energy:
U=(1/2)*C*V^2
U=(1/2)*L*I^2

Inductance is the dual of capacitance, so often when you see voltage in a capacitor equation you'll see current in an inductor equation...as current is the dual of voltage.

L is the first letter of Lenz, in his honor (almost same equation as Faraday).

C is the first letter of Capacitor, in honor of Mr. Capacitor (just kidding) :)
 
Last edited:

Thread Starter

kiroma

Joined Apr 30, 2014
40
Hi,

Faraday found:
E=N*dPhi/dt
or the signed form:
E=-N*dPhi/dt (i think attributed to Lenz)
So it's correct to say that E=-N*dPhi/dt (as my professor said, when I wrote on the board the equation N*dPhi/dt=L*di/dt, "isn't a minus missing?!" Lol) and not E=N*dPhi/dt (incorrect)
Am I right?
 

MrAl

Joined Jun 17, 2014
7,757
Hi,

Yes, and i think it was Lenz who clarified the sign of the result, not Faraday, but dont quote me on this rather look it up on the web and maybe you can make a note here in this thread.
 

MrAl

Joined Jun 17, 2014
7,757
Hi,

Yes i see they put the little minus sign with the note "Lenz's Law" pointing to it so it modifies Faraday.
 
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