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AC current on a Inductor

  • #1

Homework Statement


An alternating current is running through a serially connected inductor(L) and resistor(R). The alternating voltage causing it is:
[URL]http://upload.wikimedia.org/math/5/a/0/5a0ecaa1432c6cdce653a943b4962a21.png[/URL]

How does the voltage on the inductor change over time?

Homework Equations


Kirchhoff's cricuit laws [tex]v_{g}=v_{L}+v_{R}[/tex]

The Attempt at a Solution



If I understand the text right, overall I should be searching for the [tex]v_{L}(t)[/tex] function right?

Ok so I know that
[tex]v_{L}=L \frac{dI}{dt}[/tex]

[tex]v_{R}=IR[/tex]

therefore

[tex]\frac{dI}{dt}+\frac{R}{L}I=\frac{1}{L}\frac{dv_{g}(t)}{dt}[/tex]

and this is where I get stuck. Thanks for any help!
 
Last edited by a moderator:

Answers and Replies

  • #2
ehild
Homework Helper
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It should be 1/L vg on the right-hand side of the final equation. Assume I(t) in the form of I0sin(ωt+φ) take the derivative and substitute it for dI/dt into the equation.

ehild
 
  • #3
Integrate the whole thing? Like this:

[tex]I_{0}\omega cos(\omega t + \varphi)+I_{0}\frac{R}{L}\frac{sin(\omega t + \varphi)-\omega t cos(\omega t + \varphi)}{\omega^2}=\frac{1}{L}V_{peak}\frac{R}{L}\frac{sin(\omega t)-\omega t cos(\omega t)}{\omega^2}[/tex]



Is this right?
 
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  • #4
ehild
Homework Helper
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vg is given: vg=Vpeaksin(ωt).
Apply the sum rule to sin(ωt+φ) and cos(ωt+φ). Collect like terms. Take into account that the equation holds for any time.

There is an easier way to solve the problem, using complex impedance or phasor diagram. Have you heard about them?

ehild
 
  • #5
vg is given: vg=Vpeaksin(ωt).
Apply the sum rule to sin(ωt+φ) and cos(ωt+φ). Collect like terms. Take into account that the equation holds for any time.

ehild

Is this comment meant for the last version of the response?

I edited my response extensively before I noticed your reply, I now realized I should have just made a new post, since this ultimately confused me and makes communication difficult. Sorry, won't do it again.


There is an easier way to solve the problem, using complex impedance or phasor diagram. Have you heard about them?

ehild
I've heard of them but haven't used either before.
 
  • #6
Integrate the whole thing? Like this:

[tex]I_{0}\omega cos(\omega t + \varphi)+I_{0}\frac{R}{L}\frac{sin(\omega t + \varphi)-\omega t cos(\omega t + \varphi)}{\omega^2}=\frac{1}{L}V_{peak}\frac{R}{L}\frac{sin(\omega t)-\omega t cos(\omega t)}{\omega^2}[/tex]



Is this right?
This is just so many kinds of wrong... :(

This is what I meant to write:

[tex]I_{0}sin(\omega t + \varphi) + \frac{RI_{0}}{L}(-\frac{cos{\omega t+\varphi}{\omega}= \frac{1}{L} V_{peak} sin (\omega t) [/tex]
 
  • #7
ehild
Homework Helper
15,478
1,854
Do not integrate anything. Use the sum rules:
sin(ωt+φ)=sin(ωt)cosφ+cos(ωt)sinφ and
cos(ωt+φ)=cos(ωt)cosφ-sin(ωt)sinφ.
Collect the terms with cos(ωt), and do the same with the terms containing sin(ωt).

ehild
 
  • #8
vg is given: vg=Vpeaksin(ωt).
Apply the sum rule to sin(ωt+φ) and . Collect like terms. Take into account that the equation holds for any time.
Did you mean the trig sum rule?


Edit:
Do not integrate anything. Use the sum rules:
sin(ωt+φ)=sin(ωt)cosφ+cos(ωt)sinφ and
cos(ωt+φ)=cos(ωt)cosφ-sin(ωt)sinφ.
Collect the terms with cos(ωt), and do the same with the terms containing sin(ωt).

ehild
It dawned on me a minute too late. :) Thank you for being so patient!
 
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  • #9
V0=Vpeak

So I go from:
LI0cos(ωt+φ)+RI0sin(ωt+φ)=V0sin(ωt)

To:

LI0cos(ωt)cosφ - LI0sin(ωt)sinφ + RI0sin(ωt)cosφ + RI0cos(ωt)sinφ = V0sin(ωt)

And collect the terms:

RI0sin(ωt)cosφ - LI0sin(ωt)sinφ - V0sin(ωt)+ RI0cos(ωt)sinφ + LI0cos(ωt)cosφ = 0


sin(ωt)(RI0cosφ - LI0sinφ - V0)+ cos(ωt)(RI0sinφ + LI0cosφ) = 0


sin(ωt)(Rcosφ - Lsinφ)+ cos(ωt)(Rsinφ + Lcosφ) = sin(ωt)V0/I0


Do I now find the φ for which this equation is true? V0 seems to be causing me problems there.
 
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  • #10
Ok I've simplified it to:

[tex]Lcos(t\omega + \varphi)+Lsin(t\omega + \varphi)=sin(t\omega + \varphi)\frac{V_{0}}{I_{0}}[/tex]
 
  • #11
ehild
Homework Helper
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You forgot that ω should be together with L.
The equation has to be true for any time t, that is for such values of t when sin(ωt)=0 (and cos (ωt)=±1) and also for such values when cos(ωt) =0 (and sin(ωt) =±1). This means that

Rsinφ + Lωcosφ=0 (*)

and

Rcosφ - Lωsinφ=V0/I0 (**)


You get φ from (*). Taking the square of both (*) and (**) and adding them up, the sinφ and cosφ terms cancel and you get an equation for I0.

ehild
 
  • #12
ehild
Homework Helper
15,478
1,854
Ok I've simplified it to:

[tex]Lcos(t\omega + \varphi)+Lsin(t\omega + \varphi)=sin(t\omega + \varphi)\frac{V_{0}}{I_{0}}[/tex]
That is wrong. Go back to your previous post. I answered it.

ehild
 
  • #13
You forgot that ω should be together with L.
The equation has to be true for any time t, that is for such values of t when sin(ωt)=0 (and cos (ωt)=±1) and also for such values when cos(ωt) =0 (and sin(ωt) =±1). This means that

Rsinφ + Lωcosφ=0 (*)

and

Rcosφ - Lωsinφ=V0/I0 (**)


You get φ from (*). Taking the square of both (*) and (**) and adding them up, the sinφ and cosφ terms cancel and you get an equation for I0.

ehild

Ok I get 2*pi+pi for the value of φ

and V0/(R2+L22*cos(2*φ) for I0)
 
  • #14
I then assume put this back into the equation for VL, and derive it, I get:

V0*L/(R2+L22*cos(2*φ) times ωcos (ωt + φ)

or should I say times -ωcos (ωt) considering the value of φ.

Is this right?
 
  • #15
ehild
Homework Helper
15,478
1,854
Ok I get 2*pi+pi for the value of φ

and V0/(R2+L22*cos(2*φ) for I0)
No. φ is not 2*pi + pi, and why do you write 3*pi in this way?

Rsinφ + Lωcosφ=0. What is tanφ then?

To get an expression for I0 square the equations

Rsinφ + Lωcosφ=0
and
Rcosφ - Lωsinφ=V0/I0

and add them together. Take care of the + and - signs.

ehild
 
  • #16
SimpliciusH you need to practice trigonometry.
I recommend Trigonometry by S.L. Loney to you (It's a must for you)
 

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