How does the voltage on the inductor change over time?

In summary: Well tan φ is -Lω/R, so the sinφ is -Lω/(R2+L2ω2).And I wrote 3*pi because one of the values I got for φ is pi.So I square the equations and add them together, and I get:R2sin2φ + 2RLωsinφcosφ + L2ω2cos2φ + R2cos2φ - 2RLωsinφcosφ + L2ω2sin2φ = V02/I02simplifying I get:R2 + L2ω2 = V02/I02and from there I get:I0=V0/sqrt(R2+L2
  • #1
SimpliciusH
33
0

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!
 
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  • #2
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
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
ehild said:
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.


ehild said:
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
SimpliciusH said:
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
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
ehild said:
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:
ehild said:
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
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
SimpliciusH said:
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
ehild said:
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
SimpliciusH said:
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)
 

Related to How does the voltage on the inductor change over time?

1. What is an inductor?

An inductor is a passive electronic component that stores energy in the form of a magnetic field. It consists of a coil of wire, typically made of copper, wrapped around a core material such as iron or ferrite.

2. How does AC current affect an inductor?

When an AC current flows through an inductor, it creates a magnetic field that constantly changes direction and strength. This causes the inductor to resist changes in current flow, leading to a lag in the current and voltage waveforms.

3. What is the relationship between AC current and inductance?

The amount of inductance in an inductor is directly proportional to the amount of AC current passing through it. This means that as the current increases, so does the inductance, and vice versa.

4. Can an inductor be used to filter AC current?

Yes, inductors can be used as part of a filter circuit to remove unwanted frequencies from an AC current. The inductor will block high-frequency signals and allow lower frequencies to pass through.

5. What is the difference between AC and DC current on an inductor?

The main difference between AC and DC current on an inductor is that AC current changes direction and strength, while DC current flows in a constant direction. This results in different behaviors of the inductor, such as storing energy in a magnetic field for AC and producing a constant magnetic field for DC.

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