How does the voltage on the inductor change over time?

[SimpliciusH]

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 v_{g}=v_{L}+v_{R}

The Attempt at a Solution

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

Ok so I know that
v_{L}=L \frac{dI}{dt}

v_{R}=IR

therefore

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

and this is where I get stuck. Thanks for any help!

 

[ehild]

It should be 1/L v_g 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

 

[SimpliciusH]

Integrate the whole thing? Like this:

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}
Is this right?

 

[ehild]

v_g is given: vg=V_peaksin(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

 

[SimpliciusH]

v_g is given: vg=V_peaksin(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.

 

[SimpliciusH]

Integrate the whole thing? Like this:

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}
Is this right?

This is just so many kinds of wrong... :(

This is what I meant to write:

I_{0}sin(\omega t + \varphi) + \frac{RI_{0}}{L}(-\frac{cos{\omega t+\varphi}{\omega}= \frac{1}{L} V_{peak} sin (\omega t)

 

[ehild]

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

 

[SimpliciusH]

v_g is given: vg=V_peaksin(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!

 

[SimpliciusH]

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.

 

[SimpliciusH]

Ok I've simplified it to:

Lcos(t\omega + \varphi)+Lsin(t\omega + \varphi)=sin(t\omega + \varphi)\frac{V_{0}}{I_{0}}

 

[ehild]

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

 

[ehild]

Ok I've simplified it to:

Lcos(t\omega + \varphi)+Lsin(t\omega + \varphi)=sin(t\omega + \varphi)\frac{V_{0}}{I_{0}}

That is wrong. Go back to your previous post. I answered it.

ehild

 

[SimpliciusH]

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+L²*ω²*cos(2*φ) for I₀)

 

[SimpliciusH]

I then assume put this back into the equation for VL, and derive it, I get:

V₀*L/(R²+L²*ω²*cos(2*φ) times ωcos (ωt + φ)

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

Is this right?

 

[ehild]

Ok I get 2*pi+pi for the value of φ

and V₀/(R²+L²*ω²*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

 

[Deric Boyle]

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