# Transient Response of LR Circuit in AC

Discussion in 'Math' started by jjtjp, May 28, 2014.

1. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Mods, would you change title to: "Transient Response of LR Circuit in AC"? I couldn't get it to post with that title.

I'm taking AC circuit theory and trying to just brush up on my differential equations and math in general. Last night I tried to derive the transient response of an LR circuit, which starts as something like this:
$\frac{\mathrm{d} }{\mathrm{d} t}\left [ i(t)e^{\frac{R}{L}t} \right ]= e^{\frac{R}{L}t}\cdot V_{m}\cos(\omega t + \phi)$

and ends up something like this:
$i(t)=\frac{-V_m}{\sqrt{R^2+\omega^2L^2}}\cos(\phi-\theta)e^{\frac{-R}{L}t}+\frac{V_m}{\sqrt{R^2+\omega^2L^2}}\cos(\omega\, t+\phi-\theta)$

I know that you can do transforms to solve it, but I'm interested in the derivation without it. I know that you integrate by parts the right hand side twice then move the remaining integral from right to left. I've done this but I don't end up with anything in the denominator with a square root after reducing. If you know a site which has already worked through the problem step by step or mostly step by step, feel free to point it out. I couldn't find such a source. Thanks in advance!

EDIT: sorry, not sure why the one ω won't render right...

Last edited: May 28, 2014
2. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Your final line works for me but where does your first line come from?

At any instant the applied voltage is equal to the sum of the voltages across the resistor and the inductor

$v = iR + L\frac{{di}}{{dt}}$

Where i is the instaneous current.

further the instaneous current is the sum of the steady current and the transient current on switching.

$i = {i_s} + {i_t}$

When the transient has ceased the steady state current must still satisfy the first (differential) equation so

These equation holds for both direct and alternating currents.

$v = {i_s}R + L\frac{{d{i_s}}}{{dt}}$

also during the time when the transient exists

$v = \left( {{i_s} + {i_t}} \right)R + L\frac{{d\left( {{i_s} + {i_t}} \right)}}{{dt}}$

That is

$v = {i_s}R + L\frac{{di}}{{dt}} + {i_t}R + L\frac{{d{i_t}}}{{dt}}$

Hence

${i_t}R + L\frac{{d{i_t}}}{{dt}} = 0$

This is a simple variables separable equataion

$\int {\frac{{d{i_t}}}{t}} = \left( { - \frac{R}{L}} \right)\int {dt}$

With solution

$\int {\frac{{d{i_t}}}{t}} = \left( { - \frac{R}{L}} \right)\int {dt}$

So the complete solution for i is

$i = {i_s} + B\exp \left( { - \frac{{Rt}}{L}} \right)$

Where B is determined by the boundary conditions.

In this case the boundary conditions easier to calculate if you use a sine function, rather than a cosine function, because then i=0 at t=0.

Does this help or was there some other part of the calculation you missed?

Last edited: May 28, 2014
jjtjp likes this.
3. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Thanks for the reply. I can't give a full response right now, but my first equation comes from multiplying the entire equation by the integrating factor. Basically I am not sure how the integral gets into that final form. Is there something wrong with my original equation? Maybe I wasn't taught enough to know how to do this and I'm just way off base. That could be why they just gave us this equation, but I hate using things that I can't derive myself.

EDIT: I get that
$i_s = \frac{v_s}{\vec{Z}}$, but from pure math I just want to know how to get that integral.

Last edited: May 28, 2014
4. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
519
Sorry, which integral?

You haven't mentioned one from what I can see.

I can continue with the derivation for AC, if you like, to reach your equations (except with sines as I said) but what did you make of the story so far?

Last edited: May 28, 2014
5. ### Fibonacci New Member

May 23, 2014
25
5
Three things to mention:
First, the title must be "General response o simply Response of LR circuit in AC".
Second, your first equation need to divide by L because it comes from a non-homogeneous linear first-order differential equation acording to LKV in a LR circuit: L(di/dt)+Ri=Vmcos(ω*t+ψ), and it´s solution is given by the equation i(t)=e^(-Rt/L)[∫(Vm/L)e^(Rt/L)cos(ωt+ψ)dt+C] according to the well known theorem of this kind of equations using an integrating factor.
Solve the integral by parts, as you mentioned, and find out the next result:
i(t)=(VmωL/(R^2+ω^2*L^2))sin(ωt+ψ)+(VmR/(R^2+ω^2*L^2))cos(ωt+)+Ce^(-Rt/L).
Use the identity Acosβ+Bsinβ=sqrt(A^2+B^2)cos(β-θ), where θ=arctan(ωL/R), to convert the sum of sines and cosines into cosine alone. You will find the square root you were asking for. The equation is:
i(t)=(Vm/(sqrt(R^2+ω^2*L^2)))cos(ωt+ψ-θ)+Ce^(-Rt/L).
Third, you must obtain the value of the constant C with the initial-value problem: i(0)=0 (<< at t=0), and that´s all.

jjtjp likes this.
6. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Sorry, I figured it was obvious. I have a lot of notes in front of me and didn't think about the fact that no one else could see them.

Integrating both sides of my first equation gives...
$i(t)e^{\frac{R}{L}t}= \int e^{\frac{R}{L}t}\cdot V_{m}\cos(\omega t + \phi)\mathrm{dt}$

If you ignore the limits of integration to just get the anti derivative, that would be sufficient.

I'm really slow at math tex commands, but after performing parts twice I got:
$e^{\frac{R}{L}t}\frac{V_m}{\omega}\sin(\omega t + \phi)+\frac{V_mR}{L\omega^2}e^{\frac{R}{L}t}-\frac{R^2}{L^2\omega^2}\int e^{\frac{R}{L}t}V_m\cos(\omega t+\phi)\mathrm{dt}$

Adding the integral to both sides gave me:
$\left (\frac{R^2+L^2\omega^2}{L^2\omega^2}\right )\int e^{\frac{R}{L}t}V_m\cos(\omega t+\phi)\mathrm{dt}=e^{\frac{R}{L}t}\frac{V_m}{\omega}\sin(\omega t + \phi)+\frac{V_mR}{L\omega^2}e^{\frac{R}{L}t}$

Then:
$\int e^{\frac{R}{L}t}V_m\cos(\omega t+\phi)\mathrm{dt}=\frac{V_m\,e^{\frac{R}{L}t}( L^2\omega\sin(\omega t + \phi)+RL )}{R^2+L^2\omega^2}$

I probably have no clue what I'm doing.

7. ### jjtjp Thread Starter Member

Mar 3, 2014
30
0
Thanks Fibonacci. I guess you posted your answer while I was typing out mine and I never saw it. I just came back today to look at something else and noticed your response. Sorry I didn't see it before. Thanks for the help.