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How one does make sense of AC circuit's forced response?

  1. Feb 22, 2012 #1
    Hi people.

    I started my AC circuits course in college, but this stuff is either too tough or I probably don't have some math prerequisites.

    By the way, i'm just starting my differential equations course, but I haven't had any time to study just because AC circuits won't let me make any progress.

    So this is what happens, the first topic is all about the forced response of a circuit, but I can't make any sense of it

    Take a look at my first attachement "dontget.png"

    According yo my course's module or textbook, it says that the RL circuit's response I1coswt + I2sinwt . Just how do they know that?.

    And please take a look at the second attachement.

    I can understand where 12i and d2/dt2 came from, but I cannot understand where did di/dt came from.

    Besides, according to the textbook, the current's waveform "Obviously shoud be Ae^j3t." How is that even possible?, how do they anticipate the circuit's response?.

    Please help me people, i just can't understand this, and I've thought about cancelling the course, I just can't understand what's going in these circuits.

    By the way, this is no homework question, I just want to know how to make sense of this concept

    Attached Files:

    Last edited: Feb 22, 2012
  2. jcsd
  3. Feb 22, 2012 #2


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    This is one of the toughest things in your early college years. It is suppose to be difficult and it is a weed out course.

    Struggle through it like the rest of us. There are a few pages in your text book that describe this section. Immerse yourself in it and eventually it will start to click. I believe I spent roughly a week straight trying to figure this section out....including the entire weekend.

    You will also need to immerse yourself in your Diff EQ class. Also a weed out course....also suppose to be VERY difficult. You will also need to have your head in a book for weeks on end before this clicks as well. Put the booze, broads and friends aside if you expect to suceed.

    And yes...once you figure it out you will see all the math ties together perfectly in the circuits and diff EQ.....100% without exception.

    If you feel extremely nervous, shaken up and like you wanna puke right now....perfectly normal. We all went through it.

    Even if someone does give you an amazing explanation here....you are still going to have to work hard for weeks to get it completely.
    Last edited: Feb 22, 2012
  4. Feb 22, 2012 #3
    Yes psparky, I started reading this topic since last sunday, but I have to make hurry because I also have to read other courses' lessons and take some tests .

    I don't have too much ( or no) time because I'm working fulltime and pursing my major though distance learning, that's the reason why I can't struggle too much trying to figure out what's going on, if I can't get it by this weekend I think i'll just copy the equations for RL, RC and RLC oaralell circuits and move on to another topic.

    Differential equations seems easy ( at least the first ones, the physics and chemistry applied problems are very very tough) and I've already got the grip of the first ones ( Stewart's calculus book its so clear that makes calculus look like a joke) , but man this AC stuff really is beating me, and besides I haven't even done any DC circuit analysis in like 3 years but it's all about kirchoff law so it's not hard to remember.

    I didn't touch DC RLC topics when I was at the community college, so I guess that might be my weak point that's making me struggle in AC.

    After I can understand this response thing, I'll move to analysis with phasors which it's really easy .

    It's a shame that Boylestad's book says nothing about this.
    Last edited: Feb 22, 2012
  5. Feb 22, 2012 #4


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    Laplace transforms make this easier.....also you can think of it in terms of JωL and 1/(JωC)

    As soon as you put things in the S domain....and do KVL or KCL....all the suddent you see the differential equation perfectly.

    Even if you do the KVL or KCL without the S domain...you should see the differential equation. It's just a bit tougher without S domain.

    And yes, the phasors are much easier. Perhaps you should arrange a meeting with a more informed student....upper classman or make a meeting with your professor.
  6. Feb 22, 2012 #5


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    Come to think of it, this is the most difficult semester in your engineering college career. This was the semester I got my lowest GPA. Make it through this and you are pretty much home free.

    Keep this in mind as well....if you are struggling......90% of your fellow college students are struggling as well.

    Hopefully someone really intelligent will speak up shortly to perhaps steer you in the correct direction.
    Last edited: Feb 22, 2012
  7. Feb 22, 2012 #6


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    Looking at your circuit further....shaking out the cobwebs....

    The d2/dt2 is tricky to get. You obviously know the Cdv/dt=i(t) and that Ldi/dt=v(t).

    When you do a loop analysis like KVL and you are looking for the current for example....your loop equation looks like this

    Cdv/dt (current through capacitor) +....here's where it get tricky...you can't right the current through the inductor (unless integral) but you just write something like Cdv/dt again (because you know this equals the current)....except now you have to plug the v(t)=Ldi/dt into the Cdv/dt......taking the derivative of Ldi/dt....getting L*C*di2/dt2.

    Something like that.....look in your book....should give example. Sorry not exact....but best I can come up with at the moment.
    Last edited: Feb 22, 2012
  8. Feb 22, 2012 #7
    Problems of this kind assume: stationary state (all transients have decayed) and linearity of all elements. So if you put in a frequency, you get the same frequency out. The current may have a phase difference, but the frequency will always be the same, and the current will be sinusoidal. A complex A (Amplitude) or I1sin + I2cos are simply different ways of writing a phase-shifted sinusoidal current.
    In RLC circuits you often get an integral term (because voltage is prop. to charge = [itex]\int[/itex]i).
    By differentiating your equation you get a 2nd order differential equation. (This step isn´t documented in your attachment, but something like this might be missing)

    Don´t give up!
  9. Feb 22, 2012 #8


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    Yes...this too!

    At the time of this post....you've had 216 views and only 2 replies. That should also give you an idea of the difficulty of your question!!!!
  10. Feb 22, 2012 #9
    The way of thinking that calms me down goes like this.
    Inductors and Capacitors are some weird devices, which follows these (listed below) rules by virtue of their construction and nature of electrons/magnetic fields/electric fields/physics laws/etc
    Rule(1) : Inductors: Voltage across it (E) = L * rate of change of current through it (di/dt)
    Rule(2): Capacitors: Current through it (I) = C * rate of change of voltage across it (dv/dt)
    So, when these elements are put together in circuit, in various configuration and we put some source, we just define a bigger rule, like
    Rule(3): L*di/dt + V_capacitor + I*R = V_source*Sin(wt)
    i.e. what we are saying is,
    If inductor is to follow rule(1), capacitor is to follow rule(2), then our ckt, by logical reasoning must follow rule(3).

    Our task is to use some mathematical tools (methods) to re-define rule 3 in simpler terms, i.e. find the expression for I = f(t).
    (this is the differential equations part)

    I often view differential equations as, like
    10*di/dt + i = 5*sin(wt)
    Its purely a logical question, but we seldom use logic, but instead use some rote-learned techniques to find the ans.
    Edit: Sorry, I never saw your picture. Otherwise, I could use its example.
  11. Feb 22, 2012 #10
    Hi guys thanks for your replies and interest in this topic.

    I found this:

    http://www.virtual.unal.edu.co/cursos/ingenieria/2001601/docs_curso/contenido.html [Broken]

    If you can read spanish please have a look at chapter 6.

    Basically those are my circuits but in DC voltage levels instead of AC, however the equations look the same.

    I guess i'll have to check the response of a DC RLC circuit before jumping into AC.
    Last edited by a moderator: May 5, 2017
  12. Feb 22, 2012 #11


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    It's good you want to learn this...but I think the question you are asking is a fairly small part of AC. It's not really this bad. Eventually it's more about adding vectors in non resistive circuits and so forth. So don't get too dragged down....you'll be on to easier more interesing stuff soon.

    The vectors come in with the "J" in the Jwl and 1/Jwc. Reactances of inductors and capacitors. You will use the same rules to add reactances in series and parallel just like you did in AC...but now you obviously have vectors that are no longer at 0 degrees...so you need a little more math.

    No biggy...in the end......V=IR
    Last edited by a moderator: May 5, 2017
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