Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

RLC circuits

  1. Apr 14, 2017 #1
    I'm not sure about the physical behavior of a RLC circuit and I have to give a presentation that involves one. So I've decided to plot the current. I found a book that gives a differential equation to describe the circuit.

    ##L\frac{d^2i}{dt^2} + R\frac{di}{dt} + \frac{1}{C}i = \frac{dv}{dt}##

    So I had some questions about this equation and the notation.

    First of all, the book says ##i=\vec{I} = I_pe^{j\omega t}## and ##v = \vec{V} = V_pe^{j\omega t}##. Are the ##I_p## and ##V_p## the rms current and voltage?

    Also, they use R as resistance, but should I also include the impedance due to the capacitor and the inductor here in addition to the resistance of the resistor? If so, how should I treat the resistance since I'm not aware of a way to express it as a non-real value.

    Finally, if I am driving this with a frequency generator set at 277 kHz, should this value be ##\omega## or would I need to multiply this by ##2\pi##?
    Last edited: Apr 14, 2017
  2. jcsd
  3. Apr 14, 2017 #2
    Here's a series RLC circuit connected to an ac voltage source. I normally use Vrms cos (wt) as a voltage source rather than the exponential form.


    A more fundamental equation would be:


    Frequency is in hertz or kilohertz so


    I'm not sure what your comment means about the resistance and the non-real value. If I understand what you are asking, yes, you need to include the three impedances - the resistance, the inductive reactance and the capacitive reactance. The last two quantities include w(omega) in their calculation, so you will need to multiply the frequency by 2*pi.
  4. Apr 14, 2017 #3
    The impedance due to the inductor and capacitor appear to be the same quantity as the reactances but with a multiple of ##\sqrt{-1}##. So I wasn't sure if what to do with the equation since the book wanted to use imaginary values in the current as well as the real values. But it appears that you are only looking at the real portion, is there any reason why you omitted the imaginary part, is it just extra stuff that isn't necessary to understanding the behavior?
  5. Apr 14, 2017 #4
    I didn't mean to ignore the imaginary impedances. I was just trying to show you where the differential equation came from. What I posted follows the serices circuit that I gave you. If you are doing a steady-state rms calculation, you'll need R plus the next two. Here they are:


  6. Apr 16, 2017 #5


    User Avatar
    Science Advisor

    No, they represent the amplitude (the peak value).
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted