Plotting a RLC Circuit: Analyzing a Differential Equation

In summary: The I_p and V_p are the rms values over a period of time.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.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?No, the value is in hertz.Frequency is in hertz or kilohertz so ω = 2*pi*f.I'm not sure what your comment means
  • #1
rmiller70015
110
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:
Engineering news on Phys.org
  • #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.

upload_2017-4-14_23-40-18.png


A more fundamental equation would be:

upload_2017-4-14_23-41-21.png


Frequency is in hertz or kilohertz so

upload_2017-4-14_23-43-10.png


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.
 
  • Like
Likes rmiller70015
  • #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?
 
  • #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:

upload_2017-4-15_0-37-14.png


upload_2017-4-15_0-37-31.png
 
  • Like
Likes rmiller70015
  • #5
rmiller70015 said:
First of all, the book says i=⃗I=Ipejωti=\vec{I} = I_pe^{j\omega t} and v=⃗V=Vpejωtv = \vec{V} = V_pe^{j\omega t}. Are the IpI_p and VpV_p the rms current and voltage?
No, they represent the amplitude (the peak value).
 
  • Like
Likes rmiller70015

Similar threads

Replies
6
Views
380
Replies
8
Views
1K
Replies
22
Views
1K
Replies
1
Views
1K
Replies
5
Views
2K
Replies
3
Views
1K
Replies
13
Views
769
Replies
5
Views
2K
Back
Top