LRC equation using Poynting theorem and conservation laws

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 4K views
emeriska
Messages
17
Reaction score
0

Homework Statement


We have an ordinary LRC circuit with inductance L, capacitance C and resistance R with an oscillating voltage with low frequency (U^e). Using the energy conservation law and Poynting's theorem, find the differential equation:

$$L \frac{\partial ^2}{\partial t^2}I + R \frac{\partial }{\partial t}I + L \frac{1}{C}I = \frac{\partial }{\partial t}U^e$$

Homework Equations


I'll need to take advantage from the fact that I is defined as
$$I = \frac{\partial}{\partial t}q$$.

Knowing q is the charge.

The Attempt at a Solution


Well, I've been looking around on the web to find something but I really can't find how to connect the Poynting theorem to that kind of circuit.

If any of you have some insights of a head start to give me that'd be great!
 
on Phys.org
Hi emeriska!

Your equation looks a lot like what I would get if I wrote down Kirchoff's 2nd law for the circuit and differentiated with respect to time. Except I think you have an extra factor of [itex]L[/itex] in the 3rd term?

Poynting's theorem would give you a nice 3d snapshot of the energy flow going on in and around the circuit.
 
Poything's theorem entails the following : ## \frac{\partial u_{tot}}{\partial t} = P_{tot} = P_{in} - P_{out}##
In our case, this reduces to ##\frac{\partial u_{E}}{\partial t} + \frac{\partial u_{B}}{\partial t}= P_{U^e} - P_R##

E energy in our circuit is only "produced" in the capacitor and B only in our inductor, power is produced by our voltage and dissipated by our resistance. We know that the energy in a capacitor is ##\frac{1}{2} \frac{q^2}{C}## and, for the inductor, ##\frac{1}{2} LI^2##. The powers are given by the Joule heath law as the following: ##P = IV = I^2R##.

All together, the Poyting proprety is ...
$$\frac{\partial }{\partial t}\left(\frac{1}{2} \frac{q^2}{C} + \frac{\partial}{\partial t}\frac{1}{2} LI^2\right)= IU^e - I^2R$$ where ##I = \frac{\partial q}{\partial t}## (replace the ##I##s in the left hand side only, then when the derivation is complete, rereplace the ##\frac{\partial q}{\partial t}## by ##I##)

Developing this should give what you are looking for !
 
Last edited:
Welcome to PF.

Elm8429 said:
Developing this should give what you are looking for !
Normally in the schoolwork forums we cannot provide solutions to student problems -- the student must do the bulk of the work. We provide hints, ask probing questions, find mistakes, etc.

But since this thread is from about 8 years ago, showing a solution is okay in this case. :wink:
 
Reply
  • Like
Likes   Reactions: WWGD and Elm8429