Simple Electrical Modeling Question (w/ only C and L)

Click For Summary
SUMMARY

The discussion focuses on deriving a differential equation for a circuit involving a capacitor (C) and an inductor (L) using the variables vin and vout. The derived equation is C(d²vout/dt²) + vout = LC(d²vin/dt²), which is confirmed as valid by referencing the rule of calculus that states the derivative of a sum equals the sum of the separate derivatives. The participant expresses uncertainty about the legality of breaking up the capacitor derivative but receives confirmation that their approach is correct.

PREREQUISITES
  • Understanding of differential equations in electrical circuits
  • Knowledge of capacitor and inductor behavior in circuits
  • Familiarity with calculus, specifically derivatives
  • Basic circuit analysis techniques
NEXT STEPS
  • Study the derivation of differential equations in RLC circuits
  • Learn about the Laplace transform for circuit analysis
  • Explore the application of Kirchhoff's laws in circuit modeling
  • Investigate the behavior of second-order differential equations in electrical systems
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in modeling electrical circuits using differential equations.

Bluestribute
Messages
192
Reaction score
0

Homework Statement


Find a differential equation that describes the circuit that includes only vin and vout as variables.

Homework Equations


C(dvin - dvout)/dt = i
vin - vout = L(di/dt)

The Attempt at a Solution


So the answer I got was:

C(d2vout/dt2) + vout = LC(d2vin/dt2)

My question is, is it allowable to break up the capacitor derivate like that? After substituting in for i, I got:

vout = Ld/dt(C(dvin - dvout)/dt)

Then I broke up the capacitor into vin and vout terms and just rearranged. But I'm not 100% sure if that's legal.
 

Attachments

  • Untitled.jpg
    Untitled.jpg
    7.6 KB · Views: 412
Physics news on Phys.org
Bluestribute said:
So the answer I got was:

C(d2vout/dt2) + vout = LC(d2vin/dt2)
That first C should be LC.

A rule of the Calculus says that the derivative of a sum equals the sum of the separate derivatives, so your approach is correct.
 
  • Like
Likes   Reactions: Bluestribute
Woopsie, right. LC

Ok sweet, thanks. I just can't remember something we might have glossed over a few years ago . . . and no one recently has said "You can do this to a derivative". They just sort of do it on their own.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K