Linear approximation of a nonlinear component.

Click For Summary
SUMMARY

The discussion centers on the effective resistance of a nonlinear resistor (NLR) characterized by the equation IL = gVL² + I0. The user calculates the resistance as R = 1/(2gV0) using the derivative ∂g/∂V at V=V0 but questions the validity of the textbook solution, which states R = 1/2g. The user concludes that the coefficient g has units of amps per volt², leading to the assertion that the textbook answer is incorrect due to dimensional inconsistency.

PREREQUISITES
  • Understanding of nonlinear circuit components
  • Familiarity with calculus, specifically derivatives
  • Knowledge of electrical resistance and its units
  • Basic principles of circuit analysis
NEXT STEPS
  • Study the concept of effective resistance in nonlinear components
  • Learn about the application of derivatives in circuit analysis
  • Research dimensional analysis in electrical engineering
  • Explore advanced topics in nonlinear circuit theory
USEFUL FOR

Electrical engineers, circuit designers, and students studying nonlinear circuit analysis will benefit from this discussion.

peripatein
Messages
868
Reaction score
0
Hello,
I am trying to find the effective resistance of the NLR in the attachment (to the first order). It is given that IL = gVL2 + I0. I understand that this is normally achieved via ∂g/∂V at V=V0, but when I do so I get that R should be 1/(2gV0), and not 1/2g as shown in the solution. Could anyone please explain to me what it is I am doing wrong? Ought I to first determine V0 and then substitute it in 1/(2gV0)? But then, for my solution to be the same as that in the attachment, won't V0 have to be 1? I'd sincerely appreciate some guidance.
 

Attachments

  • Untitled4.jpg
    Untitled4.jpg
    5.6 KB · Views: 495
Engineering news on Phys.org
Your equation indicates that the coefficient g has units of amps per volt2
The reciprocal of this must have units of volt2 per ampere. This is not Ohms, nor Ohms-1.

Therefore, the textbook answer cannot be correct.
 

Similar threads

Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 1 ·
Replies
1
Views
940
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
10
Views
3K