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For the given nonlinear circuit, ????x = [???? ????x−1]/ 10 in which the units of ????x and ????x are V and A respectively. The source voltage is 3V and the source resistance is 1Ω.
Use graphical load-line techniques to solve for ????x and ????x. [Use a computer program to plot the characteristic curve and load line.]

Answer :

[tex]V_{x}[/tex] = 0.75 V, [tex]I_{x}[/tex]= 0.56 A

Explanation:

Given data:

[tex]V_{s}[/tex] = 3 V

[tex]R_{s}[/tex] = 1 Ω

[tex]I_{x}[/tex] = [tex]\frac{3V_{X} }{4}[/tex]

Solution:

Using Kirchhoff's voltage law.

The law states that the algebraic addition of potential difference in loop is similar to zero.

[tex]V_{s}[/tex] - [tex]I_{x}[/tex][tex]R_{s}[/tex] - [tex]V_{x}[/tex] = 0

[tex]V_{s}[/tex] - [tex]I_{x}[/tex][tex]R_{s}[/tex] = [tex]V_{x}[/tex]

3 - [tex]\frac{3V_{X} }{4}[/tex] × 1 = [tex]V_{x}[/tex]

12 - 3[tex]V_{x}[/tex]  = 4

3 - 3[tex]V_{x}[/tex]  =

[tex]V_{x}[/tex] = 3/4

[tex]V_{x}[/tex] = 0.75 V

[tex]I_{x}[/tex] = [tex]\frac{3V_{X} }{4}[/tex]

[tex]I_{x}[/tex] = [tex]\frac{3}{4}[/tex] × 0.75

[tex]I_{x}[/tex] = [tex]\frac{2.25}{4}[/tex]

[tex]I_{x}[/tex] = 0.56 A.

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