- #1

recon

- 401

- 1

Consider an infinite network of resistors of resistances [tex]R_0[/tex] and [tex]R_1[/tex] shown in the figure.

Show that the equivalent resistance of this network is

[tex]R_{eq} = R_1 + \sqrt{{R_1}^2 + 2R_1R_0}[/tex]

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter recon
- Start date

- #1

recon

- 401

- 1

Consider an infinite network of resistors of resistances [tex]R_0[/tex] and [tex]R_1[/tex] shown in the figure.

Show that the equivalent resistance of this network is

[tex]R_{eq} = R_1 + \sqrt{{R_1}^2 + 2R_1R_0}[/tex]

- #2

Justin Lazear

- 290

- 1

For an alternative view, replace all resistors to the right of the leftmost 3 (top, bottom, and middle) as a single resistor with resistance Req. Do you see why yuo can do that? Now, calculate the resistance of this new network in terms of Req, R0, and R1. Shouldn't the resistance of this network be equal to Req? Do you understand why this is also true? Use this to set up and equation and solve for Req.

--J

- #3

recon

- 401

- 1

Through the steps you outlined, I managed to derive the following:

[tex]R_{eq} = R_1 \pm \sqrt {{R_1}^2 + 2R_1R_0}[/tex]

However, since [tex] \sqrt {R_1(R_1 + 2R_0)} \geq R_1 [/tex], this means that [tex]R_{eq} = R_1 - \sqrt {{R_1}^2 + 2R_1R_0}[/tex] is not valid and must be rejected.

Hence, [tex]R_{eq} = R_1 + \sqrt {{R_1}^2 + 2R_1R_0}[/tex]

Share:

- Replies
- 3

- Views
- 287

- Replies
- 24

- Views
- 808

- Last Post

- Replies
- 6

- Views
- 293

- Replies
- 4

- Views
- 456

- Last Post

- Replies
- 7

- Views
- 814

- Replies
- 9

- Views
- 524

- Replies
- 12

- Views
- 377

- Last Post

- Replies
- 8

- Views
- 356

- Last Post

- Replies
- 0

- Views
- 275

- Replies
- 4

- Views
- 359