Electrónica Aplicada 2

Jesús Alfonso López Lomelí

Tarea 4

∆ = 1 +2

=2

∆ − 1

=

=2

∆ − 1

∆ = 1.5

=2

1.5 − 1 = 4

Si = 1Ω

= 4Ω

∆ = 3

=2

3 − 1 =

Si = 1Ω

= 1Ω

∆ = 10

=2

10 − 1 =

2

9

Si = 9Ω

= 2Ω