Tarea4 Ea2
4
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
Transcript of Tarea4 Ea2
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
