2020_ATIVIDADE 1_ANDRY GABRIELE_resolvido

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Lista 1: - Andry Gabrielle 
1) 
𝑉𝐴 = 
𝑉𝐴
𝑉𝐵
𝑉𝐶
= [ 
120∠0°
380∠ − 90°
380∠90°
] (𝑉) 
 
 
𝑉𝑎0
𝑉𝑎1
𝑉𝑎2
=
1
3
∗ [ 
1 1 1
1 ∝ ∝ ²
1 ∝ ² ∝
] ∗ [ 
120∠0°
380∠ − 90°
380∠90°
] 
 
𝑉𝑎0 = 40 𝑉 
𝑉𝑎1 = 259,39 𝑉 
𝑉𝑎2 = 178,39 𝑉 
 
OU: 
 
𝑉𝐴0 =
1
3
∗ (𝑉𝐴 + 𝑉𝐵 + 𝑉𝐶) =
1
3
∗ (120 + 380∠-90° + 380<90°) = 40 V 
𝑉𝐴1 =
1
3
∗ (𝑉𝐴+∝ 𝑉𝐵+∝ ²𝑉𝐶) =
1
3
∗ (120 + (1∠-120°)*(380∠-90°) +(1∠-120°)*(380∠-90°) ) = 259,39 V 
𝑉𝐴2 =
1
3
∗ (𝑉𝐴 +∝2 𝑉𝐵+∝ 𝑉𝐶) =
1
3
∗ (120 + (1∠120°)*(380∠-90°) +(1∠-120°)*(380∠-90°) ) = 179,39 V 
 
2) 
 
𝐼0
𝐼1
𝐼2
= [ 
3,61∠146,31°
13,11∠26,80°
4,12∠ ∗ 71,61°
] (𝐴) 
 
 
𝐼0
𝐼1
𝐼2
= [ 
1 1 1
1 ∝ ∝ ²
1 ∝ ² ∝
] ∗ [ 
3,61∠146,31°
13,11∠26,80°
4,12∠ ∗ 71,61°
] 
𝐼0 = 10 𝐴 
𝐼1 = 12,05∠ − 94,76°𝐴 
𝐼2 = 18,98∠161,56° 𝐴 
 
𝐼0 = 𝐼0+ 𝐼1 + 𝐼2 = (3,61∠ − 146,31°) + (13,11∠26,80°) + (4,12∠ − 71,61°) = 10 𝐴 
𝐼0 = 𝐼0+ ∝ ²𝐼1 + ∝ 𝐼2 = (3,61∠146,31°) + ((1∠ − 120°) ∗ (13,11∠26,80°)) + ((1∠120°) ∗ (4,12∠ − 71,61°)) = 12,05∠ − 94,76°𝐴 
𝐼0 = 𝐼0+ ∝ 𝐼1 + ∝ ²𝐼2 = (3,61∠146,31°) + ((1∠120°) ∗ (13,11∠26,80°)) + ((1∠ − 120°) ∗ (4,12∠ − 71,61°)) = 18,98∠161,56° 𝐴 
 
3) V- F-F-F-F