Zasilanie z transformatora B

I = I(cosφ – jsinφ)

cosφ=0,9 sinφ=0,44

I₄₅ =42 [A]

I₃₄ = I₄ + I₄₅ = 42 +47 =89 [A]

I₂₃ = I₃ + I₃₄ = 47 +89 =136 [A]

I₁₂ = I₃ + I₃₄ = 57 +136 =193 [A]

I_B1 = I₁ + I₁₂ = 44,75 +193 =237,75 [A]

I = I(cosφ – jsinφ)

I₄₅ = I₄₅(cosφ – jsinφ)=42(0,9 - j0,44)=37,8 - j18,48 [A]

I₃₄ = 80,1 – 39,16 [A]

I₂₃ = 122,4 - j59,84 [A]

I₁₂ = 173,7 – j84,92 [A]

I_B1 = 213,975 - j104,61 [A]

Części czynne i bierne prądów :

- części czynne: - części bierne:

I′₄₅ = 37,8 [A] I″₄₅ =18,48 [A]

I′₃₄ = 80,1 [A] I″₃₄ = 39,16 [A]

I′₂₃ = 122,4 [A] I″₂₃ = 59,84 [A]

I′₁₂ = 173,7 [A] I″₁₂ =84,92[A]

I′_B1 = 213,975 [A] I″_B1 =104,61 [A]

Spadki napięć na elementach sieci:

ΔU₄₅ = √3·(37,8·0,44·8,5+18,48·0,4·8,5)=353,27 [V]

ΔU₃₄ = √3·(80,1·0,44·7,2+39,16·0,4·7,2)= 634,11[V]

ΔU₂₃ = √3·(122,4·0,44·4+59,84·0,4·4)= 538,3[V]

ΔU₁₂ = √3·(173,7·0,44·4+84,92·0,4·4)= 763,94[V]

ΔU_B1 =√3·(213,975·0,44·5,5+104,61·0,4·5,5)= 1293,97[V]

Napięcia u poszczególnych odbiorców:

U₁ = 15000 - ΔU_A1 = 13706,03 [V]

U₂ = 15000 - ΔU_A1 - ΔU₁₂ = 12942,09[V]

U₃ = 15000 - ΔU_A1 - ΔU₁₂ - ΔU₂₃ = 12403,79[V]

U₄= 15000 - ΔU_A1 - ΔU₁₂ - ΔU₂₃ - ΔU₃₄ = 11769,68[V]

U₅ = 15000 - ΔU_A1 - ΔU₁₂ - ΔU₂₃ - ΔU₃₄ - ΔU₄₅ = 11416,41[V]