[tex]\frac{1}{R_{123}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{20}+\frac{1}{10}+\frac{1}{10}=\frac{1+2+2}{20}=\frac{5}{20} \ \ \rightarrow \ \ R_{123}=\frac{20}{5}=4 \ \Omega[/tex]
[tex]R=R_w+R_{123}=2+4=6 \ \Omega[/tex]
[tex]I=\frac{E}{R}=\frac{120}{6}=20 \ A[/tex]
[tex]I=I_{123}=I_w[/tex]
[tex]U_w=I*R_w=20*2=40 \ V[/tex]
U=E-Uw
U=120-40=80 V
U=U₁=U₂=U₃
[tex]I_1=\frac{U}{R_1}=\frac{80}{20}=4 \ A\\ \\I_2=\frac{U}{R_2}=\frac{80}{10}=8 \ A\\ \\I_3=\frac{U}{R_3}=\frac{80}{10}=8 \ A[/tex]
Sprawdzenie z praw Kirchhoffa
[tex]E=U_w+U=40+80=120 \ V[/tex]
[tex]I=I_w=I_1+I_2+I_3=4+8+8=20 \ A[/tex]
