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Ta có :R₁₂=R₁+R₂=10+10=20\(\Omega\)

Có :(R₁nt R₂)//R₃ :

\(\Rightarrow\)R₁₂₃=\(\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{20.5}{20+5}=4\Omega\)

Có : R₄nt(R₁ntR₂)//R₃):

\(\Rightarrow\)R_tđ=R₄+R₁₂₃=6+4=10\(\Omega\)

\(\Rightarrow\)I_c=\(\frac{U}{R_{tđ}}=\frac{12}{10}=1,2A\)

\(\Rightarrow\)I_c=I₄=I₁₂₃=1,2A

\(\Rightarrow\)U₄=I₄.R₄=1,2.6=7,2V

Có :R₄nt((R₁ntR₂)//R₃)

\(\Rightarrow\)U=U₄+U₁₂₃

\(\Rightarrow\)U₁₂₃=U-U₄=12-7,2=4,8V

mà (R₁ntR₂)//R₃

\(\Rightarrow\)U₁₂=U₃=U₁₂₃=4,8V

\(\Rightarrow\)I₁₂=\(\frac{U_{12}}{R_{12}}=\frac{4,8}{20}=0,24A\)\(\Rightarrow\)I₁=I₂=I₁₂=0,24A\(\Rightarrow\)\(\left\{{}\begin{matrix}U_1=R_1.I_1=10.0,24=2,4V\\U_2=R_2.I_2=10.0,24=2,4V\end{matrix}\right.\)

\(\Rightarrow\) I₃=\(\frac{U_3}{R_3}=\frac{4,8}{5}=0,96\)A

\(=>R1nt\left(R2//R3\right)nt\left[R4//\left(R5ntR6\right)\right]\)

\(=>RTd=R1+\dfrac{R2R3}{R2+R3}+\dfrac{R4\left(R5+R6\right)}{R4+R5+R6}\)

\(=12,6+\dfrac{4.6}{4+6}+\dfrac{30\left(15+15\right)}{30+15+15}=30\left(om\right)\)

\(=>Im=\dfrac{U}{Rtd}=\dfrac{30}{30}=1A=I1=I23=I456\)

\(=>U23=I23.R23=1.\dfrac{4.6}{4+6}=2,4V=U2=U3\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{2,4}{4}=0,6A=>I3=I23-I2=0,4A\)

\(=>U456=I456.R456=\dfrac{30\left(15+15\right)}{30+15+15}=15V=U4=U56\)

\(=>I4=\dfrac{U4}{R4}=\dfrac{15}{30}=0,5A\)

\(=>I56=I5=I6=\dfrac{U56}{R56}=\dfrac{15}{15+15}=0,5A\)