R₁nt(R₂//R₃) 

a) \(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=2\left(\Omega\right)\)

\(\rightarrow R_{td}=R_1+R_{23}=4+2=6\left(\Omega\right)\)

b) Ta có : \(I_1=I_{23}=I=\dfrac{U}{R_{tđ}}=\dfrac{6}{2}=3A\)

\(U_{23}=U_2=U_3=I_{23}.R_{23}=3.2=6V\)

 \(\rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{6}{6}=1A\)

a/ \(\frac{1}{R_{234}}=\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{10}+\frac{1}{6}+\frac{1}{9}=\frac{17}{45}\)

\(\Leftrightarrow R_{234}=\frac{45}{17}\left(Ôm\right)\)

\(R_m=R_1+R_{234}=5+\frac{45}{17}=\frac{130}{17}\left(Ôm\right)\)

b/ \(I_m=\frac{U}{R_m}=\frac{15}{\frac{130}{17}}=\frac{51}{26}\left(A\right)=I_1=I_{234}\)

\(U_{234}=I_{234}.R_{234}=\frac{51}{26}.\frac{45}{17}=\frac{135}{26}\left(V\right)=U_2=U_3=U_4\)

\(I_2=\frac{U_2}{R_2}=\frac{\frac{135}{26}}{10}=\frac{27}{52}\left(A\right)\)

\(I_3=\frac{U_3}{R_3}=\frac{\frac{135}{26}}{6}=\frac{45}{52}\left(A\right)\)

\(I_4=\frac{U_4}{R_4}=\frac{\frac{135}{26}}{9}=\frac{15}{26}\left(A\right)\)

Vậy...

Có : R₁//R₂//R₃:

\(\Rightarrow\)R₁₂₃=\(\frac{R_1.R_2.R_3}{R_2.R_3+R_1.R_3+R_1.R_2}=\frac{40.20.40}{20.40+40.40+40.20}=10\Omega\)

Vì R₄nt (R₁//R₂//R₃)

\(\Rightarrow\)R_tđ=R₄+R₁₂₃=10+10=20\(\Omega\)

\(\Rightarrow\)I_c=\(\frac{U_{AB}}{R_{tđ}}=\frac{20}{20}=1A\)

\(\Rightarrow\)I_c=I₄=I₁₂₃=1A \(\Rightarrow\)U₄=I₄.R₄=1.10=10(V)

Có : R₄nt(R₁//R₂//R₃)\(\Rightarrow U_{AB}=U_4+U_{123}\)

\(\Rightarrow\)U₁₂₃=U_AB-U₄=20-10=10(V)

mà R₁//R₂//R₃ nên :

\(\Rightarrow\)U₁=U₂=U₃=U₁₂₃=10(V)

Khi đó : I₁=\(\frac{U_1}{R_1}=\frac{10}{40}=0,25A\)

I₂=\(\frac{U_2}{R_2}=\frac{10}{20}=0,5A\)

I₃=\(\frac{U_3}{R_3}=\frac{10}{40}=0,25A\)

Vậy ....

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