\(R_{12}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{4\cdot4}{4+4}=2\left(\Omega\right)\)

\(R_{34}=R_3+R_4=3+5=8\left(\Omega\right)\)

\(R_{345}=\dfrac{R_5R_{34}}{R_5+R_{34}}=\dfrac{8\cdot8}{8+8}=4\left(\Omega\right)\)

\(R_{tđ}=R_{12}+R_{345}=2+4=6\left(\Omega\right)\)

\(I_{12}=I_{345}=I=\dfrac{U}{R_{tđ}}=\dfrac{12}{6}=2\left(A\right)\)

\(U_1=U_2=U_{12}=I_{12}\cdot R_{12}=2\cdot2=4\left(V\right)\)

\(U_5=U_{34}=U_{345}=I_{345}\cdot R_{345}=2\cdot4=8\left(V\right)\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{4}{4}=1\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{4}{4}=1\left(A\right)\)

\(I_3=I_4=I_{34}=\dfrac{U_{34}}{R_{34}}=\dfrac{8}{8}=1\left(A\right)\)

\(I_5=\dfrac{U_5}{R_5}=\dfrac{8}{8}=1\left(A\right)\)

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

Bài dễ mà bn, ADCT là ra :))

Có \(R_{tđ}=\frac{R_1.R_2}{R_1+R_2}+\frac{R_3.R_4}{R_3+R_4}=\frac{30.60}{30+60}+\frac{60.R_3}{60+R_3}\)\(=20+\frac{60.R_3}{60+R_3}\)

Có \(R_{tđ}=\frac{U_{AB}}{I}=\frac{22}{0,5}=44\left(\Omega\right)\)

\(\Rightarrow20+\frac{60R_3}{60+R_3}=44\Leftrightarrow\frac{60R_3}{60+R_3}=24\)

\(\Leftrightarrow R_3=40\left(\Omega\right)\)

b/ Có I=I₁₂=I₃₄= 0,5(A)

\(\Rightarrow U_1=U_2=U_{12}=I_{12}.R_{12}=0,5.20=10\left(V\right)\)

\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{10}{30}=\frac{1}{3}\left(A\right)\)

\(\Rightarrow I_2=0,5-\frac{1}{3}=\frac{1}{6}\left(A\right)\)

\(\Rightarrow U_3=U_4=U_{34}=I_{34}.R_{34}=0,5.24=12\left(V\right)\)

\(\Rightarrow I_3=\frac{U_3}{R_3}=\frac{12}{40}=0,3\left(A\right)\)

\(\Rightarrow I_4=\frac{U_4}{R_4}=\frac{12}{60}=0,2\left(A\right)\)

Rồi hình vẽ ở đâu v bạn??

Ta có R1nt(R2//R3)=>RTđ=R1+R23=10+15=25\(\Omega\)

b)\(I=\dfrac{U}{Rtđ}=\dfrac{12}{25}=0,48A=>I1=I23=I=0,48A;\)

Vì R2//R3=>U2=U3=U23=I23.R23=0,48.15=7,2V

=>\(I1=\dfrac{U1}{R1}=\dfrac{7,2}{30}=0,024A=>I2=\dfrac{7,2}{30}=0,024A\)

Ta có đoạn mạch: R₁ nt (R₂ // R₃)

a/ R_td= R₁ + (\(\dfrac{R_2.R_3}{R_2+R_3}\)) = 10 + \(\dfrac{30.30}{30+30}\)= 25 \(\)ôm

b/ I = I₁ = I₂₃ = \(\dfrac{U}{R_{td}}\)= \(\dfrac{12}{25}\) = 0,48 A

U₂₃ = U₂ = U₃ = \(\)I₂₃ . R₂₃ = 0,48 . 15 = 7,2 V

I₂ = \(\dfrac{U_2}{R_2}\)= \(\dfrac{7,2}{30}\)= 0,24 A

I₃ = \(\dfrac{U_3}{R_3}\)=\(\dfrac{7,2}{30}\)= 0,24A