\(R_{tđ}=R_1+R_2+R_3=1+2+2=5\Omega\)

\(I_1=I_2=I_3=I=\dfrac{U}{R}=\dfrac{16}{5}=3,2A\)

\(U_1=I_1\cdot R_1=1\cdot3,2=3,2V\)

\(U_2=U_3=3,2\cdot2=6,4V\)

Khi mắc R₁ nt R₂ ntR₃

=> R_td=R₁+R₂+R₃=\(\dfrac{U}{I_1}=\dfrac{110}{2}=55\left(\Omega\right)\)(1)

Khi mắc R₁ntR₂

=>R'_td=R₁+R₂=\(\dfrac{U}{I_2}=\dfrac{110}{5,3}=\dfrac{1100}{53}\approx20,75\left(\Omega\right)\)(2)

Khi mắc R₁ntR₃

=>R''_td=\(\dfrac{U}{I_3}=\dfrac{110}{2,2}=50\left(\Omega\right)\)(3)

Thay (2) vào (1)

Ta có:R₁+R₂+R₃=55(Ω)

=>20,75+R₃=55

=> R₃=55-20,75=32,25(Ω)

Thay R₃ vào (3) Ta được R₁=50-32,25=17,75(Ω)

=> R₂=27,25-17,75=9,5(Ω)

Do các điện trở được mắc nối tiếp với nhau nên ta có:

\(I_{AB}=I_{AD}=I_{CB}=1,5A\)

\(R_{AB}=R_1+R_2+R_3=\dfrac{U_{AB}}{I_{AB}}=\dfrac{100}{1,5}=\dfrac{200}{3}\Omega\) (1)

\(R_{AD}=R_1+R_2=\dfrac{U_{AD}}{I_{AD}}=\dfrac{50}{1,5}=\dfrac{100}{3}\Omega\) (2)

\(R_{CB}=R_2+R_3=\dfrac{U_{CB}}{I_{CB}}=\dfrac{70}{1,5}=\dfrac{140}{3}\Omega\) (3)

Từ (1), (2), (3) Ta tìm được: \(R_1=20\Omega,R_2=\dfrac{40}{3}\Omega,R_3=\dfrac{100}{3}\Omega\)

Từ th2 ta có R1ntR2=>Rtđ'=R1+R2=\(\dfrac{U}{I'}=\dfrac{6}{0,24}=25\Omega\)=>R1=25-R2 (1)

Từ th1 ta có R1//R2=>\(Rt\text{đ}=\dfrac{R1.R2}{R1+R2}=\dfrac{U}{I}=\dfrac{6}{1}=6\Omega\) (2)

Thay 1 vào 2 ta có \(\dfrac{\left(25-R2\right).R2}{25-R2+R2}=6=>R2=10\Omega=>R1=15\Omega\)

Vì điện trở của ampe kế ko đáng kể

Nên M trùng N

MCD:R₁nt(R₂//R₄)nt(R₃//R₅)

a,\(R_{24}=\dfrac{R_2\cdot R_4}{R_2+R_4}=\dfrac{4\cdot5}{4+5}=\dfrac{20}{9}\left(\Omega\right)\)

\(R_{35}=\dfrac{R_3\cdot R_5}{R_3+R_5}=\dfrac{6\cdot10}{6+10}=3,75\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{24}+R_{35}=2+\dfrac{20}{9}+3,75=\dfrac{287}{36}\left(\Omega\right)\)

\(I_1=I_{24}=I_{35}=I=\dfrac{U}{R_{tđ}}=\dfrac{40}{\dfrac{287}{36}}=\dfrac{1440}{287}\left(A\right)\)

\(U_2=U_4=U_{24}=I_{24}\cdot R_{24}=\dfrac{1440}{287}\cdot\dfrac{20}{9}=\dfrac{3200}{287}\left(V\right)\)

\(U_3=U_5=U_{35}=I_{35}\cdot R_{35}=\dfrac{1440}{287}\cdot3,75=\dfrac{5400}{287}\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{\dfrac{3200}{287}}{4}=\dfrac{800}{287}\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{\dfrac{5400}{287}}{6}=\dfrac{900}{287}\left(A\right)\)

\(I_4=\dfrac{U_4}{R_4}=\dfrac{\dfrac{3200}{287}}{5}=\dfrac{640}{287}\left(A\right)\)

\(I_5=\dfrac{U_5}{R_5}=\dfrac{\dfrac{5400}{287}}{10}=\dfrac{540}{287}\left(A\right)\)

\(U_1+U_2+U_{MN}+U_5=U\Leftrightarrow R_1I_1+U_2+U_{MN}+U_5=U\)

\(\Rightarrow2\cdot\dfrac{1440}{287}+\dfrac{3200}{287}+U_{MN}+\dfrac{3200}{287}=40\Leftrightarrow U_{MN}=\dfrac{2200}{287}\left(V\right)\)

b, Chọn chiều dòng điện MN là từ N đến M

\(I_A=I_3-I_2=\dfrac{900}{287}-\dfrac{800}{287}=\dfrac{100}{287}\left(A\right)\)