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

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

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

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

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

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

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

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

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

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

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

Yêu cầu của đề là gì vậy em ?

Tiếp nè

Do \(R_{12}//R_3\)

\(\Rightarrow U_{12}=U_3=U_{123}=\frac{48a}{7}\left(V\right)\)

\(\rightarrow I_3=\frac{U_3}{R_3}=\frac{\frac{48a}{7}}{4a}=\frac{12}{7}\left(A\right)\)

\(\rightarrow I_{12}=\frac{U_{12}}{R_{12}}=\frac{\frac{48a}{7}}{3a}=\frac{16}{7}\left(A\right)\)

Do \(R_1ntR_2\)

\(\Rightarrow I_1=I_2=I_{12}=\frac{16}{7}\left(A\right)\)

Vậy:

\(I_4=4\left(A\right)\)

\(I_3=\frac{12}{7}\left(A\right)\)

\(I_1=I_2=\frac{16}{7}\left(A\right)\)

Ta có : [(R\(_1\) nt R\(_2\) ) \(//\) R\(_3\) ] nt R\(_4\)

\(\Rightarrow\) R\(_{12}\) = R\(_1\) + R\(_2\) = a+2a = 3a (\(\Omega\) )

\(\Rightarrow R_{234}=\frac{R_{12}.R_3}{R_{12}+R_3}=\frac{3a.4a}{3a+4a}=\frac{12a^2}{7a}=\frac{12a}{7}\left(\Omega\right)\)

\(\Rightarrow R_{td}=R_{123}+R_4=\frac{12a}{7}+a=\frac{19a}{7}\left(\Omega\right)\)

Do \(R_{123}ntR_4\)

\(\Rightarrow I_{123}=I_4=I=4\left(A\right)\)

\(\Rightarrow U_{123}=I_{123}.R_{123}=\frac{12a}{7}.4=\frac{48a}{7}\left(V\right)\)

Do

\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{12\cdot6}{12+6}=4\Omega\)

\(R_{34}=R_{tđ}-R_{12}=10-4=6\Omega\)

\(\dfrac{1}{R_{34}}=\dfrac{1}{R_3}+\dfrac{1}{R_4}=\dfrac{1}{24}+\dfrac{1}{R_4}=\dfrac{1}{6}\)

\(\Rightarrow R_4=8\Omega\)

theo hình vẽ giả sử dòng điện từ A có chiều từ C đến D qua R5

\(=>I4=I2+I5=I2+2\left(A\right)\)

\(=>I3=I1-I5=I1-2\left(A\right)\)

\(=>Uab=U1+U3=U2+U4\)

\(=>2.I1+2.I3=10.I2+2.I4\)

\(< =>\)\(2I1+2\left(I1-2\right)=10I2+2\left(I2+2\right)\)

\(< =>2I1+2I1-4=10I2+2I2+4\)

\(< =>4I1=12I2+8=>I1=\dfrac{12I2+8}{4}=3I2+2\left(A\right)\)

\(=>60=2I1+2I3+3,2\left(I1+I2\right)\)

\(=>60=2I1+2\left(I1-2\right)+3,2\left(I1+I2\right)\)

\(=>60=2\left(3I2+2\right)+3,2\left(3I2+2+I2\right)=>I2=4\left(om\right)\)

\(=>I1=3.2+2=8\left(om\right)\)

\(=>U5=10.2-8.2=4V=>R5=\dfrac{4}{2}=2\left(om\right)\)

*TK: http://nghiachi.com/16179/611.pdf

\(\)

K đóng I₂= 2A

Giả sử dòng điện đi qua R₅ có chiều từ C đến D

*Tại C: I₃=I₁-I₅ = I₁-2

*Tại D: I₄= I₂ + 2

*Ta có : U_AE=U_AC+U_CE = U_AD + U_DE

 ⇔I₁R₁ + I₃R₃ = I₂R₂ + I₄R₄

  ⇔ 2I₁ + 2(I₁ - 2)= 10I₂ + 2(I₂ + 2)

  ⇔4I₁ = 12I₂ + 8

  ⇔I₁ = 3I₂ + 2

*CĐDĐ qua R₆ : I₆ = I₁ + I₂ = 4I₂ + 2

*Lại có : U = U_AE + U_EB = U_AD + U_DE + U_EB

 ⇔ 60 = 10I₂ + 2(I₂ + 2) +3,2(4I₂ +2)

 ⇒I₂ = 2 (A)

 ⇒I₁ = 3I₂ +2 = 8 A

*Hđt giữa 2 đầu R₅ là : U₅ = U_CD = U_CA + U_AD

                                             = -I₁R₁ + I₂R₂

                                             = -8.2 + 10.2 = 4 V

Giá trị của R₅ là: R₅ = U₅/I₅ = 2 Ω

Đáp án B