Question

Cho mach dien co so do nhu hinh ve trong do cac dien tro R₁ = 14Ω , R₂ = 14Ω , R₃ = 24Ω dong dien di qua R₁ co cuong do la I₁ = 0,4A 

a) Tinh cac cuong do dong dien I₂ , I₃ tuong ung di qua cac dien tro R₂ , R₃

b) Tinh cac hieu dien the U_AC ; U_CB va U_AB 

                    

  

Answer 1

a) \(R_1nt(R_2//R_3)\)

\(R_1=\dfrac{U_1}{I_1}\Rightarrow U_1=0,4.14=5,6\left(V\right)\)

\(I_1=I_{AB}=0,4A\)

Có \(R_{AB}=R_1+R_{23}=14+\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{434}{19}\left(\Omega\right)\)

\(\Rightarrow U_1+U_{23}=U_{AB}=R_{AB}.I_{AB}=\dfrac{439}{19}.0,4=\dfrac{868}{95}\left(V\right)\)

\(\Rightarrow U_{23}=\dfrac{868}{95}-5,6=\dfrac{336}{95}\left(V\right)\)

\(\Rightarrow U_2=U_3=\dfrac{336}{95}\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{95}\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{14}{95}\left(V\right)\)

b) \(U_{AB}=\dfrac{868}{95}\left(V\right)\)

\(U_{AC}=I_1.R_1=0,4.14=5,6\left(V\right)\)

\(U_{CB}=I_{23}.R_{23}=0,4.\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{336}{95}\left(V\right)\)

Vậy...

Answer 2

Giup minh voi