Tính điện trở tương đương của mạch có cường độ dòng điện đi qua mạch chính là I = 2A ?

có 5 TH mắc mạch dien:

TH1:a, \(R_{tđ}=R_1+R_2+R_3=6+2+4=12\left(\Omega\right)\)

b, \(U=I.R_{tđ}=2.12=24\left(V\right)\)

vì tất cả đều noi tiep nhau nên I=I₁=I₂=I₃=2A

TH2: a,Vì (R1 // R2) nt R3 => \(R_{tđ}=R_3+\dfrac{R_1.R_2}{R_1+R_2}=4+\dfrac{6.2}{6+2}=4+1,5=5,5\left(\Omega\right)\)

b, \(U=I.R_{tđ}=2.5,5=11\left(V\right)\)

Vì nt R3 nên I = I₁₊₂= I₃= 2A

hieu dien the R3 la: \(U_3=I_3.R_3=2.4=8\left(V\right)\)

=> \(U_{1+2}=U-U_3=11-8=3\left(V\right)\)

Vì R1 // R2 => U₁=U₂=U₁₊₂= 3V

cuong do dong dien qua R1; R2:

\(I_1=\dfrac{U_1}{R_1}=\dfrac{3}{6}=0,5\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{3}{2}=1,5\left(A\right)\)

TH3: a, Vì (R1 // R3) nt R2 =>

\(R_{tđ}=R_2+\dfrac{R_1.R_3}{R_1+R_3}=2+\dfrac{6.4}{6+4}=2+2,4=4,4\left(\Omega\right)\)

b, \(U=I.R_{tđ}=2.4,4=8,8\left(V\right)\)

Vì nt R2 nên I = I₁₊₃= I₂= 2A

hieu dien the R2 la: \(U_2=I_2.R_2=2.2=4\left(V\right)\)

\(\Rightarrow U_{1+3}=8,8-4=4,8\left(V\right)\)

Vì R1 // R3 => U₁=U₃=U₁₊₃= 4,8V

cuong do dong dien qua R1; R3:

\(I_1=\dfrac{U_1}{R_1}=\dfrac{4,8}{6}=0,8\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{4,8}{4}=1,2\left(A\right)\)

TH4: a, Vì (R2 // R3) nt R1 =>

\(R_{tđ}=R_1+\dfrac{R_2.R_3}{R_2+R_3}=6+\dfrac{2.4}{2+4}=6+\dfrac{4}{3}=\dfrac{22}{3}\left(\Omega\right)\)

b, \(U=I.R_{tđ}=2.\dfrac{22}{3}=\dfrac{44}{3}\left(V\right)\)

Vì nt R1 nên I = I₂₊₃= I₁= 2A

hieu dien the R1 la: \(U_1=I_1.R_1=2.6=12\left(V\right)\)

\(\Rightarrow U_{2+3}=\dfrac{44}{3}-12=\dfrac{8}{3}\left(V\right)\)

Vì R2 // R3 => U₂=U₃=U₂₊₃= \(\dfrac{8}{3}\)V

cuong do dong dien qua R2; R3:

\(I_2=\dfrac{U_2}{R_2}=\dfrac{8}{3}:2=\dfrac{4}{3}\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{8}{3}:4=\dfrac{2}{3}\left(A\right)\)

TH5: a,vì R1 // R2 // R3 =>

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{6}+\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{11}{12}\)

\(\Rightarrow11R_{tđ}=12\Rightarrow R_{tđ}=\dfrac{12}{11}\left(\Omega\right)\) b, \(U=I.R_{tđ}=2.\dfrac{12}{11}=\dfrac{24}{11}\left(V\right)\) => U = U₁=U₂=U₃ = \(\dfrac{24}{11}\left(V\right)\) cuong do dong dien qua R1,2,3: \(I_1=\dfrac{U_1}{R_1}=\dfrac{24}{11}:6=\dfrac{4}{11}\left(A\right)\) \(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{11}:2=\dfrac{12}{11}\left(A\right)\) \(I_3=\dfrac{U_3}{R_3}=\dfrac{24}{11}:4=\dfrac{6}{11}\left(A\right)\)