Tóm tắt:

\(R_1=2\Omega\\ R_2=6\Omega\\ R_3=4\Omega\\ R_4=10\Omega\\ U_{AB}=28V\\ a,R_{tđ}=?\\ b,I_1?\\ I_2=?\\ I_3=?\\ I_4=?\\ c,U_1=?\\ U_2=?\)

Giải:

Cấu tạo: R₁nt[R₄//(R₂ntR₃)]

\(R_{23}=R_2+R_3=6+4=10\left(\Omega\right)\)

\(R_{234}=\dfrac{R_4\cdot R_{23}}{R_4+R_{23}}=5\Omega\)

\(R_{tđ}=R_1+R_{234}=7\Omega\)

b,\(I_1=I=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{28}{7}=4\left(A\right)\)

\(I_{234}=I_1=4\left(A\right)\)

\(\Rightarrow U_{23}=U_4=U_{234}=I_{234}\cdot R_{234}=4\cdot5=20\left(V\right)\)

\(I_4=\dfrac{U_4}{R_4}=\dfrac{20}{10}=2\left(A\right)\)

\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{20}{10}=2\left(A\right)\)

c,\(U_1=R_1\cdot I_1=2\cdot4=8\left(V\right)\)

\(U_2=I_2\cdot R_2=2\cdot6=12\left(V\right)\)

R1 mắc nối tiếp với r2 ạ

\(a.R_{tđ}=R_1+R_2=4+6=10\Omega\\ b.R_{tđ}'=R_1+\dfrac{R_2.R_3}{R_2+R_3}=4+\dfrac{6.12}{6+12}=8\Omega\\ I=\dfrac{U_{AB}}{R_{tđ}'}=\dfrac{18}{8}=2,25A\\ Vì.R_1ntR_{23}\\ \Rightarrow I=I_1=I_{23}=2,25A\\ U_1=I_1.R_1=4.2,25=9V\\ U_{23}=U_{AB}-U_1=18-9=9V\\ Vì.R_2//R_3\Rightarrow U_{23}=U_2=U_3=9V\\ I_3=\dfrac{U_3}{R_3}=\dfrac{9}{12}=0,75A\)

a) \(R_1ntR_2\Rightarrow R_{tđ}=R_1+R_2=4+6=10\Omega\)

\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{18}{10}=1,8A\)

b) CTM: \(R_1nt\left(R_2//R_3\right)\)

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{6\cdot12}{6+12}=4\Omega\)

\(R_{tđ}=R_1+R_{23}=4+4=8\Omega\)

c)\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{18}{8}=2,25A\)

\(R_1nt\left(R_2//R_3\right)\Rightarrow I_{23}=I_1=I_m=2,25A\)

\(U_{23}=I_{23}\cdot R_{23}=2,25\cdot4=9V\Rightarrow U_3=9V\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{9}{12}=0,75A\)

\(R_{12}=\dfrac{15.30}{15+30}=10\left(\Omega\right)\)

\(R_m=R_{12}+R_3=10+30=40\left(\Omega\right)\)

\(I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{40}=0,3\left(A\right)\)

\(b,I_{12}=I_3=0,3\left(A\right)\)

\(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{30}{15}=\dfrac{2}{1}\)

\(\rightarrow I_1=0,2\left(A\right);I_2=0,1\left(A\right)\)

\(a,R_{23}=R_2+R_3=30+30=60\left(\Omega\right)\)

\(R_m=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{60.15}{60+15}=12\left(\Omega\right)\)

\(b,I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{12}=1\left(A\right)\)

\(I_1+I_{23}=1\left(A\right)\)

\(\dfrac{I_1}{I_{23}}=\dfrac{R_{23}}{R_1}=\dfrac{60}{15}=\dfrac{4}{1}\)

\(\rightarrow I_1=0,8\left(A\right);I_{23}=0,2\left(A\right)\)

\(\rightarrow I_2=I_3=0,2\left(A\right)\)

a,\(R1nt\left(R2//R3\right)=>Rtd=R1+\dfrac{R2R3}{R2+R3}=4+\dfrac{6.3}{6+3}=6\left(om\right)\)

b,\(=>I1=I23=\dfrac{Uab}{Rtd}=\dfrac{9}{6}=1,5A\)

\(=>U23=I23.R23=1,5.\dfrac{6.3}{6+3}=3V=U2=U3\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{3}{6}=0,5A,=>I3=\dfrac{U3}{R3}=\dfrac{3}{3}=1A\)

c,\(=>Im=Ix=I23=\dfrac{1}{3}.1,5=0,5A\)

\(=>RTd=Rx+\dfrac{R2.R3}{R2+R3}=Rx+\dfrac{6.3}{6+3}=\dfrac{U}{Im}=\dfrac{9}{0,5}=18\)

\(=>Rx=16\left(om\right)\)

MCD: R1 nt(R2//R3)

a, ĐIện trở tương đương của đoạn mạch

\(R_{23}=\dfrac{R_2R_3}{R_2+R_3}=\dfrac{30\cdot20}{30+20}=12\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{23}=18+12=30\left(\Omega\right)\)

b,Cường độ dòng điện qua mỗi điện trở

\(I_1=I_{23}=I=\dfrac{U}{R_{tđ}}=\dfrac{60}{30}=2\left(A\right)\)

\(U_2=U_3=U_{23}=I_{23}\cdot R_{23}=2\cdot12=24\left(V\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{24}{30}=0,8\left(A\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{24}{20}=1,2\left(A\right)\)

CTM: \(R_1nt\left(R_2//R_3\right)\)

a)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{2\cdot6}{2+6}=1,5\Omega\)

\(R_{tđ}=R_1+R_{23}=3+1,5=4,5\Omega\)

\(I=\dfrac{U}{R_{tđ}}=\dfrac{9}{4,5}=2A\)

b)Công đoạn mạch sản ra trong thời gian \(t=5min=300s\) là:

\(A=UIt=9\cdot2\cdot300=5400J\)

c)\(I_1=I=2A\)

Điện năng tiêu thụ trên điện trở \(R_1\) trong thời gian \(t=5min=300s\) là:

\(A_1=U_1.I_1.t=I_1^2.R_1.t=2^2\cdot3\cdot300=3600J\)