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

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

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

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

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

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

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

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

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

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

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

a, \(=>\dfrac{1}{R1234}=\dfrac{1}{\left(R1+R2\right)}+\dfrac{1}{R3}+\dfrac{1}{R4}=>\dfrac{1}{R1234}=\dfrac{1}{24}+\dfrac{1}{12}+\dfrac{1}{8}=>R1234=4\left(om\right)\)

\(=>Rab=R7+\dfrac{R6\left(R5+R1234\right)}{R6+R5+R1234}=8+\dfrac{12\left(20+4\right)}{12+20+4}=16\left(om\right)\)

b.,\(=>I7=123456=\dfrac{U}{Rab}=\dfrac{48}{16}=3A=>U7=I7.R7=24V=>U123456=48-24=24V\)

\(=>U123456=U6=U12345=24V=>I6=\dfrac{U6}{R6}=2A=>I12345=\dfrac{24}{R12345}=\dfrac{24}{24}=1A=I5=I1234\)

\(=>U5=I5R5=20V,=>U1234=24-20=4V=U12=U3=U4\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{4}{12}=\dfrac{1}{3}A,=>I4=\dfrac{U4}{R4}=\dfrac{4}{8}=0,5A\)

\(=I12=I1=I2=\dfrac{4}{24}=\dfrac{1}{6}A=>U1=I1R1=\dfrac{2}{3}V=>U2=I2R2=\dfrac{10}{3}V\)

(bài dài nên bn bấm lại máy tính ktra cho chắc)

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

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

Bạn có thể giúp mik giải bài này đc ko mai mik thi r ko bt phải làm sao. Mong các bạn giúp mik

a) Do \(R_2ntR_3\)

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

Do \(R_{23}//R_1\)

\(\Rightarrow R_{tđ}=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{10.10}{10+10}=5\left(\Omega\right)\)

b) Hiệu điện thế giữa 2 đầu mạch chính:

 \(U=I.R=1,5.5=7,5\left(\Omega\right)\)

Do mạch song song nên \(U=U_1=U_{23}=7,5\Omega\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{7,5}{10}=0,75\left(A\right)\)

\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{7,5}{10}=0,75\left(A\right)\)