Đáp án+Giải thích các bước giải:

Tóm tắt

`R_1=2,5\Omega`

`R_2=6\Omega`

`R_3=10\Omega`

`R_4=1,2\Omega`

`R_5=5\Omega`

`U_(AB)=6V`

`I_1=?`

`I_2=?`

`I_3=?`

`I_4=?`

`I_5=?`

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Giải

Vì $R_1//R_2//R_3$ nên

`1/R_(123)=1/R_1+1/R_2+1/R_3=1/(2,5)+1/6+1/10=2/3`
`=>R_(123)=3/2=1,5(\Omega)`

Vì $R_4//R_5$ nên

`1/R_(45)=1/R_4+1/R_5=1/(1,2)+1/5=31/30`
`=>R_(45)=30/31(\Omega)`

Vì `R_(123)` nối tiếp `R_(45)` nên

`R_(tđ)=R_(123)+R_(45)=1,5+30/31=153/62(\Omega)`

`=>I=U/R_(tđ)=6:  153/62=124/51(A)`

`=>I_(123)=I_(45)=I=124/51(A)`

`=>U_(123)=I_(123).R_(123)=124/51 . 1,5=62/17(V)`

Vì $R_1//R_2//R_3$ nên

`U_1=U_2=U_3=U_(123)=62/17(V)`

`=>I_1=U_1/R_1=62/17 : 2,5=124/85(A)`

     `I_2=U_2/R_2=62/17 : 6=31/51(A)`

     `I_3=U_3/R_3=62/17 : 10=31/85(A)`

Ta có: `U_(45)=I_(45).R_(45)=124/51 . 30/31=40/17(V)`

Vì $R_4//R_5$ nên

`U_4=U_5=U_(45)=40/17(V)`

`=>I_4=U_4/R_4=40/17 : 1,2=100/51(A)`

     `I_5=U_5/R_5=40/17 : 5=8/17(A)`