Đáp án:

a) \(\dfrac{{240}}{{49}}V\)

b) \(\dfrac{{96}}{{49}}A\)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 20 + 20 = 40\Omega \\
R = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{5.40}}{{5 + 40}} = \dfrac{{40}}{9}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{\dfrac{{40}}{9} + 1}} = \dfrac{{108}}{{49}}A\\
{I_2} = \dfrac{{{R_1}}}{{{R_1} + {R_{23}}}}.I = \dfrac{5}{{5 + 40}}.\dfrac{{108}}{{49}} = \dfrac{{12}}{{49}}A\\
{U_2} = {I_2}{R_2} = \dfrac{{12}}{{49}}.20 = \dfrac{{240}}{{49}}V
\end{array}\)

b) Ta có: 

\({I_1} = I - {I_2} = \dfrac{{108}}{{49}} - \dfrac{{12}}{{49}} = \dfrac{{96}}{{49}}A\)

Đáp án: `a) 240/49 (V)`

`b)96/49 (A)`

 ta có $R_1 // ( R_2 nt R_3) $

`R_(td)=(R_1.R_(23))/(R_1+R_(23))`

`=(5.(20+20))/(5+(20+20))`

`=>R_(td)=40/9 Omega` 

 `I_(dm)=E / (R_(dm)+r)=12/(40/9 + 1) =108/49 (A)`

`U=I_(dm).R_(dm)=108/49 .40/9 =480/49 (V)`

vì $R_1// R_(23)$

`=> U=U_1=U_(23)=480/49(V)`

mà `R_2 nt R_3 `

`=> I_(23)=I_2=I_3=U_(23)/R_(23)=(480/49)/(20+20)=12/49 (A)`

`=U_2=I_2.R_2=12/49 .20= 240/49 (V)`

vì $R_1// R_(23)$

`=> U=U_1=U_(23)=480/49(V)`

`=>I_1=U_1/R_1=(480/49)/5=96/49 (A)`