Đáp án:

a) \(5\Omega \)

b) \({I_1} = 3A\)

Giải thích các bước giải:

Mạch gồm: R4 // (R1 nt (R5 // (R2 nt (R3 // R6))))

a) Ta có:

\(\begin{array}{l}
{R_{36}} = \dfrac{{{R_3}{R_6}}}{{{R_3} + {R_6}}} = \dfrac{{10.10}}{{10 + 10}} = 5\Omega \\
{R_{236}} = {R_2} + {R_{36}} = 5 + 5 = 10\Omega \\
{R_{5236}} = \dfrac{{{R_5}.{R_{236}}}}{{{R_5} + {R_{236}}}} = \dfrac{{10.10}}{{10 + 10}} = 5\Omega \\
{R_{15236}} = {R_1} + {R_{5236}} = 5 + 5 = 10\Omega \\
{R_{td}} = \dfrac{{{R_4}.{R_{15236}}}}{{{R_4} + {R_{15236}}}} = \dfrac{{10.10}}{{10 + 10}} = 5\Omega 
\end{array}\)

b) Ta có:

\(\begin{array}{l}
{I_4} = \dfrac{U}{{{R_4}}} = \dfrac{{30}}{{10}} = 3A \Rightarrow {I_{{A_1}}} = 3A\\
{I_{15236}} = \dfrac{U}{{{R_{15236}}}} = \dfrac{{30}}{{10}} = 3A \Rightarrow {I_1} = 3A\\
{I_5} = \dfrac{{{R_{236}}}}{{{R_5} + {R_{236}}}}.{I_{15236}} = \dfrac{{10}}{{10 + 10}}.3 = 1,5A\\
{I_{{A_2}}} = {I_{{A_1}}} + {I_5} = 3 + 1,5 = 4,5A\\
{I_2} = {I_1} - {I_5} = 3 - 1,5 = 1,5A\\
{I_6} = \dfrac{{{R_3}}}{{{R_3} + {R_6}}}.{I_2} = \dfrac{1}{2}.1,5 = 0,75A\\
{I_3} = {I_2} - {I_6} = 1,5 - 0,75 = 0,75A\\
{I_{{A_3}}} = {I_{{A_2}}} + {I_6} = 4,5 + 0,75 = 5,25A
\end{array}\)