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Đáp án:

a> Rtd=4

I=6A

b> 
\(\left\{ \begin{align}
  & {{I}_{2}}={{I}_{5}}=1,5A \\ 
 & {{I}_{3}}={{I}_{6}}=0,75A \\ 
 & {{I}_{4}}={{I}_{1}}=3A \\ 
\end{align} \right.\)

\(\left\{ \begin{align}
  & {{U}_{4}}=12V \\ 
 & {{U}_{5}}={{U}_{1}}=6V \\ 
 & {{U}_{2}}={{U}_{3}}=3V \\ 
\end{align} \right.\)

Giải thích các bước giải:

Mạch tương đương
\((\left[ (\left[ ({{R}_{6}}//{{R}_{3}})nt{{R}_{2}} \right]//{{R}_{5}})nt{{R}_{1}} \right]//{{R}_{4}})\)

a> điện trở tương đương:
\({{R}_{36}}=\dfrac{{{R}_{3}}.{{R}_{6}}}{{{R}_{3}}+{{R}_{6}}}=\dfrac{4}{2}=2\Omega \Rightarrow {{R}_{236}}={{R}_{2}}+{{R}_{36}}=2+2=4\Omega \)

\({{R}_{5236}}=\dfrac{{{R}_{5}}.{{R}_{236}}}{{{R}_{5}}+{{R}_{236}}}=\dfrac{4.4}{4+4}=2\Omega \Rightarrow {{R}_{15236}}={{R}_{2}}+{{R}_{36}}=2+2=4\Omega \)

\({{R}_{td}}=\dfrac{{{R}_{4}}.{{R}_{15236}}}{{{R}_{4}}+{{R}_{15236}}}=\dfrac{4.4}{4+4}=2\Omega \)

Số chỉ của ampe kế: \(I=\dfrac{U}{{{R}_{td}}}=\frac{12}{2}=6A\)

b> ta có:
\(U={{U}_{4}}={{U}_{15236}}=12V\)
\({{I}_{4}}=\dfrac{{{U}_{4}}}{{{R}_{4}}}=\dfrac{12}{4}=3A\)

\({{I}_{15236}}={{I}_{1}}={{I}_{5236}}=\dfrac{U}{{{R}_{15236}}}=\dfrac{12}{4}=3A\)

\({{U}_{1}}={{I}_{1}}.{{R}_{1}}=3.2=6V\)
\({{U}_{5236}}={{U}_{5}}={{U}_{236}}={{I}_{5236}}.{{R}_{5236}}=3.2=6V\)
\({{I}_{5}}=\dfrac{{{U}_{5}}}{{{R}_{5}}}=\dfrac{6}{4}=1,5A\)

\({{I}_{236}}={{I}_{2}}={{I}_{36}}=\dfrac{{{U}_{236}}}{{{R}_{236}}}=\dfrac{6}{4}=1,5A\)
\({{U}_{2}}={{I}_{2}}.{{R}_{2}}=1,5.2=3V\)
\({{U}_{36}}={{U}_{3}}={{U}_{6}}={{I}_{36}}.{{R}_{36}}=1,5.2=3V\)

\({{I}_{3}}={{I}_{6}}=\dfrac{{{U}_{36}}}{{{R}_{3}}}=\dfrac{3}{4}=0,75A\)