Đáp án:

a) \(8\Omega \)

b) \(4A\)

c) \(6V\)

Giải thích các bước giải:

a) Mạch gồm: R0 nt ((R4 nt R5) // (R1 nt (R2 // R3)))

Ta có:

\(\begin{array}{l}
{R_{45}} = {R_4} + {R_5} = 3 + 12 = 15\Omega \\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{30.15}}{{30 + 15}} = 10\Omega \\
{R_{123}} = {R_1} + {R_{23}} = 5 + 10 = 15\Omega \\
{R_{12345}} = \dfrac{{{R_{123}}.{R_{45}}}}{{{R_{123}} + {R_{45}}}} = \dfrac{{15.15}}{{15 + 15}} = 7,5\Omega \\
{R_{AB}} = {R_0} + {R_{12345}} = 0,5 + 7,5 = 8\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{I_{{A_1}}} = \dfrac{U}{{{R_{AB}}}} = \dfrac{{48}}{8} = 6A\\
{I_{123}} = \dfrac{{{R_{45}}}}{{{R_{123}} + {R_{45}}}}.{I_{{A_1}}} = \dfrac{1}{2}.6 = 3A\\
{I_3} = \dfrac{{{R_2}}}{{{R_2} + {R_3}}}.{I_{123}} = \dfrac{{30}}{{30 + 15}}.3 = 2A\\
{I_{{A_2}}} = {I_{{A_1}}} - {I_3} = 6 - 2 = 4A
\end{array}\)

c) Ta có: 

\(\begin{array}{l}
{I_1} = {I_{123}} = 3A\\
{U_1} = {I_1}{R_1} = 3.5 = 15V\\
{I_4} = I - {I_1} = 6 - 3 = 3A\\
{U_4} = {I_4}{R_4} = 3.3 = 9V\\
{U_{MN}} =  - {U_4} + {U_1} =  - 9 + 15 = 6V
\end{array}\)