Đáp án:

a) \(6\Omega \) ; \(12\Omega \)

b) \(\dfrac{{18}}{{23}}A\)

Giải thích các bước giải:

a) Mạch gồm: R1 nt R2 nt (R3 // R5)

Ta có:

\(\begin{array}{l}
{U_2} = {U_V} = 9V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{9}{6} = 1,5A\\
 \Rightarrow {I_1} = {I_2} = 1,5A\\
{U_1} = {I_1}{R_1} = 1,5.2 = 3V\\
{U_{35}} = U - {U_1} - {U_2} = 18 - 3 - 9 = 6V\\
{I_3} = {I_2} - {I_A} = 1,5 - 0,5 = 1A\\
{R_3} = \dfrac{{{U_{35}}}}{{{I_3}}} = \dfrac{6}{1} = 6\Omega \\
{R_5} = \dfrac{{{U_{35}}}}{{{I_A}}} = \dfrac{6}{{0,5}} = 12\Omega 
\end{array}\)

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

Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 6 + 6 = 12\Omega \\
{R_{45}} = {R_4} + {R_5} = 6 + 12 = 18\Omega \\
{R_{CB}} = \dfrac{{{R_{23}}.{R_{45}}}}{{{R_{45}} + {R_{23}}}} = \dfrac{{12.18}}{{12 + 18}} = 7,2\Omega \\
{R_{AB}} = {R_1} + {R_{CB}} = 2 + 7,2 = 9,2\Omega \\
I = \dfrac{{{U_{AB}}}}{{{R_{AB}}}} = \dfrac{{18}}{{9,2}} = \dfrac{{45}}{{23}}A\\
{I_A} = \dfrac{{{R_{23}}}}{{{R_{45}} + {R_{23}}}}.I = \dfrac{{12}}{{30}}.\dfrac{{45}}{{23}} = \dfrac{{18}}{{23}}A\\
{U_4} = {I_A}.{R_4} = \dfrac{{18}}{{23}}.6 = \dfrac{{108}}{{23}}A\\
{I_2} = I - {I_A} = \dfrac{{45}}{{23}} - \dfrac{{18}}{{23}} = \dfrac{{27}}{{23}}A\\
{U_2} = {I_2}{R_2} = \dfrac{{27}}{{23}}.6 = \dfrac{{162}}{{23}}V\\
{U_V} = \left| { - {U_2} + {U_4}} \right| = \left| { - \dfrac{{162}}{{23}} + \dfrac{{108}}{{23}}} \right| = \dfrac{{54}}{{23}}V
\end{array}\)