Đáp án:

a) \(24\Omega \)

b) \(0,75A\)

Giải thích các bước giải:

a) K mở, mạch gồm: R1 nt R2 nt (R3 // R5)

Ta có:

\(\begin{array}{l}
{U_5} = {I_5}{R_5} = 1,125.8 = 9V\\
{U_{12}} = U - {U_5} = 36 - 9 = 27V\\
{I_{12}} = \dfrac{{{U_{12}}}}{{{R_1} + {R_2}}} = \dfrac{{27}}{{6 + 12}} = 1,5A\\
{I_3} = {I_{12}} - {I_5} = 1,5 - 1,125 = 0,375A\\
{R_3} = \dfrac{{{U_5}}}{{{I_3}}} = \dfrac{9}{{0,375}} = 24\Omega 
\end{array}\)

b) K đóng, mạch gồm: R1 nt (R2 // R4) nt (R3 // R5)

Ta có:

\(\begin{array}{l}
{R_{24}} = \dfrac{{{R_2}{R_4}}}{{{R_2} + {R_4}}} = \dfrac{{12.24}}{{12 + 24}} = 8\Omega \\
{R_{35}} = \dfrac{{{R_3}{R_5}}}{{{R_3} + {R_5}}} = \dfrac{{24.8}}{{24 + 8}} = 6\Omega \\
R = {R_1} + {R_{24}} + {R_{35}} = 6 + 8 + 6 = 20\Omega \\
I = \dfrac{U}{R} = \dfrac{{36}}{{20}} = 1,8A\\
{I_2} = \dfrac{{{R_4}}}{{{R_2} + {R_4}}}.I = \dfrac{{24}}{{12 + 24}}.1,8 = 1,2A\\
{I_3} = \dfrac{{{R_5}}}{{{R_3} + {R_5}}}.I = \dfrac{8}{{24 + 8}}.1,8 = 0,45A\\
{I_A} = {I_2} - {I_3} = 1,2 - 0,45 = 0,75A
\end{array}\)