Đáp án:

Mạch gồm: (R//R//R1) nt (R//R2) nt R3

Ta có:

\(\begin{array}{l}
U = {U_{AB}} + {U_{BO}} \Rightarrow 12 = {U_{AB}} + 6 \Rightarrow {U_{AB}} = 6\\
{U_{AC}} = {U_{AB}} + {U_{BC}} \Rightarrow 10 = 6 + {U_{BC}} \Rightarrow {U_{BC}} = 4\\
U = {U_{AB}} + {U_{BC}} + {U_{CO}} \Rightarrow 12 = 6 + 4 + {U_{CO}} \Rightarrow {U_{CO}} = 2
\end{array}\)

Mặt khác:

\(\begin{array}{l}
I = \dfrac{{{U_{AB}}}}{{{R_3}}} = \dfrac{{{U_{BC}}}}{{{R_2}}} + \dfrac{{{U_{BC}}}}{R} = \dfrac{{{U_{CO}}}}{{{R_1}}} + \dfrac{{{U_{CO}}}}{R} + \dfrac{{{U_{CO}}}}{R}\\
 \Rightarrow I = \dfrac{6}{{{R_3}}} = \dfrac{4}{{{R_2}}} + \dfrac{4}{{10}} = \dfrac{2}{{{R_1}}} + \dfrac{2}{{10}} + \dfrac{2}{{10}}\\
 \Rightarrow \dfrac{4}{{{R_2}}} = \dfrac{2}{{{R_1}}} \Rightarrow {R_2} = 4{R_1}
\end{array}\)

Nếu \({R_1} = {R_3} \Rightarrow \dfrac{6}{{{R_1}}} = \dfrac{2}{{{R_1}}} + \dfrac{4}{{10}} \Rightarrow {R_1} = {R_3} = 10\Omega  \Rightarrow {R_2} = 40\Omega \)

Nếu \({R_2} = {R_3} \Rightarrow \dfrac{6}{{{R_2}}} = \dfrac{4}{{{R_2}}} + \dfrac{4}{{10}} \Rightarrow {R_2} = {R_3} = 5\Omega  \Rightarrow {R_1} = 1,25\Omega \)