Cho mạch điện như hình vẽ. R1=R3=R5=2Ω;R2=4Ω;R4=6Ω.

* Hướng dẫn giải

Cấu tạo mạch điện: \(\left[ {{R_4}//\,\left( {{R_1}\,nt\,{R_2}} \right)} \right]\,\,nt\,\left( {{R_3}//{R_5}} \right)\)

Ta có:

\({R_1}\,nt\,{R_2} \Rightarrow {R_{12}} = {R_1} + {R_2} = 2 + 4 = 6{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)\)

\(\begin{array}{*{20}{l}}{{R_4}//{R_{12}} \Rightarrow {R_{124}} = \frac{{{R_4}.{R_{12}}}}{{{R_4} + {R_{12}}}} = \frac{{6.6}}{{6 + 6}} = 3{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)}\\{{R_3}//{R_5} \Rightarrow {R_{35}} = \frac{{{R_3}.{R_5}}}{{{R_3} + {R_5}}} = \frac{{2.2}}{{2 + 2}} = 1{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)}\\{{R_{124}}nt{R_{35}} \Rightarrow {R_{td}} = {R_{124}} + {R_{35}} = 3 + 1 = 4{\mkern 1mu} {\mkern 1mu} \left( \Omega \right)}\end{array}\)

Chọn C