Bài 64.
a)\(R_1//R_2\Rightarrow R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{20\cdot20}{20+20}=10\Omega\)
b)\(R_1//R_2//R_3\)
\(\Rightarrow\dfrac{1}{R_{AC}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{20}+\dfrac{1}{20}+\dfrac{1}{15}=\dfrac{1}{6}\Rightarrow R_{AC}=6\Omega\)
Nhận thấy: \(R_{AC}=0,3R_1=0,3R_2\) và \(R_{AC}=0,4R_3\)