Cấu tạo mạch: \(R_1//R_2//R_3\)
a)\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{9}+\dfrac{1}{18}+\dfrac{1}{24}=\dfrac{5}{24}\Rightarrow R_{tđ}=4,8\Omega\)
b)\(I_A=\dfrac{U_{mạch}}{R_{tđ}}=\dfrac{3,6}{4,8}=0,75A\)
\(U_{12}=U_{mạch}=3,6V\)
\(\dfrac{1}{R_{12}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{1}{9}+\dfrac{1}{18}=\dfrac{1}{6}\Rightarrow R_{12}=6\Omega\Rightarrow I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{3,6}{6}=0,6A\)
\(MCD:\left(R1//R2\right)//R3\)
\(\Rightarrow R_{td}=\dfrac{\left(\dfrac{9\cdot18}{9+18}\right)\cdot24}{\left(\dfrac{9\cdot18}{9+18}\right)+24}=4,8\Omega\)
\(U=U12=U3=3,6V\Rightarrow\left[{}\begin{matrix}I_A=\dfrac{U}{R}=\dfrac{3,6}{4,8}=0,75A\\I12=\dfrac{U12}{R12}=\dfrac{3,6}{\dfrac{9\cdot18}{9+18}}=0,6A\end{matrix}\right.\)
