\(R_1//R_2\Rightarrow U_1=U_2=U_{mạch}\)
\(R_{tđ}=\dfrac{U}{I}=\dfrac{18}{2}=9\Omega\Rightarrow\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{1}{R_1}+\dfrac{1}{60}=\dfrac{1}{9}\)
\(\Rightarrow R_1=\dfrac{180}{17}\Omega\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{U}{R_1}=\dfrac{18}{\dfrac{180}{17}}=1,7A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{18}{60}=0,3A\)
\(MCD:R_1//R_2\)
\(\Rightarrow R_{td}=\dfrac{U}{I}=\dfrac{18}{2}=9\Omega\)
Ta có: \(\dfrac{1}{R_{td}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\)
\(\Rightarrow\dfrac{1}{R_1}=\dfrac{1}{R_{td}}-\dfrac{1}{R_2}=\dfrac{1}{9}-\dfrac{1}{60}=\dfrac{17}{180}\)
\(\Rightarrow R_1\approx10,58\Omega\)
Ta có: \(U=U_1=U_2=18V\)
\(\Rightarrow\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{18}{\dfrac{180}{17}}=1,7A\\I_2=I-I_1=2-1,7=0,3A\end{matrix}\right.\)
