Ta có: \(R_1+R_2=\dfrac{U}{I}=\dfrac{12}{0,3}=40\left(1\right)\)\(\Rightarrow R_1=40-R_2\)
\(\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{12}{1,6}=7,5\)
\(\Rightarrow R_1\cdot R_2=7,5\cdot40=300\)
\(\Rightarrow\left(40-R_2\right)\cdot R_2=300\)\(\Rightarrow\left[{}\begin{matrix}R_2=30\Omega\\R_2=10\Omega\end{matrix}\right.\)
Ta có 2 giá trị \(R_1\)tương ứng\(\Rightarrow\left[{}\begin{matrix}R_1=40-30=10\Omega\\R_1=40-10=30\Omega\end{matrix}\right.\)
Tóm tắt:
\(U=12V\)
\(I=0,3A\)
\(I'=1,6A\)
\(R_1=?\)
\(R_2=?\)
Giải:
Vì: \(R_1ntR_2\)
\(R_{tđ}=R_1+R_2=\dfrac{U}{I}=\dfrac{12}{0,3}=40\Omega\)
\(\Leftrightarrow R_1=40-R_2\left(1\right)\)
Vì: \(R_1\text{/ /}R_2\)
\(R'_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{U}{I'}=\dfrac{12}{1,6}=7,5\Omega\left(2\right)\)
Thế (1) và (2)
\(\dfrac{\left(40-R_2\right)R_2}{40}=7,5\)
\(\Leftrightarrow\left(40-R_2\right)R_2=7,5.40\)
\(\Leftrightarrow-R^2_2+40R_2-300=0\)
\(\Leftrightarrow R^2_2-40R_2+300=0\)
\(\Leftrightarrow R^2_2-10R_2-30R_2+300=0\)
\(\Leftrightarrow R_2\left(R_2-10\right)-30\left(R_2-10\right)=0\)
\(\Leftrightarrow\left(R_2-30\right)\left(R_2-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}R_2-30=0\\R_2-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}R_2=30\\R_2=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}R_1=40-30=10\\R_2=30\end{matrix}\right.\\\left\{{}\begin{matrix}R_1=40-10=30\\R_2=10\end{matrix}\right.\end{matrix}\right.\)
Vậy \(R_1=10\Omega\) , \(R_2=30\Omega\) hoặc \(R_1=30\Omega\), \(R_2=10\Omega\)
