Đáp án:
$\begin{array}{l} R_1=4 \ \Omega \\ R_2=6 \ \Omega \end{array}$
Giải:
Khi `R_1 \ nt \ R_2`:
`R_{td}=\frac{U}{I}=\frac{12}{1,2}=10 \ (\Omega)`
→ `R_1+R_2=R_{td}=10` (1)
Khi $R_1 \ // \ R_2$:
$R'_{td}=\dfrac{U}{I'}=\dfrac{12}{5}=2,4 \ (\Omega)$
→ $\dfrac{R_1R_2}{R_1+R_2}=R'_{td}=2,4$ (2)
Thay (1) vào (2)
→ `\frac{R_1(10-R_1)}{10}=2,4`
→ `10R_1-R_1^2=24`
→ $\left [\begin{array}{l} R_1=4 \ (\Omega) & (nhan) \\ R_1=6 \ (\Omega) & (loai) \end{array} \right. \ (R_1<R_2)$
→ `R_2=6 \ \Omega`