解:閉合開關(guān)$S_{1}$、$S_{2}$時,$R_{2}$短路,$R_{1}=\frac{U}{I}=\frac{9\ \text{V}}{1.5\ \text{A}}=6\ \Omega;$斷開$S_{2}$后,$R_{1}$與$R_{2}$串聯(lián),電壓表示數(shù)6 V為$R_{1}$兩端電壓,電流$I=\frac{6\ \text{V}}{6\ \Omega}=1\ \text{A},$$R_{2}$兩端電壓$9\ \text{V}-6\ \text{V}=3\ \text{V},$$R_{2}=\frac{3\ \text{V}}{1\ \text{A}}=3\ \Omega。$