解：?$(1)R_{1}=\frac {U_{1}}{I_{1}}=\frac {2\ \text {V}}{ 0.2\ \text {A}}=10 \ Ω$?
?$(2)U_{2}=U-U_{1}=6\ \text {V}- 2\ \text {V}=4\ \text {V}$?
?$P_{2}=U_{2}I= 4\ \text {V}×0.2\ \text {A}=0.8\ \text {W}$?
?$(3)$?當(dāng)?$I=0.6\ \text {A}$?時，?$U_{1}=IR_{1}=0.6\ \text {A}×10 \ Ω=6\ \text {V}，$?超出電壓表量程
所以?$R_{1}$?功率最大時，?$U_{1}=3\ \text {V}，$?
?$P_{大}=\frac {U_1^{2}}{R}=\frac {(3\ \text {V})^2}{10 \ Ω}=0.9\ \text {W}$?
當(dāng)?$R_{2}=50 \ \mathrm {Ω}$?時，?$R_{1}$?功率最小，
此時?$I_{ 小}=\frac {U_{ }}{R_{1 }+R_{2}}=\frac {6\ \text {V}}{10 \ Ω+ 50 \ Ω}=0.1\ \text {A}$?
?$P_{小}=I_{小}^2R_{1}=(0.1\ \text {A})^2×10 \ Ω=0.1\ \text {W}$?
所以?$R_{1}$?功率的范圍為?$0.1\sim 0.9\ \text {W}$?
