解：$(1)R_{1}=\frac {U_{1}}{I_{1}}=\frac {2\ \text {V}}{ 0.2\ \text {A}}=10 \ Ω$
$(2)U_{2}=U-U_{1}=6\ \text {V}- 2\ \text {V}=4\ \text {V}$
$P_{2}=U_{2}I= 4\ \text {V}×0.2\ \text {A}=0.8\ \text {W}$
$(3)$当$I=0.6\ \text {A}$时，$U_{1}=IR_{1}=0.6\ \text {A}×10 \ Ω=6\ \text {V}，$超出电压表量程
所以$R_{1}$功率最大时，$U_{1}=3\ \text {V}，$
$P_{大}=\frac {U_1^{2}}{R}=\frac {(3\ \text {V})^2}{10 \ Ω}=0.9\ \text {W}$
当$R_{2}=50 \ \mathrm {Ω}$时，$R_{1}$功率最小，
此时$I_{ 小}=\frac {U_{ }}{R_{1 }+R_{2}}=\frac {6\ \text {V}}{10 \ Ω+ 50 \ Ω}=0.1\ \text {A}$
$P_{小}=I_{小}^2R_{1}=(0.1\ \text {A})^2×10 \ Ω=0.1\ \text {W}$
所以$R_{1}$功率的范围为$0.1\sim 0.9\ \text {W}$
