解：$(1)U=IR_{1}+U_L=0.6\ \text {A}×10 \ Ω+12\ \text {V}=18\ \text {V}$
$ (2)I_{1}=\frac {U}{R_{1}}=\frac {18\ \text {V}}{10 \ Ω}=1.8\ \text {A}$
$ I_{2}=I'-I_{1}=1.9\ \text {A}-1.8\ \text {A}=0.1\ \text {A}$
$ R_{2}=\frac {U}{I_{2}}=\frac {18\ \text {V}}{0.1\ \text {A}}=180 \ Ω$
(3)R_{2\ \text {min}}=\frac {$U_{2\ \text {min}}$}${I_{max}}=\frac {(18-12)\ \text {V}}{0.6\ \text {A}}=10 \ Ω$
$ R_L=\frac {U_L}{I_L}=\frac {12\ \text {V}}{0.6\ \text {A}}=20 \ Ω$
$ R_{2\ \text {m}ax}∶R_L=U_2\ \text {m}ax∶U_L=15\ \text {V}∶(18-15)\ \text {V}=5$，得$R_{2\ \text {m}ax}=5R_L=100\ Ω$
$ $所以阻值范围是$10\sim 100 \ \mathrm {Ω}$
