解：$(1)I_{1}=\frac {U}{R_{1}}=\frac {15\ \mathrm {V}}{10 \ \mathrm {Ω}}=1.5\ \mathrm {A}$
$(2)①$并联，理由如下：
$I_{\mathrm {\mathrm {min}}}=\frac {P_{\mathrm {\mathrm {min}}}}{U}=\frac {27\ \mathrm {W}}{15\ \mathrm {V}}=1.8\ \mathrm {A}>1.5\ \mathrm {A}$
$②I_{2\mathrm {\mathrm {min}}} =I_{\mathrm {\mathrm {min}}}-I_{1} =1.8\ \mathrm {A}-1.5\ \mathrm {A}=0.3\ \mathrm {A}$
$R_{2\ \mathrm {max}}=\frac {U}{I_{2\mathrm {\mathrm {min}}}}=\frac {15\ \mathrm {V}}{0.3\ \mathrm {A}}=50 \ \mathrm {Ω}$
$③I_{\mathrm {max}}=I_{1}+I_{2\ \mathrm {max}}=1.5\ \mathrm {A}+0.6\ \mathrm {A}=2.1\ \mathrm {A}$
$P_{\mathrm {max}}=UI\mathrm {max}=15\ \mathrm {V}×2.1\ \mathrm {A}=31.5\ \mathrm {W}$
