解?$：(1)I=\frac {U}{R}=\frac {15\ \mathrm {V}}{10 \ \mathrm {Ω}}=1.5\ \mathrm {A}$?
?$(2)R_{總}(cāng)=R+R_{L}=10\ \mathrm {Ω}+R_{L}$?
?$I=\frac {U}{R_{總}(cāng)}=\frac {15\ \mathrm {V}}{10\ \mathrm {Ω}+R_{L}}①$?
由于燈泡正常工作?$，$?根據(jù)?$P=I^2R $?可知
?$I=I_{L}=\sqrt {\frac {P_{L}}{R_{L}}}=\sqrt {\frac {5\ \mathrm {W}}{R_{L}}}②$?
聯(lián)立?$①②$?解得?$R_{L}=5\ \mathrm {Ω}$?或?$20\ \mathrm {Ω}$?
當(dāng)?$R_{L{1}}=20\ \mathrm {Ω}$?時(shí)?$，I_{L{1}}=I=\frac {U}{R_{總}(cāng)}=\frac {15\ \mathrm {V}}{10 \ \mathrm {Ω}+20 \ \mathrm {Ω}}=0.5\ \mathrm {A}$?
此時(shí)燈泡電功率?$P_{L{1}}=I_{L{1}}^2R_{L}=(0.5\ \mathrm {A})^2×20\ \mathrm {Ω}=5\ \mathrm {W}$?
當(dāng)?$R_{L{2}}=5\ \mathrm {Ω}$?時(shí)?$，I_{L{2}}=I=\frac {U}{R_{總}(cāng)}=\frac {15\ \mathrm {V}}{10 \ \mathrm {Ω}+5 \ \mathrm {Ω}}=1\ \mathrm {A}$?
此時(shí)燈泡電功率?$P_{L{2}}=I_{L{2}}^2R_{L}=(1\ \mathrm {A})^2×5\ \mathrm {Ω}=5\ \mathrm {W}$?
所以兩個(gè)阻值都符合題意
根據(jù)?$I=\frac {U}{R}$?可得
當(dāng)?$R_{L{1}}=5\ \mathrm {Ω}$?時(shí)?$U_{L{1}}=I_{L{1}}R_{L}=1\ \mathrm {A}×5 \ \mathrm {Ω}=5\ \mathrm {V}$?
當(dāng)?$R_{L{2}}=20\ \mathrm {Ω}$?時(shí)?$U_{L{2}}=I_{L{2}}R_{L}=0.5\ \mathrm {A}×20 \ \mathrm {Ω}=10\ \mathrm {V}$?
所以燈泡的額定電壓為?$5\ \mathrm {V}{或}10\ \mathrm {V}$?
