(1)根據(jù)串聯(lián)電路總電阻$R=R_{L}+R=10\ \Omega+20\ \Omega=30\ \Omega，$電流$I=\frac{U}{R}=\frac{6\ \text{V}}{30\ \Omega}=0.2\ \text{A}；$
(2)總電阻$R'=10\ \Omega+10\ \Omega=20\ \Omega，$電流$I'=\frac{6\ \text{V}}{20\ \Omega}=0.3\ \text{A}，$燈泡兩端電壓$U_{L}=I'R_{L}=0.3\ \text{A}\times10\ \Omega=3\ \text{V}。$
