(1)根据串联电路总电阻$R=R_{L}+R=10\ \Omega+20\ \Omega=30\ \Omega，$电流$I=\frac{U}{R}=\frac{6\ \text{V}}{30\ \Omega}=0.2\ \text{A}；$
(2)总电阻$R'=10\ \Omega+10\ \Omega=20\ \Omega，$电流$I'=\frac{6\ \text{V}}{20\ \Omega}=0.3\ \text{A}，$灯泡两端电压$U_{L}=I'R_{L}=0.3\ \text{A}\times10\ \Omega=3\ \text{V}。$
