解：$(1)$当闭合开关$S $和$S_{1}$，断开开关$S_{2}$时，$R_{1}$被短路，$R_{2}$为断路，电路为$R_{3}$的简单电路，则$R_{3}$两端的电压$U_{3}=U=9\ \mathrm {V}$
$(2)$当开关$S$、$S_{1}$和$S_{2}$均闭合时，$R_{2}$、$R_{3}$并联，电流表测干路电流，由并联电路的电压规律可知，$R_{2}$、$R_{3}$两端电压$U_{2}=U_{3}=U=9\ \mathrm {V}$，则通过$R_{2}$的电流$I_{2}=\frac {U_{2}}{R_{2}}=\frac {9\ \mathrm {V}}{45\ \mathrm {Ω}}=0.2\ \mathrm {A}$，通过$R_{3}$的电流$I_{3}=\frac {U_{3}}{R_{3}}=\frac {9\ \mathrm {V}}{30\ \mathrm {Ω}}=0.3\ \mathrm {A}$，则电流表的示数$I=I_{2}+I_{3}=0.2\ \mathrm {A}+0.3\ \mathrm {A}=0.5\ \mathrm {A}$
$(3)$当闭合开关$S$，断开开关$S_{1}$和$S_{2}$时，$R_{1}$和$R_{3}$串联，电压表测$R_{3}$两端电压，$R_{1}$两端$ $电压$U_{1}=U-U_{3}'=9\ \mathrm {V}-6\ \mathrm {V}=3\ \mathrm {V}$，通过$R_{1}$的电流$I_{1}=\frac {U_{3}'}{R_{3}}=\frac {6\ \mathrm {V}}{30\ \mathrm {Ω}}=0.2\ \mathrm {A}$，$R_{1}$消耗的电功率$P=U_{1}×I_{1}=3\ \mathrm {V}×0.2\ \mathrm {A}=0.6\ \mathrm {W}$
