解：$(1) $由图可知，闭合开关，$R_{1}$、$R_{2} $并联，电流表测量干路电流，电压表测量电源电压.
$R_{1} $两端的电压$U_{1}=U_{2}=U=6\ \mathrm {V}$，通过$ R_{1} $的电流$ I_{1}=\frac {U_{1}}{R_{1}}=\frac {6\ \mathrm {V}}{30\ \mathrm {Ω}}=0.2\ \mathrm {A}.$
$(2) $通过$ R_{2} $的电流$ I_{2}=I-I_{1} =0.5\ \mathrm {A}-0.2\ \mathrm {A}=0.3\ \mathrm {A}$，电阻$R_{2} 1 \mathrm {\mathrm {min}}$消耗的电能$W_{2}=U_{2}I_{2}t=6\ \mathrm {V}×0.3\ \mathrm {A}×60\ \mathrm {s}=108\ \mathrm {J}.$
