解：$(1)$当开关$S $闭合时，$R_{1}$与$R_{2}$串联，$R_{1}$两端的电压$U_{1}=IR_{1}=0.2\ \mathrm {A}×20\ \mathrm {Ω}=4\ \mathrm {V}$；
$R_{2}$两端的电压$U_{2}=U-U_{1}=6\ \mathrm {V}-4\ \mathrm {V}=2\ \mathrm {V}$，由$I=\frac {U}{R}$得，$R_{2}$的阻值$R_{2}=\frac {U_{2}}{I}=\frac {2\ \mathrm {V}}{0.2\ \mathrm {A}}=10\ \mathrm {Ω}$
$(2)$通电$1 \mathrm {\mathrm {min}}$，电流通过$R_{1}$做的功$W_{1}=U_{1}It=4\ \mathrm {V}×0.2\ \mathrm {A}×60\ \mathrm {s}=48\ \mathrm {J}$
$(3)$电路消耗的电功率$P=UI=6\ \mathrm {V}×0.2\ \mathrm {A}=1.2\ \mathrm {W}$
