解：$(1) S_{1}$闭合$，S_{2}$断开
$I_{ }=\frac {U_{ }}{R_{ 1}+R_{2 }}=\frac {{ 6 }\ \text {V}}{{ 10 }Ω+{ 20}Ω}={ 0.2}\ \text {A}$
$P_{ 低 }=I_{ }^2R_{ 2 }=({ 0.2 }\ \text {A})^2×{ 20 }Ω={ 0.8}\ \text {W}$
$(2) S_{1}、$$S_{2}$均闭合
$R_{ }=\frac {R_{2}R_{3}}{R_{2}+R_{3}}=\frac {{ 20 }\ \mathrm {Ω}×{10 }\ \mathrm {Ω}}{{ 20 }\ \mathrm {Ω}+{10 }\ \mathrm {Ω}}={ \frac {20}3 }\ \mathrm {Ω}$
$P_{ 高 }=\frac {U_{ }^2}{R_{ }}=\frac {({6 }\ \text {V})^2}{{ \frac {20}3 }Ω}={ 5.4 }\ \text {W}$
$(3)$较好的电路是$(\mathrm {b})，$因为在电路$(\mathrm {a})$中$，$低温挡时$R_{1}$要发热$，$浪费电能
