解：?$(1)$?應(yīng)滑到?$a$?端
?$I_{ }=\frac {U_{ }}{R_{3 }+R_{ 1}}=\frac {{ 12 }\ \text {V}}{{ 20 }Ω+{ 10 }Ω}={ 0.4 }\ \text {A}$?
?$U_{ 3}=I_{ }R_{ 3 }={0.4 }\ \text {A}×{ 20 }\ Ω={ 8 }\ \text {V}$?
?$Q_{ 1 }=I_{ }^2R_{1 }t_{ }=({0.4 }\ \text {A})^2×{ 10 }Ω×{ 10×60 }\ \text {s}={ 960 }\ \text {J}$?
?$(2)P_{ 最大 }=U_{ }I_{最大 }={ 12 }\ \text{V}×{ 0.6 }\ \text{A}={ 7.2 }\ \text{W}$?
