?$ 解：(1)R_{ L }=\frac {U_{L }^2}{P_{ L }}=\frac {({ 6 }\ \text {V})^2}{{ 3 }\ \text {W}}={ 12 }Ω$?
?$ (2)由圖可知，電源電壓U=6\ \text {V}，U=6\ \text {V}時(shí)，I_R=0.5\ \text {A}$?
?$ I_{ L }=\frac {P_{L }}{U_{ L}}=\frac {{ 3 }\ \text {W}}{{ 6}\ \text {V}}={0.5 }\ \text {A}$?
?$ I=I_R+I_L=0.5\ \text {A}+0.5\ \text {A}=1\ \text {A}$?
?$ (3)由圖知I=0.35\ \text {A}時(shí)，U_R=2.5\ \text {V}，$?
?$ U_{P }=I_{ }R_{ P }={ 0.35 }\ \text {A}×{ 2 }\ Ω={ 0.7}\ \text {V}$?
?$ U_L=U-U_R-U_P=6\ \text {V}-2.5\ \text {V}-0.7\ \text {V}=2.8\ \text {V}$?
?$ P_{ L }=U_{ L }I_{ }={ 2.8 }\ \text{V}×{ 0.35 }\ \text{A}={ 0.98 }\ \text{W}$
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