解：$(1)U=U_{额}=6\ \text {V}$
$(2)I_1=I-I_L=1.2\ \text {A}-1\ \text {A}=0.2\ \text {A}$
$R_{ 1 }=\frac {U_{ }}{I_{ 1}}=\frac {{ 6 }\ \text {V}}{{ 0.2 }\ \text {A}}={ 30 }Ω$
$(3)I_{ }=\frac {U_{ }}{R_{ 1}+R_2}=\frac {{ 6 }\ \text {V}}{{ 30 }Ω+3Ω}≈{0.18 }\ \text {A}$
