解：?$(1)U_{0}=IR_{0}=0.015\ \text {A}×400 \ Ω=6\ \text {V}$?
?$P_{0}=U_{0}I=6\ \text {V}×0.015\ \text {A}=0.09\ \text {W}$?
?$(3)R'=\frac {U}{I}=\frac {7.5\ \text {V}}{0.015\ \text {A}}=500 \ Ω$?
?$R=R'-R_{0}=500 \ \mathrm {Ω}- 400 \ \mathrm {Ω}= 100 \ \mathrm {Ω}$?
根據(jù)?$R $?和?$F $?的關(guān)系可知：量程為?$0\sim 500\ \text {N}$?
?$(4)$?方法一：換?$R_{0} ，$?增大?$R_{0}$?的阻值；
方法二：換電源，減小電源電壓
