帮我解决一道物理题...矩形金属框,可动边AB长为0.10m,电阻为0.20Ω,CD边电阻为0.80Ω,导轨电阻不计,匀
2个回答

(1)E=Blv=0.5*0.1*15=0.75v

(2) I=E/(R1+R2)=0.75/(0.2+0.8)=0.75A

(3) U=IR2=0.75*0.8=0.6v

(4) p=EI=0.75*0.75=0.56w

注意哦,AB相当于电源,接了一个电阻CD,AB两端电压就是路端电压,第(3)问中可别代入了电源内阻算路端电压.