设灯的电阻R1,电阻为R2
串联 I 相等 I²R1=81,I²R2=9所以R1=9R2
P=U²/(R1+R2)=U²/10R2=81+9=90,即U²/10R2=90所以U²/R2=900
进而可知U²/9R2=100,也即U²/R1=100W,把电阻去掉后,电源电压全加到R1上
故:灯泡的P额=100W
设灯的电阻R1,电阻为R2
串联 I 相等 I²R1=81,I²R2=9所以R1=9R2
P=U²/(R1+R2)=U²/10R2=81+9=90,即U²/10R2=90所以U²/R2=900
进而可知U²/9R2=100,也即U²/R1=100W,把电阻去掉后,电源电压全加到R1上
故:灯泡的P额=100W