Question

如图所示电路,电源电压不变,R1=12 Ω,小灯泡标有“6V 12W”（电阻不变）．求：（1）灯

Answer 1

（1）RL=U额^2/P额=36/12=3Ω（2）U=I(RL+R1)=0.8(3+12)=12V（3）P1=[U/(RL+R2)]^2*R2 (1)P2=U^2/0.25*R2 (2)(1)/(2)且将RL=3Ω带入R2=6Ω则P1=[U/(RL+R2)]^2*R2 =[12/9]^2*6=10.66W