Question

一道物理题!!!在线等!!!

额定电压为U，额定功率为P1的电灯L1与额定电压为U，额定功率为P2电灯L2串联在电源电压为U的电路中，证明电路的总功率P等于P1P2/(P1+P2)

Answer 1

L1电阻 R1 = U*U/P1
L2电阻 R2 = U*U/P2
总功率 P = U*U/(R1+R2) = U*U/(U*U/P1+U*U/P2) = P1P2/(P1+P2)

Answer 2

R1=U²/P1，R2=U²/P2
R=R1+R2=U²(1/P1+1/P2) = U²(P1+P2)/P1P2
P=U²/R = P1P2/(P1+P2)

Answer 3

R1=U²/P1，R2=U²/P2
R=R1+R2=U²(1/P1+1/P2) = U²(P1+P2)/P1P2
P=U²/R = P1P2/(P1+P2)

追问

你是复制的还是咋滴- -

Answer 4

首先算电阻R1=U²/P1，R2=U²/P2，总功率P=U²/(R1+R2),将R1，R2代入，得：
P=P1P2/(P1+P2)。

Answer 5

P=UI
=U^2/R
=U^2/(R1+R2)
=U^2/(U^2/P1+U^2/P2)
=1/(1/P1+1/P2)
P1P2/(P1+P2)