依题意可算出L1的电流为5W/6V=0.83A,电阻为6V/0.83A=7.23Ω
L2的电流为1W/3V=0.33A,电阻为3V/0.33A=9.09Ω
∵串联电路电流处处相等,为防止电灯烧坏
∴电流最大为0.33A
又∵L1、L2串联
∴R总=R1+R2
=7.23Ω+9.09Ω=16.32Ω
由欧姆定律I=U/R得U=IR
=0.33×16.32Ω
=5.386V
≈5.4V
依题意可算出L1的电流为5W/6V=0.83A,电阻为6V/0.83A=7.23Ω
L2的电流为1W/3V=0.33A,电阻为3V/0.33A=9.09Ω
∵串联电路电流处处相等,为防止电灯烧坏
∴电流最大为0.33A
又∵L1、L2串联
∴R总=R1+R2
=7.23Ω+9.09Ω=16.32Ω
由欧姆定律I=U/R得U=IR
=0.33×16.32Ω
=5.386V
≈5.4V