Question

如图，灯L1标有‘6v 3w‘，灯L2标有“12V 4W”，将开关闭合后L1恰好正常发光。求:

Answer 1

解
1.
开关闭合后L1与L2并联

电压表测电源电压
即
L1也是
L2两端电压

而L1恰好正常发光

∴U=U01=6V

2.RL2=U02^2/p=(12V)^2/4W=36Ω
I2=U/R2=6V/36Ω=0.17A
3.P2实=I2^2R
2=(1/6A)^2*36Ω=1W

W=P2实t=1W*60s=60J
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Answer 2

(1)开关闭合后，电路为并联电路
∵L1恰好正常发光
∴U1=U2=U=6V
(2)R2=（U额）²/P额=(12V)²/4W=36Ω
I2=U2/R2=6V/36Ω≈0.17A
(3)P实=U2I2=0.17A×6V=1.02W
Q=I²Rt=(0.17A)²×36Ω
×
60S=62.424J