将一个标有“6V 3W”的灯泡和一个18Ω的电阻串联后接到12V的电源上，此时电路中的电流是______A，灯泡的实际电功率为______W．
根据P=
可得：U2 R
灯泡的电阻RL=
=UL额2 PL额
=12Ω，(6V)2 3W
灯泡和电阻串联后接到12V的电源上时，电路中的电流：
I=
=U R+RL
=0.4A，12V 18Ω+12Ω
根据P=I2R可得：
灯泡的实际电功率为PL实=I2RL=（0.4A）2×12Ω=1.92W．
故答案为：0.4；1.92．
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