∵P= U2 R ,∴灯泡电阻:RL= U PL额 = (8V)2 2W =32Ω,由电路图可知,当开关S闭合时只有灯泡接入电路,此时灯泡正常发光,则电源电压:U=UL额=8V,当开关S断开时,灯泡与电阻串联接入电路,电路电流I= U R+RL = 8V 8Ω+32Ω =0.2A,灯泡实际功率:P=I2RL=(0.2A)2×32Ω=1.28W;故答案为:1.28.