条件不足,
设灯泡阻值为R,于是灯泡分压为6R/(R+4)
于是有
2 = [6R/(R+4)]²/R
解方程得
2 = 36R/(R+4)²
(R+4)² = 18R
R² - 10R + 16 = 0
R = 2或R = 8
因此灯泡电阻为2Ω或8Ω