(1)灯丝的电阻:R= U P额 = (220V)2 100W =484Ω;(2)通过灯丝的电流:∵P=UI,∴I额= P额 U额 = 100W 220V ≈0.45A;(3)t=2h=7200s,灯泡消耗的电能:W1=Pt1=P额t1=100W×7200s=7.2×105J.答:(1)灯丝的电阻为484Ω;(2)通过灯丝的电流为0.45A;(3)灯泡消耗的电能为7.2×105J.