甲灯的电阻R₁=

U 21

P1

=

(220V)2

40W

=1210Ω，乙灯的电阻R₂=

U 22

P2

=

(220V)2

60W

≈807Ω，
串联后电路电流I=

U

R1+R2

=

220V

1210Ω+807Ω

≈0.1A，
乙灯的实际功率P_实=I²R₂=（0.1A）²×807Ω=8.07W，

P实

P额

=

8.07W

60W

≈0.13．
故答案为：0.13．