I=I_额=

P额

U额

=

40W

36V

≈1.1A．
R=

U 2额

P额

=

(36V)2

40W

=32.4Ω．
P_实=

U 2实

R

=

(40V)2

32.4Ω

≈49.4W．
答：灯泡正常发光时通过它的电流约1.1A．灯丝的电阻约32.4Ω；若接到40V的电路上时，实际功率约是49.4W．