|Two Bulbs in Series|
Lamp A by itself has power of P = I x V ; 60=I x 220 ; so I = 60/220 = 0.27 amperes.
Thus the lamp has a resistance V = I x R ; R=220/0.27 = 807 ohms.
Lamp B by itself has a power of P = I x V ; so I = 100/220 = 0.45 amperes.
Thus the lamp has a resistance V=I x R ; R=220/0.45 = 489 ohms.
Since the resistance in series is added up, the total resistance is 807+489 =1296 ohms and the current
I = V/R = 220/1296 = 0.17 Ampere
Since P= I x I x R
For 60W lamp A, P = 0.17 x 0.17 x 807 = 23.3 watts.
For 100W lamp B, P = 0.17 x 0.17 x 489 = 14.1 watts.
So, when they are wired in series, the bright one and the dim one appear to be reversed as to their respective "normal rated powers".