The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e = 1.6 10-19 C)
- 1.25 1019
- 31.25 1017
- 6.25 1018
- 6.25 1017
Answer
(2) 31.25 1017
Solution
P = VI I = 110220 = 0.5 A
I = qt = nte (n = number of electrons):
nt = Ie = 0.51.6 10-19
= 3.125 1018 = 31.25 1017 per second
Answer: (2)