The charge that passes through the filament of a certain light bulb in 5.00 s is 3.0 C. a. What is the current in the light bulb? b. How many electrons pass through the filament of the light bulb in a time interval of 1.0 min?

(a) I = 0.6 A

(b) 22.5 * 10^19 electrons

Explanation:

Parameters given:

(a) Current is given as the time Tate of charge flow. Mathematically,

I = Q/t

Charge, Q = 3 C

Time, t = 5 secs

I = 3/5 = 0.6A

(b) The current flowing through the filament is 0.6A. In 1 min (60 secs), the charge will be:

Q = I*t

Q = 0.6 * 60 = 36 C

An electron has a charge, Qe, of 1.6022 * 10^(-19) C. Hence, the number of electrons will be:

N = Q/Qe

N = 36/(1.6022 * 10^(-19))

N = 22.5 * 10^19 electrons