A 9.0-v battery is connected to a bulb whose resistance is 2.5 ω. part a how many electrons leave the battery per minute?

Solution:

According to the formula I=V/R

so 9.0v/2.5ohm=3.6A

since 3.6A=3.6c/s

therefore for per minute

3.6*60=216c leaving the battery.

Now,

Each electron carries a charge of 1.6*10^ (-19) c

so 216 / 1.6*10^ (-19) = 1.35*10^21c

1.35*10^21c this is the required electrons leaving the battre per minute.