A9.0-v battery is connected to a bulb whose resistance is 2.5 ω . part a how many electrons leave the battery per minute?

As the problem depicts the 9v battery is connected to the bulb of resistance 2.5ω, the electrons leaving the battery per minute are electrons.

Ohms law states that the voltage across a conductor is directly proportional to the current flowing through it.

According to the ohms law,

where,

V= Voltage

I= Current

R= Resistance

Putting the given data we get,

Now we have to find the electrons leaving per minute so,

As we know each electron has a charge of

So no. of electrons leaving per minute =

= electrons/minute

To know more about Ohms law,

link

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.

i would say