That makes more sense. But it just leaves me with one question. How does this all relate to columbs?.
Like how is multiplying the number of electrons by the amount of electrons giving you columbs?
Your so observant I checked the question it is 0.1s. So it would be
0.48C/0.1s=4.8A
And my last question is would you happen to know the reason for this relationship. How Q(Charge in coloumbs or the number of electrons) = # of electrons*charge of one electron?
Well I guess one of my mistakes was I mixed up calculations. So this means the final answer should be
I=Q/t
0.48C/10s=0.048A
So that means 0.048A is the current/answer.
Right?
I am sorry again I made a mistake in the question I stated it but did not show it.
I multiplied the charge one electron by the number of electrons in in the circuit. So (1.60*10^-19 C)*( 3.0*10^18 ). This is how I got 0.00048 C.
This is what I meant.
Sorry about all this.
FOR VOKO
Sorry for the lack of information. I guess I need more sleep :) By 1.60*10^-19 I meant that is the charge of one electron in coulombs. So 1 electron=1.60*10^-19 C,
For 6.24*10^18 I meant that this is the amount of electrons that are in one coloumb. So 1 C = 6.24*10^18 .
The numbers I...
The question is:
What current is flowing in a metal wire if 3.0*10^18 electrons pass a given point in a time of 0,10s?
I know that 1 electron has a charge of 1.60*10^-19
I also know that one coulomb is 6.24*10^18
Lastly I know to find current I need to use the formula Q=I*t, but change it...