Current in DC Circuit: 8.41 x 10^19 Electrons in 2s

[jst]

Homework Statement

In a DC circuit 8.41 x 10^19 electrons travel through a point in the circuit in two seconds. What current is in the circuit?

Homework Equations

I'm not sure at all on this. I was thinking it would be something like (charge*#electrons)/time, but I don't have a value for the charge

The Attempt at a Solution

Would it possibly be: (6.241 509 629 152 65×10^18 * 8.41 x 10^19)/2


Thanks,

Jason

 

[hage567]

6.241 509 629 152 65×10^18

This is how many electrons are needed for one coulomb of charge. What you want is to find the charge of one electron, if you are going to use your equation the way it is.

 

[rdx]

As I understand the problem you need to convert the number of electrons to amperes. The ampere is a coulomb/sec and 1 electron is 1.602e-19 coul. I get 6.75 amps.

 

[Kurdt]

1 ampere is 6.24x10¹⁸ electrons passing a point in a second. Since you know how many pass in two seconds you need to divide half the number given in the question by the number given in the definition.

EDIT: All posted in the same minute and I'm last Figures

 

[jst]

Thanks to everyone! It's funny, I think I might have figured out before reading your posts.

See what you think:

=((# of electrons)*(charge of electron))/time

=
((8.41*10^19)*(1.6*10^-19))/2

=6.728


...at least it maches up with one of the possible solutions :)

Thanks again,

Jason

 

[jst]

It was right...big thanks to everyone.

Jason