00:01
So in this problem, we have some junction, and we are told that wire 1 will go into the junction with a current of 0 .4 amps, and then we have wire 2 coming out of the junction with the current i2 as 0 .65 amps.
00:26
And for part a, we want to know the number of electrons flowing through some wire 3, which we don't know where wire 3 is that.
00:37
But we do know that when we're dealing with currents flowing in through a junction, the current flowing into the junction, i -n, has to equal the current coming out of the junction.
00:50
We'll call i -out.
00:51
So we know that the current in is 0 .4, and the current out, this point six five so to find the wire three since we know that both sides of the equation have to be equal then we just need to subtract i out from i in so we get 0 .65 minus 0 .4 and this will give us the current in the wire 3 which will call i3 so we get 0 .25 amps so that's the current and now we need to know the number of electrons it's passing through that wire.
01:36
But we know that current point for wire 3, 0 .25 amps.
01:43
Amps is really cool homes per second.
01:47
So the amount of charge flowing through the wire per second.
01:51
So if we just multiply by one over the charge of the electrons, we get the number of electrons...