We will have the following:
1)
[tex]I=\frac{V}{R}\Rightarrow I=\frac{15V}{120000\Omega}\Rightarrow I=1.25\cdot10^{-4}A[/tex][tex]\Rightarrow I=0.125mA[/tex]2)
First we determine the quantity of coulombs:
[tex]Q=(1.25\cdot10^{-4}A)(1s)\Rightarrow Q=1.25\cdot10^{-4}C[/tex]Now, we know that 1 coulomb represents 6.24*10^18 electrons, so:
[tex]1.25\cdot10^{-4}C\cdot\frac{6.24\cdot10^{18}\text{electrons}}{1C}=7.8\cdot10^{14}\text{electrons}[/tex]So, approximately 7.8*10^14 electrons pass each second.
