## College Physics (4th Edition)

(a) The current flows from the Anode to the Filament. (b) $I = 9.6\times 10^{-7}~A$
(a) The electrons move from the Filament to the Anode. However, we think of the direction of current as the movement of positive charges. Since electrons have negative charges, the current flows from the Anode to the Filament. (b) We can find the current: $I = (6.0\times 10^{12}~electrons/s)(1.6\times 10^{-19}~C/electron)$ $I = 9.6\times 10^{-7}~A$