Question 21.74: A 100 W lamp has a steady current of 0.83 A in its filament.......

Electric current i is the rate dq/dt at which charge passes a point. With i = 0.83 A, the time it takes for one mole of electron to pass through the lamp is

\Delta t=\frac{\Delta q}{i}=\frac{N_A e}{i}=\frac{\left(6.02 \times 10^{23}\right)\left(1.6 \times 10^{-19} C \right)}{0.83\,A }=1.16 \times 10^5 s \approx 1.3 \,\text { days. }