JEE Main 20246 Apr, Evening shiftPhysicsEasyCurrent Electricity150× since 2002The number of electrons flowing per second in the filament of a 110 W110 \mathrm{~W}110 W bulb operating at 220 V220 \mathrm{~V}220 V is : (Given e=1.6×10−19C\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}e=1.6×10−19C)AOption A. 1.25×10191.25 \times 10^{19}1.25×1019BOption B. 31.25×101731.25 \times 10^{17}31.25×1017COption C. 6.25×10186.25 \times 10^{18}6.25×1018DOption D. 6.25×10176.25 \times 10^{17}6.25×1017CheckSkip · show solution
