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Current Electricity question

2024 · 6 Apr · Shift 2 · Q64
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Current Electricity question

2024 · 6 Apr · Shift 2 · Q64

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The 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)
  1. A
    1.25×10191.25 \times 10^{19}1.25×1019
  2. B
    31.25×101731.25 \times 10^{17}31.25×1017
  3. C
    6.25×10186.25 \times 10^{18}6.25×1018
  4. D
    6.25×10176.25 \times 10^{17}6.25×1017
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