Verified Solution • 2025–26 Session moderate

QJEEM24Phf2f1: JEE Main 2024 Physics - Class 12 JEEM24Phf2f1 Solution: The number of electrons flowing per second in the filament of a …

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is : (Given e=1.6×10−19C)

Multiple Choice Options

  • A 1.25×1019
  • B 31.25×1017
  • C 6.25×1018
  • D 6.25×1017

Step-by-Step Verified Solution

P=v×i⇒i=110220=12 Ai=ne⇒n=12×1.6×10−19=31.25×1017

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