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NEET UG PHYSICS MCQ: CURRENT ELECTRICITY

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Application +4 | -1
The number of electrons flowing per second in the filament of a 110 W110 \mathrm{~W} bulb operating at 220 V220 \mathrm{~V} is :

(Given e=1.6×10−19C\mathrm{e}=1.6 \times 10^{-19} \mathrm{C})
View answer and explanation
Correct option: A
31.25×101731.25 \times 10^{17}
Option A (31.25×101731.25 \times 10^{17}) is correct because it is calculated by dividing the power by the voltage to find the current, then dividing the current by the elementary charge to find the number of electrons per second.

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