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

Asked in JEE Main 6th April 2nd Shift 2024 · Rated bulbs and appliances

(1) 1.25×10¹⁹
(2) 6.25×10¹⁸
(3) 6.25×10¹⁷
(4) 31.25×10¹⁷

Answer: (4) 31.25×10¹⁷

Step-by-step solution

I=P/V=(110)/(220)=0.5 A.

n=I/e=(0.5)/(1.6×10⁻¹⁹)=3.125×10¹⁸ s⁻¹.

This is 31.25×10¹⁷ per second.

Why the other options are wrong

(1) Inverts the ratio, taking I=V/P=2 A.
(2) Takes I=1 A, i.e. just 1/e.
(3) Power-of-ten slip on 1/e (would need I=0.1 A).

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