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) (a) 1.25 × 10¹⁹ (b) 31.25 × 10¹⁷ (c) 6.25 × 10¹⁸ (d) 6.25 × 10¹⁷

Answer: (b). Current in the bulb I = P/V = (110)/(220) = 0.5 A Electrons per second n = I/e = 0.51.6 × 10⁻¹⁹ = 3.125 × 10¹⁸ = 31.25 × 10¹⁷ So the answer is (b) 31.25 × 10¹⁷.

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Electric Current, Drift Velocity & Current Density

The number of electrons flowing per second in the filament of a $110\text{ W}$ bulb operating at $220\text{ V}$ is: (Given $\text{e} = 1.6 \times 10^{-19}\text{ C}$ )

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Answer: (b)

Current in the bulb $I = \dfrac{P}{V} = \dfrac{110}{220} = 0.5\text{ A}$
Electrons per second $n = \dfrac{I}{e} = \dfrac{0.5}{1.6 \times 10^{-19}} = 3.125 \times 10^{18}$
$= 31.25 \times 10^{17}$
So the answer is (b) $31.25 \times 10^{17}$ .

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