The number of electrons flowing per second in the filament of a $110 \mathrm{~W}$ bulb operating at $220 \mathrm{~V}$ is : (Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ )
JEE MAIN 2024, Diffcult
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$\text { Power }(P)=V . I$
$\Rightarrow 110=(220)(I)$
$\Rightarrow I=0.5 \mathrm{~A}$
$\text { Now, } I=\frac{\mathrm{n} \cdot \mathrm{e}}{\mathrm{t}}$
$\Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right)$
$\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}}$
$\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17}$
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