Current Electricity

JEE Main level, single correct, medium

230 of 380

Question 230 of 380

The number of electrons flowing per second in the filament of a 110 W110 \mathrm{~W} bulb operating at 220 V220 \mathrm{~V} is : (\left(\right. Given e=1.6×10−19C)\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)

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Answer: (D) 31.25×101731.25 \times 10^{17}

Power (P)=V.I(\mathrm{P})=\mathrm{V} . \mathrm{I}
$\begin{aligned}
& \Rightarrow 110=(220)(\mathrm{I}) \\
& \Rightarrow \mathrm{I}=0.5 \mathrm{~A}
\end{aligned}$
Now, I=n⋅etI=\frac{n \cdot e}{t}
$\begin{aligned}
& \Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right) \\
& \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}} \\
& \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17}
\end{aligned}$

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