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