JEE MAIN - Physics (2024 - 6th April Evening Shift - No. 5)
The number of electrons flowing per second in the filament of a $$110 \mathrm{~W}$$ bulb operating at $$220 \mathrm{~V}$$ is : (Given $$\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$$)
$$1.25 \times 10^{19}$$
$$31.25 \times 10^{17}$$
$$6.25 \times 10^{18}$$
$$6.25 \times 10^{17}$$
Explanation
$$\begin{aligned}
& P=v \times i \Rightarrow i=\frac{110}{220}=\frac{1}{2} \mathrm{~A} \\
& i=n e \Rightarrow n=\frac{1}{2 \times 1.6 \times 10^{-19}}=31.25 \times 10^{17}
\end{aligned}$$
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