The number of electrons flowing per second in the filament of a $110 \mathrm{~W}$ bulb operating at $220 \mathrm{~V}$ is : (Given $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$)
Solution
<p>$$\begin{aligned}
& P=v \times i \Rightarrow i=\frac{110}{220}=\frac{1}{2} \mathrm{~A} \\
& i=n e \Rightarrow n=\frac{1}{2 \times 1.6 \times 10^{-19}}=31.25 \times 10^{17}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Electrical Power
This question is part of PrepWiser's free JEE Main question bank. 120 more solved questions on Current Electricity are available — start with the harder ones if your accuracy is >70%.