The number of electrons flowing per second in the filament of a $110 \mathrm{~W}$ bulb operating at $220…

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 $\left.\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right)$
  1. $6.25 \times 10^{17}$
  2. $1.25 \times 10^{19}$
  3. $6.25 \times 10^{18}$
  4. $31.25 \times 10^{17}$

Solution

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

Asked in: JEE Main 2024 (06 Apr Shift 2)

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