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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2024
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(31.25 \times 10^{17}\)

Step-by-step Solution

Detailed explanation

\(\text { Power }(P)=V . I\)

\(\Rightarrow 110=(220)(I)\)

\(\Rightarrow I=0.5 \mathrm{~A}\)

\(\text { Now, } I=\frac{\mathrm{n} \cdot \mathrm{e}}{\mathrm{t}}\)

\(\Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right)\)

\(\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{0.5}{1.6 \times 10^{-19}}\)

\(\Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=31.25 \times 10^{17}\)

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