JEE Mains · Physics · STD 12 - 3. current electricity
The number of electrons flowing per second in the filament of a \(110 \mathrm{~W}\) bulb operating at \(220 \mathrm{~V}\) is _______. (Given \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) )
- A \(31.25 \times 10^{17}\)
- B \(6.25 \times 10^{18}\)
- C \(6.25 \times 10^{17}\)
- D \(1.25 \times 10^{19}\)
Answer & Solution
Correct Answer
(A) \(31.25 \times 10^{17}\)
Step-by-step Solution
Detailed explanation
\(\text { Power }(P)=V . I\) \(\Rightarrow 110=(220)(I)\) \(\Rightarrow I=0.5 \mathrm{~A}\) \(\text { Now, } I=\frac{\mathrm{n} \cdot \mathrm{e}}{\mathrm{t}}\) \(\Rightarrow 0.5=\left(\frac{\mathrm{n}}{\mathrm{t}}\right)\left(1.6 \times 10^{-19}\right)\)…
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