KCET · Physics · Current Electricity
The number of electrons moving per second through the filament of a lamp of \(60\) W operating at \(120\) V is nearly \((e = 1.6 \times 10^{-19}\text{ C})\) _____________.
- A \(6.2 \times 10^{18}\)
- B \(6.2 \times 10^{19}\)
- C \(3.1 \times 10^{18}\)
- D \(3.1 \times 10^{19}\)
Answer & Solution
Correct Answer
(C) \(3.1 \times 10^{18}\)
Step-by-step Solution
Detailed explanation
Power of the lamp, \(P = 60\) W
Operating voltage, \(V = 120\) V
Current through the filament, \(I = \dfrac{P}{V} = \dfrac{60}{120} = 0.5\) A
Let \(n\) be the number of electrons flowing per second. Then the current is given by \(I = ne\), where \(e\) is the elementary charge.
\(n = \dfrac{I}{e} = \dfrac{0.5}{1.6 \times 10^{-19}}\)
\(n = \dfrac{5}{16} \times 10^{19} = 3.125 \times 10^{18}\)
The number of electrons moving per second is nearly \(3.1 \times 10^{18}\).
Answer: \(3.1 \times 10^{18}\)
Operating voltage, \(V = 120\) V
Current through the filament, \(I = \dfrac{P}{V} = \dfrac{60}{120} = 0.5\) A
Let \(n\) be the number of electrons flowing per second. Then the current is given by \(I = ne\), where \(e\) is the elementary charge.
\(n = \dfrac{I}{e} = \dfrac{0.5}{1.6 \times 10^{-19}}\)
\(n = \dfrac{5}{16} \times 10^{19} = 3.125 \times 10^{18}\)
The number of electrons moving per second is nearly \(3.1 \times 10^{18}\).
Answer: \(3.1 \times 10^{18}\)
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