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Question

Physics Question on Current electricity

A 60 W60\, W incandescent lamp operates at 120 V120\, V. The number of electrons passing through the filament per second will be

A

1.61×10121.61 \times 10^{12}

B

3.12×10183.12 \times 10^{18}

C

7.21×10127.21 \times 10^{12}

D

12.40×101312.40 \times 10^{13}

Answer

3.12×10183.12 \times 10^{18}

Explanation

Solution

The power of the incandescent lamp P = 60 W Operating voltage V = 120 V ? Current through the filament I=PV=60120=12I=\frac{P}{V}=\frac{60}{120}=\frac{1}{2} A Now Q = I ? t For t = 1 sec, Q=I=12Q=I=\frac{1}{2} C ⇒ne=12\Rightarrow ne =\frac{1}{2} where e = electronic charge, and n = number of electrons. ∴n=12e=12×1.6×1019=3.12×1018\therefore n=\frac{1}{2e}=\frac{1}{2\times1.6\times10^{19}}=3.12\times10^{18}