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Physics Question on Electric Current

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

A

31.25×101731.25 \times 10^{17}

B

6.25×10186.25 \times 10^{18}

C

6.25×10176.25 \times 10^{17}

D

1.25×10191.25 \times 10^{19}

Answer

31.25×101731.25 \times 10^{17}

Explanation

Solution

Given:
Power (P)=V⋅I\text{Power (P)} = V \cdot I
110=220×I110 = 220 \times I
I=0.5 AI = 0.5 \, \text{A}
Now, we know:
I=n⋅etI = \frac{n \cdot e}{t}
Substitute the values:
0.5=n×(1.6×10−19)t0.5 = \frac{n \times (1.6 \times 10^{-19})}{t}
Rearrange to solve for nn:
nt=0.51.6×10−19\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}}
nt=31.25×1017\frac{n}{t} = 31.25 \times 10^{17}