Question

The number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V is: (Given $e = 1.6 \times 10^{-19}$ 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 1

The Correct Option is A

Given:
\[\text{Power (P)} = V \cdot I\]
\[110 = 220 \times I\]
\[I = 0.5 \, \text{A}\]
Now, we know:
\[I = \frac{n \cdot e}{t}\]
Substitute the values:
\[0.5 = \frac{n \times (1.6 \times 10^{-19})}{t}\]
Rearrange to solve for \( n \):
\[\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}}\]
\[\frac{n}{t} = 31.25 \times 10^{17}\]