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

To find the number of electrons flowing per second in the filament of a 110 W bulb operating at 220 V, we first need to determine the current flowing through the circuit using the power formula:

\(P = V \times I\), where \(P\) is the power, \(V\) is the voltage, and \(I\) is the current.

Given:

• Power \(P = 110\) W
• Voltage \(V = 220\) V

We calculate the current \(I\) using:

\(I = \frac{P}{V}\)

Substituting the given values:

\(I = \frac{110}{220} = 0.5\) A

This current is the flow of charge per second. The charge \((Q)\) flowing per second is given by:

\(Q = I \times t\)

In SI units, time \(t = 1\) second, so:

\(Q = 0.5 \times 1 = 0.5\) C

The number of electrons \((n)\) can be calculated by dividing the total charge \((Q)\) by the charge of one electron \((e = 1.6 \times 10^{-19} \text{ C})\):

\(n = \frac{Q}{e} = \frac{0.5}{1.6 \times 10^{-19}}\)

\(n = \frac{0.5 \times 10^{19}}{1.6} = \frac{5 \times 10^{18}}{1.6}\)

\(n = 3.125 \times 10^{18}\)

Rewriting it in terms of the given options:

\(n = 31.25 \times 10^{17}\)

Therefore, the correct answer is \(31.25 \times 10^{17}\), which matches the given option.

Answer 2

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 \( n \):
\[\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}}\]
\[\frac{n}{t} = 31.25 \times 10^{17}\]