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)
- $31.25 \times 10^{17}$
- $6.25 \times 10^{18}$
- $6.25 \times 10^{17}$
- $1.25 \times 10^{19}$
The Correct Option is A
Approach Solution - 1
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.
Approach Solution -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 \( n \):
\[\frac{n}{t} = \frac{0.5}{1.6 \times 10^{-19}}\]
\[\frac{n}{t} = 31.25 \times 10^{17}\]
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