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 ascertain the quantity of electrons traversing the filament of a 110 W bulb operating at 220 V per second, the circuit's current must first be established using the power equation: \(P = V \times I\), where \(P\) denotes power, \(V\) represents voltage, and \(I\) signifies current.

Provided data:

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

The current \(I\) is computed as:

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

With the given values substituted:

\(I = \frac{110}{220} = 0.5\) A

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

\(Q = I \times t\)

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

\(Q = 0.5 \times 1 = 0.5\) C

The count of electrons \(n\) is obtained by dividing the total charge \(Q\) by the elementary charge \(e = 1.6 \times 10^{-19}\) C:

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

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

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

Expressed in relation to the provided options:

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

Consequently, the result is \(31.25 \times 10^{17}\), aligning with the presented option.