Chapter 5.3, Problem 13P

To determine

To Find: How many electrons flow through the filament.

Answer to Problem 13P

  $1.94184\times {10}^{22}$ electrons

Explanation of Solution

Given information:

For $1$ Ampere of current $6.2\times {10}^{18}$ electrons flow per second.

Filament of a $100$ watt, $115$ volt light bulb for one hour.

Watts $=$ Amperes $\times$ Volts

Filament of a $100$ watt, $115$ volt light bulb for one hour.

Watts $=$ Amperes $\times$ Volts On substituting values

We get

  $\begin{array}{l}\Rightarrow 100=A\times 115\\ \Rightarrow \frac{100}{115}=\frac{A\times 115}{115}\\ \Rightarrow A=0.87\end{array}$

Therefore,

Current $=A=0.87$

For 1 second

  $1$ Ampere $\to 6.2\times {10}^{18}$ electrons.

Multiplying with $0.87$ on both sides

We get

  $\Rightarrow 1\times 0.87$ Ampere $\to 6.2\times {10}^{18}\times 0.87$ electrons

  $\Rightarrow 1\times 0.87$ Ampere $\to 5.394\times {10}^{18}$ electrons

Therefore,

For 1 second for $0.87$ Ampere of current $5.394\times {10}^{18}$ electrons.

For $0.87$ amperes

  $1$ second $\to 5.394\times {10}^{18}$ electrons.

Multiplying with $3600$ on both sides

We get

  $\Rightarrow 1\times 3600$ seconds $\to 5.394\times {10}^{18}\times 3600$ electrons.

  $\Rightarrow 3600$ seconds $\to 1.94184\times {10}^{22}$

Therefore,

For $0.87$ Amperes of Current $1.94184\times {10}^{22}$ electrons.