Q53P - Fundamentals Of Physics

Question

A 120 V potential difference is applied to a space heater that dissipates 500 W  during operation. (a) What is its resistance during operation? (b) At what rate do electrons flow through any cross section of the heater element?

Step-by-Step Solution

Answer

1. The resistance during operation is $28.8\mathrm{\Omega }$.
2. The rate of electron flow through any cross section of the heater element is $2.60\times {10}^{19}/s$.

Step 1: Identification of the given data

The given data can be listed below as:

• The applied potential difference is V = 120 V.
• The dissipated power is, P = 500W.

Step 2: Understanding the concept of the resistance and electron transfer rate

In this problem, by using equation (i), we can find the resistance during operation. Using equation (iii) in the formula for the rate of electron transport, we can find the rate of electron flow through any cross-section of the heater element.

 

Formulae:

The electric power generated during the energy transfer is, 

 $P=\frac{V^2}{R}$                                                                                                                      … (i)

 

The current flowing due to energy transfer is,

 $i=\frac{P}{V}$                                                                                                                        … (ii)

 

The rate of electron transport due to the transfer is,

$\mathrm{Electron} \mathrm{transport} \mathrm{rate}=\frac{i}{e}$                                                                                       … (iii)

Step 3: (a) Calculation of the resistance during operation

Using the given data in equation (i), we can get the value of the resistance during the operations as follows:

$R=\frac{V^2}{P}$

 

Substitute the values in the above equation.

$$\begin{gathered} R=\frac{{(120 \mathrm{V})}^2}{500 \mathrm{W}} \\ =28.8\mathrm{\Omega } \end{gathered}$$

 

Hence, the value of the resistance is $28.8\mathrm{\Omega }$.

Step 4: (b) Calculation of the electron flow rate

Using equation (ii) in equation (iii) with the given values, we can get the value of the electron flow rate through any cross-section of the heater element as follows:

$\mathrm{Electron} \mathrm{transport} \mathrm{rate}=\frac{P}{eV}$

Here, e is the electric charge whose value is $e=1.60\times {10}^{-19}\mathrm{C}$.

 

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{Electron} \mathrm{transport} \mathrm{rate} =\frac{500\mathrm{W}}{1.60\times {10}^{-19}\mathrm{C}\times 120\mathrm{V}} \\ =2.60\times {10}^{19}/\mathrm{s} \end{gathered}$$

 

Hence, the value of the electron flow rate is $2.60\times {10}^{19}/\mathrm{s}$.