Q65P - Fundamentals Of Physics

Question

A potential difference V is applied to a wire of cross-sectional area A , length L, and resistivity p. You want to change the applied potential difference and stretch the wire so that the energy dissipation rate is multiplied by 30.0 and the current is multiplied by 4.00 . Assuming the wire’s density does not change, what are (a) the ratio of the new length to L and (b) the ratio of the new cross-sectional area to  A?

Step-by-Step Solution

Answer

1. The ratio of the new length to L is 1.37 .
2. The ratio of new cross-sectional area to A is 0.730 .

Step 1: The given data:

A potential difference V  is applied to a wire of cross-sectional area A , length  L, and resistivity p. 

 

By changing the applied potential difference and stretching the wire, the energy dissipation rate is multiplied by 30.0 and the current is multiplied by 4.00. 

 

The resistivity, $\mathrm{p}=3.5\times {10}^{-5}\mathrm{\Omega }.\mathrm{m}$  

Step 2: Understanding the concept of the energy rate and potential difference:

Power can be written as resistance and current. In this equation, you can substitute resistance using the formula for resistance in terms of resistivity, length, and cross-sectional area. Using this resulting equation and the given conditions, you can find the required ratios. 

Formulas:

The current density of a material having uniform current flow is defined by, 

 $\mathrm{J}=\frac{\mathrm{i}}{\mathrm{A}}$                                                                                                                       ….. (1)              

Here,   is the current and   is the area.

 

The electric power in terms of potential difference is, 

P = iV                                                                                                                        ….. (2)                   

Here, V is the potential difference.

The rate of energy dissipation from the system, 

$\mathrm{P}={\mathrm{i}}^2\mathrm{R}$                                                                                                                      ….. (3)                    

 

The resistance of a material is given by, 

  $\mathrm{R}=\frac{\mathrm{pL}}{\mathrm{A}}$                                                                                                                    ….. (4)                                

Here, p is the resistivity and L is the length.

 

The cross-sectional area of a circular surface is, 

$A=\pi r^2$                                                                                                                      ….. (5)

Step 3: (a) Calculation of the ratio of the new length to the original length:

Substituting the value of resistance of equation (4) in equation (3), you can get an equation of length to area ratio by rearranging the power equation as follows:

$$\begin{gathered} \mathrm{P}={\mathrm{i}}^2\frac{\mathrm{pL}}{\mathrm{A}} \\ \frac{\mathrm{L}}{\mathrm{A}}=\frac{\mathrm{P}}{{\mathrm{i}}^2\mathrm{p}} \end{gathered}$$

The ratio of length to area in the form of new to old current and resistance using above equation can be given as:

 $$\begin{gathered} {\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{new}}={\left(\frac{\mathrm{P}}{{\mathrm{i}}^2\mathrm{p}}\right)}_{\mathrm{new}} \\ =\frac{30}{4^2}{\left(\frac{\mathrm{P}}{{\mathrm{i}}^2\mathrm{p}}\right)}_{\mathrm{odd}} \end{gathered}$$

Substitute $\frac{\mathrm{L}}{\mathrm{A}}$ for $\frac{\mathrm{P}}{{\mathrm{i}}^2\mathrm{p}}$ in the above equation.

 

 $$\begin{gathered} {\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{new}}=\frac{30}{16}{\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{old}} \\ =1.875{\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{old}} \\ {\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{new}}=1.875{\left(\frac{\mathrm{L}}{\mathrm{A}}\right)}_{\mathrm{old}} \end{gathered}$$                                                                                              ….. (6)

But as known that the total volume of the wire is the same after the stretch and before the stretch. 

Thus, the volume equation for both cases can be related using length and area as,

$$\begin{gathered} {\left(\mathrm{LA}\right)}_{\mathrm{new}}={\left(\mathrm{LA}\right)}_{\mathrm{old}} \\ {\mathrm{A}}_{\mathrm{new}}=\frac{{\mathrm{L}}_{\mathrm{old}}{\mathrm{A}}_{\mathrm{old}}}{{\mathrm{L}}_{\mathrm{new}}} \end{gathered}$$                                                                                                            ….. (7)

Now, substituting this above value in equation (6) and rearranging it, you can get the required new length to old length ratio as follows:

$$\begin{gathered} {\mathrm{L}}_{\mathrm{new}}=\frac{{\mathrm{L}}_{\mathrm{old}}{\mathrm{A}}_{\mathrm{old}}}{{\mathrm{L}}_{\mathrm{new}}}\times 1.875\times \frac{{\mathrm{L}}_{\mathrm{old}}}{{\mathrm{A}}_{\mathrm{old}}} \\ =\sqrt{1.875}{\mathrm{L}}_{\mathrm{old}} \\ =1.37\times {\mathrm{L}}_{\mathrm{old}} \\ \frac{{\mathrm{L}}_{\mathrm{new}}}{{\mathrm{L}}_{\mathrm{old}}}=1.37 \end{gathered}$$ 

 

 Hence, the value of the length ratio is 1.37 .

Step 4: (b) Calculation of the ratio of the new area to the original area:

Substituting the length ratio in equation (7), you can obtain the required new to old area ratio as follows:

 $$\begin{gathered} \frac{{\mathrm{L}}_{\mathrm{new}}}{{\mathrm{L}}_{\mathrm{old}}}=\frac{{\mathrm{A}}_{\mathrm{old}}}{{\mathrm{A}}_{\mathrm{new}}} \\ \frac{{\mathrm{A}}_{\mathrm{old}}}{{\mathrm{A}}_{\mathrm{new}}}=1.37 \\ \frac{{\mathrm{A}}_{\mathrm{new}}}{{\mathrm{A}}_{\mathrm{old}}}=\frac{1}{1.37} \\ =0.730 \end{gathered}$$

 

 

Hence, the value of the area ratio is 0.730 .