Chapter 18, Problem 82GP

To determine

The power dissipated in the wire.

Answer to Problem 82GP

  $P_f=0.25P_0$ .

Explanation of Solution

Given info:

A length of wire is double. $L_f=2L_0$

Formula used:

It is known that resistance:

  $R=\frac{\rho L}{A}=\frac{\rho L}{\pi r^2}=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}$

Where,

  $\rho$ is the resistivity.

  $L$ is the length

  $A$ is the cross-sectional area of the wire.

  $\pi$ is constant

  $r$ is a radius.

  $d$ is a diameter.

  $R$ is resistance.

Calculation:

By using the given formula, $R=\frac{\rho L}{\pi {\left(\frac{d}{2}\right)}^2}=\frac{4\rho L}{\pi d^2}$

Now, solve for $\frac{4\rho }{\pi }=cons\tan t$

Therefore, $\frac{R_f{d}_f^2}{L_f}=\frac{R_0{d}_0^2}{L_0}$

So, $\frac{R_f{d}_f^2}{2.0L_0}=\frac{R_0{d}_0^2}{L_0}$

  $\frac{R_f{d}_f^2}{2.0}=R_0{d}_0^2$

  $R_f=\frac{2.0R_0{d}_0^2}{d_f^2}$

According to the question, the wire is stretched uniformly which means that the value of the volume of the wire will not change. So, by using the formula of volume:

  $Volume=\frac{\pi d^{2}L}{4}$

Where, $L$ is the length

  $\pi$ is constant

  $d$ is a diameter.

Since the volume will not change, then $Volum{e}_f=Volum{e}_0$

Therefore, $\frac{\pi d_f^2{L}_f}{4}=\frac{\pi d_0^2{L}_0}{4}$

  $d_f^2{L}_f=d_0^2{L}_0$

  $d_f^2\times 2.0L_0=d_0^2{L}_0$

  $2.0d_f^2=d_0^2$

  $R_f=\frac{2.0R_0\times 2.0D_f^2}{D_f^2}=4R_0$

Expression for electrical power is $P=\frac{V^2}{R}$

Where, $V$ is voltage, $I$ is current and $R$ is resistance.

So, $P_f=\frac{V^2}{R_f}$ and $P_0=\frac{V^2}{R_0}$

Therefore,

  $\frac{P_f}{P_0}=\frac{\frac{V^2}{R_f}}{\frac{V^2}{R_0}}$

Substitute the value $R_f=4R_0$ :

  $\frac{P_f}{P_0}=\frac{R_0}{4R_0}$

So, $\frac{P_f}{P_0}=\frac{1}{4}$

  $P_f=\frac{1}{4}{P}_0$

Conclusion: The power decreases by a factor of $4$ . So, $P_f=0.25P_0$