Chapter 17, Problem 68AP

A man wishes to vacuum his car with a canister vacuum cleaner marked 535 W at 120. V. The car is parked far from the building, so he uses an extension cord 15.0 m long to plug the cleaner into a 120.-V source. Assume the cleaner has constant resistance. (a) If the resistance of each of the two conductors of the extension cord is 0.900 Ω, what is the actual power delivered to the cleaner? (b) If, instead, the power is to be at least 525 W, what must be the diameter of each of two identical copper conductors in the cord the young man buys? (c) Repeat part (b) if the power is to be at least 532 W. Suggestion: A symbolic solution can simplify the calculations.

To determine

The actual power delivered to vacuum cleaner.

Answer to Problem 68AP

The actual power delivered to vacuum cleaner is $470\text{W}$

Explanation of Solution

Given Info: A vacuum cleaner utilizes $535\text{W}$ power when connected to $120\text{V}$ supply. The extension cord from vacuum cleaner which used to clean the car consists of two conducting wires. The total length of extension cord is $15.0\text{m}$. The resistance of each conductor is $0.900\Omega$.

Explanation:

Formula to calculate the total current in the circuit is,

$I=\frac{\Delta V}{R_{\text{total}}}$ (I)

• $I$ is the current in the circuit,
• $\Delta V$ is the voltage of power supply,
• $R_{\text{total}}$ is the total resistance in the path of current,

Formula to calculate the total resistance in the path of current is,

$R_{\text{total}}=R_{\text{vaccum}}+2R_{\text{c}}$ (II)

• $R_{\text{vaccum}}$ is the resistance offered by the vacuum cleaner,
• $R_{\text{c}}$ is the resistance offered by each conductor in the extension cord,

Formula to calculate the resistance offered by vacuum cleaner is,

$R_{\text{vaccum}}=\frac{{\left(\Delta V\right)}^2}{P}$ (III)

• $P$ is the rated power of vacuum cleaner,

Substitute equation (III) in equation (II) and rewrite equation (II).

$R_{\text{total}}=\frac{{\left(\Delta V\right)}^2}{P}+2R_{\text{c}}$ (IV)

Substitute equation (IV) in equation (I) and rewrite equation (I).

$I=\frac{\Delta V}{\left(\frac{{\left(\Delta V\right)}^2}{P}+2R_{\text{c}}\right)}$

Substitute $120\text{V}$ for $\Delta V$, $535\text{W}$ for $P$ and $0.900\Omega$ for $R_{\text{c}}$ in the above equation to find $I$.

$\begin{array}{c}I=\frac{120\text{V}}{\left(\left(\frac{{\left(120\text{V}\right)}^2}{535\text{W}}\right)+2\left(0.900\Omega \right)\right)}\\ =4.18\text{A}\end{array}$

The total current in the circuit is $4.18\text{A}$

Conclusion

Formula to calculate the actual power delivered to vacuum cleaner is,

$P_{\text{delivered}}=I^2{R}_{\text{vaccum}}$ (V)

• $P_{\text{delivered}}$ is the actual power delivered to vacuum cleaner

Formula to calculate the resistance offered by vacuum cleaner is,

$R_{\text{vaccum}}=\frac{{\left(\Delta V\right)}^2}{P}$

Use ${\left(\Delta V\right)}^2/P$ the above equation for $R_{\text{vaccum}}$ in the above equation to rewrite $P$.

$P_{\text{delivered}}=I^2\frac{{\left(\Delta V\right)}^2}{P}$

Substitute $4.18\text{A}$ for $I$, $120\text{V}$ for $\Delta V$ and $535\text{W}$ for $P$ in the above equation to find $P$.

$\begin{array}{c}P={\left(4.18\text{A}\right)}^2\left(\frac{{\left(120\text{V}\right)}^2}{535\text{W}}\right)\\ =470.28\text{W}\\ \approx 470\text{W}\end{array}$

Therefore, the actual power delivered to vacuum cleaner is $470\text{W}$

To determine

The diameter of each of identical copper wire in the extension cord when minimum power is $525\text{W}$.

Answer to Problem 68AP

The diameter of each of identical copper wire in the extension cord is $1.59\times {10}^{-3}\text{m}$

Explanation of Solution

Given Info: A vacuum cleaner utilizes $535\text{W}$ power when connected to $120\text{V}$ supply Vacuum cleaner utilizes at least $525\text{W}$ power. The extension cord from vacuum cleaner which used to clean the car consists of two conducting wires. The total length of extension cord is $15.0\text{m}$.

Explanation:

Formula to calculate the actual minimum power delivered to vacuum cleaner is,

$P_{\min }=I^2{R}_{\text{vaccum}}$

• $P_{\text{min}}$ is the minimum acceptable power delivered to vacuum cleaner,

Use $\left(\Delta V/R_{\text{total}}\right)$ for I and $\left({\left(\Delta V\right)}^2/P\right)$ for $R_{\text{vaccum}}$ to rewrite $P_{\text{delivered}}$.

$P_{\min }={\left(\frac{\Delta V}{R_{\text{total}}}\right)}^2\left(\frac{{\left(\Delta V\right)}^2}{P}\right)$

Rewrite the above equation in terms of $R_{\text{total}}$

$R_{\text{total}}=\frac{{\left(\Delta \text{V}\right)}^2}{\sqrt{P_{\text{min}}P}}$ (VI)

Equate (IV) and equation (VI)

$\frac{{\left(\Delta V\right)}^2}{P}+2R_{\text{c}}=\frac{{\left(\Delta \text{V}\right)}^2}{\sqrt{P_{\text{min}}P}}$

Use $R_{\text{c,max}}$ for $R_{\text{c}}$ in the above relation to rewrite it.

• $R_{\text{c,max}}$ is the maximum value of resistance of wire when power delivered to vacuum cleaner is minimum.

$\begin{array}{c}{R}_{\text{c,max}}=\frac{1}{2}\left[\frac{{\left(\Delta \text{V}\right)}^2}{\sqrt{P_{\text{min}}P}}-\frac{{\left(\Delta V\right)}^2}{P}\right]\\ R_{\text{c,max}}=\frac{{\left(\Delta V\right)}^2}{2}\left[\frac{1}{\sqrt{P_{\text{min}}P}}-\frac{1}{P}\right]\end{array}$

Substitute $120\text{V}$ for $\Delta V$, $525\text{W}$ for $P_{\text{min}}$ and $535\text{W}$ for $P$ in the above equation to find $R_{\text{c,max}}$.

$\begin{array}{c}{R}_{\text{c,max}}=\frac{{\left(120\text{V}\right)}^2}{2}\left[\frac{1}{\sqrt{\left(525\text{W}\right)\left(535\text{W}\right)}}-\frac{1}{\left(535\text{W}\right)}\right]\\ =0.1276\Omega \\ \approx 0.128\Omega \end{array}$

The maximum resistance of copper wire is $0.128\Omega$.

Conclusion:

Formula to calculate the diameter of copper wire is,

$d=\sqrt{\frac{4\rho L}{\pi R_{\text{c,max}}}}$

• $d$ is the diameter of copper wire,
• $L$ is the length of each copper wire,
• $\rho$ is the resistivity of copper,

Substitute $1.7\times {10}^{-8}\text{Ω}\text{.m}$ for $\rho$, $15.0\text{m}$ for $L$, $3.14$ for $\pi$ and $0.128\Omega$ for $R_{\text{c,max}}$ in the above equation to find $d$.

$\begin{array}{c}d=\sqrt{\frac{4\left(1.7\times {10}^{-8}\text{Ω}\text{.m}\right)\left(15.0\text{m}\right)}{\left(3.14\right)\left(0.128\Omega \right)}}\\ =1.59\times {10}^{-3}\text{m}\\ =1.59\text{mm}\end{array}$

Therefore, the diameter of each of identical copper wire in the extension cord is $1.59\text{mm}$

To determine

The diameter of each of identical copper wire in the extension cord when minimum power is $532\text{W}$.

Answer to Problem 68AP

The diameter of each of identical copper wire in the extension cord is $2.93\text{mm}$.

Explanation of Solution

Formula to find the maximum resistance of copper wire is,

$R_{\text{c,max}}=\frac{{\left(\Delta V\right)}^2}{2}\left[\frac{1}{\sqrt{P_{\text{min}}P}}-\frac{1}{P}\right]$

Substitute $120\text{V}$ for $\Delta V$, $532\text{W}$ for $P_{\text{min}}$ and $535\text{W}$ for $P$ in the above equation to find $R_{\text{c,max}}$.

$\begin{array}{c}{R}_{\text{c,max}}=\frac{{\left(120\text{V}\right)}^2}{2}\left[\frac{1}{\sqrt{\left(532\text{W}\right)\left(535\text{W}\right)}}-\frac{1}{\left(535\text{W}\right)}\right]\\ =0.03789\Omega \\ \approx 0.0379\Omega \end{array}$

Formula to calculate the diameter of copper wire is,

$d=\sqrt{\frac{4\rho L}{\pi R_{\text{c,max}}}}$

Substitute $1.7\times {10}^{-8}\text{Ω}\text{.m}$ for $\rho$, $15.0\text{m}$ for $L$ $3.14$ for $\pi$ and $0.0379\Omega$ for $R_{\text{c,max}}$ in the above equation to find $d$.

$\begin{array}{c}d=\sqrt{\frac{4\left(1.7\times {10}^{-8}\text{Ω}\text{.m}\right)\left(15.0\text{m}\right)}{\left(3.14\right)\left(0.0379\Omega \right)}}\\ =2.93\times {10}^{-3}\text{m}\\ =2.93\text{mm}\end{array}$

Conclusion:

Therefore, the diameter of each of identical copper wire in the extension cord is $2.93\text{mm}$.