Chapter 21, Problem 42P

(a)

To determine

The actual power delivered to the vacuum cleaner.

Answer to Problem 42P

The actual power delivered to the vacuum cleaner is $\underset{\_}{\frac{{\left(\Delta V\right)}^{2}R}{{\left(R+2r\right)}^2}}\text{and it is }\underset{\_}{470\text{W}}$.

Explanation of Solution

Write the expression for the power delivered to the vacuum cleaner.

    $P=I\Delta V$        (I)

Here, $P$ is the power delivered, $I$ is the current, $\Delta V$ is the voltage.

Write the expression for the $I$.

    $I=\frac{\Delta V}{R}$        (II)

Here, $R$ is the intrinsic resistance of the vacuum cleaner.

Use equation (II) in (I) to solve for $P$.

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

Use equation (III) to solve for $R$.

    $R=\frac{{\left(\Delta V\right)}^2}{P}$        (IV)

Write the expression for the total resistance in the circuit.

    $R_{\text{Tot}}=R+2r$        (V)

Here, $r$ is the contact resistance.

Write the expression for current throughout the circuit.

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

Use equation (II) to solve for $\Delta V$.

    $\Delta V=IR$        (VII)

Write the expression for the power of the cleaner.

    $P_{\text{cleaner}}=I{\left(\Delta V\right)}_{\text{cleaner}}$        (VIII)

Here, $P_{\text{cleaner}}$ is the power of the cleaner.

Use equation (VII) in (VIII) to solve for $P_{\text{cleaner}}$.

    $P_{\text{cleaner}}=I^{2}R$        (IX)

Use equation (V) in (VI) to solve for $I$.

    $I=\frac{\Delta V}{R+2r}$        (X)

Use equation (X) in (IX) to solve for $P_{\text{cleaner}}$.

    $\begin{array}{c}{P}_{\text{cleaner}}={\left(\frac{\Delta V}{R+2r}\right)}^{2}R\\ =\frac{{\left(\Delta V\right)}^{2}R}{{\left(R+2r\right)}^2}\end{array}$        (XI)

Conclusion:

Substitute $120\text{V}$ for $\Delta V$, $535\text{W}$ for $P$ in equation (IV) to find $R$.

    $\begin{array}{c}R=\frac{{\left(120\text{V}\right)}^2}{535\text{W}}\\ =26.9\Omega \end{array}$

Substitute $0.9\Omega$ for $r$, $26.9\Omega$ for $R$, $120\text{V}$ for $\Delta V$ in equation (VI) to find $I$.

    $\begin{array}{c}I=\frac{120\text{V}}{26.9\Omega +2\left(0.9\Omega \right)}\\ =\frac{120\text{V}}{28.7\Omega }\\ =4.18\text{A}\end{array}$

Substitute $4.18\text{A}$ for $I$, $26.9\Omega$ for $R$ in equation (IX) to find $P_{\text{cleaner}}$.

    $\begin{array}{c}{P}_{\text{cleaner}}={\left(4.18\text{A}\right)}^2\left(26.9\Omega \right)\\ =470\text{W}\end{array}$

Therefore, the actual power delivered to the vacuum cleaner is $\underset{\_}{\frac{{\left(\Delta V\right)}^{2}R}{{\left(R+2r\right)}^2}}\text{and it is }\underset{\_}{470\text{W}}$.

(b)

To determine

The required diameter of the copper conductors for a power of at least $525\text{W}$.

Answer to Problem 42P

The required diameter of the copper conductors for a power of at least $525\text{W}$ is $\underset{\_}{1.60\text{mm or more}}$.

Explanation of Solution

Use equation (XI) to solve for $r$.

    $r=\frac{\Delta V}{2}{\left(\frac{R}{P_{\text{cleaner}}}\right)}^{1/2}-\frac{R}{2}$        (XII)

Write the expression for the $r$.

    $r=\frac{\rho l}{A}$        (XIII)

Here, $\rho$ is the resistivity, $l$ is the length of the wire, $A$ is the area of cross-section.

Write the expression for $A$.

    $A=\pi {\left(\frac{d}{2}\right)}^2$        (XIV)

Here, $d$ is the diameter of the wire.

Use equation (XIV) in (XIII) to solve for $d$.

    $d={\left(\frac{4\rho l}{\pi r}\right)}^{1/2}$        (XV)

Conclusion:

Substitute $120\text{V}$ for $\Delta V$, $26.9\Omega$ for $R$, $525\text{W}$ for $P_{\text{cleaner}}$ in equation (XII) to find $r$.

    $\begin{array}{c}r=\frac{120\text{V}}{2}{\left(\frac{26.9\Omega }{525\text{W}}\right)}^{1/2}-\frac{26.9\Omega }{2}\\ =0.128\Omega \end{array}$

Substitute $0.128\Omega$ for $r$, $1.7\times {10}^{-8}\Omega \cdot \text{m}$ for $\rho$, $15.0\text{m}$ for $l$ in equation (XV) to find $d$.

    $\begin{array}{c}d={\left[\frac{4\left(1.7\times {10}^{-8}\Omega \cdot \text{m}\right)\left(15.0\text{m}\right)}{\pi \left(0.128\Omega \right)}\right]}^{1/2}\\ =1.60\text{mm}\end{array}$

Therefore, the required diameter of the copper conductors for a power of at least $525\text{W}$ is $\underset{\_}{1.60\text{mm or more}}$

(c)

To determine

The required diameter of the copper conductors for a power of at least $535\text{W}$.

Answer to Problem 42P

The required diameter of the copper conductors for a power of at least $535\text{W}$ is $\underset{\_}{2.93\text{mm or more}}$.

Explanation of Solution

Use equation (XII) to solve for $r$.

Use equation (XV) to solve for $d$.

Conclusion:

Substitute $120\text{V}$ for $\Delta V$, $26.9\Omega$ for $R$, $535\text{W}$ for $P_{\text{cleaner}}$ in equation (XII) to find $r$.

    $\begin{array}{c}r=\frac{120\text{V}}{2}{\left(\frac{26.9\Omega }{535\text{W}}\right)}^{1/2}-\frac{26.9\Omega }{2}\\ =0.0379\Omega \end{array}$

Substitute $0.0379\Omega$ for $r$, $1.7\times {10}^{-8}\Omega \cdot \text{m}$ for $\rho$, $15.0\text{m}$ for $l$ in equation (XV) to find $d$.

    $\begin{array}{c}d={\left[\frac{4\left(1.7\times {10}^{-8}\Omega \cdot \text{m}\right)\left(15.0\text{m}\right)}{\pi \left(0.0376\Omega \right)}\right]}^{1/2}\\ =2.93\text{mm}\end{array}$

Therefore, the required diameter of the copper conductors for a power of at least $535\text{W}$ is $\underset{\_}{2.93\text{mm or more}}$.