Chapter 18, Problem 103P

(a)

To determine

Find the resistance of the heater.

Answer to Problem 103P

The resistance of the heater is $6.5\Omega$.

Explanation of Solution

Write the equation for resistance from power.

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

Here, $P$  is the power, $V$ is the potential difference and $R$ is the resistance.

Conclusion:

Substitute $\text{120 V}$ for $V$ and $2200\text{ W}$ for $P$ in equation I.

$\begin{array}{c}R=\frac{{\left(\text{120 V}\right)}^2}{2200\text{ W}}\\ =6.5\Omega \end{array}$

Therefore, the resistance of the heater is $6.5\Omega$.

(b)

To determine

Find the current in the wire.

Answer to Problem 103P

The current in the wire is $\text{18 A}$.

Explanation of Solution

Write the equation for current from power.

$I=\frac{P}{V}$ (II)

Here, $I$ is the current.

Conclusion:

Substitute $\text{120 V}$ for $V$ and $2200\text{ W}$ for $P$ in equation II.

$\begin{array}{c}I=\frac{2200\text{ W}}{\text{120 V}}\\ =18\text{ A}\end{array}$

Therefore, the current in the wire is $\text{18 A}$.

(c)

To determine

Find the diameter of the wire.

Answer to Problem 103P

The diameter of the wire is $\text{0}\text{.86 mm}$.

Explanation of Solution

The wire has a circular cross section, then write the equation for area of this cross section.

$A=\frac{1}{4}\pi d^2$ (III)

Here, $A$ is the area of the wire.

Write the equation for resistance of the wire.

$R=\rho \frac{L}{A}$ (IV)

Here, $R$ is the resistance, $\rho$ is the resistivity and $L$ is the length of the wire.

Write the equation for resistivity.

$\rho ={\rho }_0\left(1+\alpha \Delta T\right)$ (V)

Here, $\rho$ is the resistivity, ${\rho }_0$ is the initial resistivity, $\alpha$ is the temperature coefficient of resistivity and $\Delta T$ is the change in temperature.

Substitute equation III and V in equation IV.

$R={\rho }_0\left(1+\alpha \Delta T\right)\frac{L}{\frac{1}{4}\pi d^2}$

Rewrite the above equation to get diameter of the wire,

$d=\sqrt{\frac{4{\rho }_0\left(1+\alpha \Delta T\right)L}{\pi R}}$ (VI)

Conclusion:

Substitute $\text{108}\times {\text{10}}^{-8}\Omega \text{m}$ for ${\rho }_0$, $0.00040°{\text{C}}^{-1}$ for $\alpha$, $420°\text{C}-20°\text{C}$ for $\Delta T$, $3.0\text{ m}$ for $L$ and $\text{6}\text{.5 }\Omega$ for $R$ in equation VI.

$\begin{array}{c}d=\sqrt{\frac{4\left(\text{108}\times {\text{10}}^{-8}\Omega \text{m}\right)\left(1+\left(0.00040°{\text{C}}^{-1}\right)\left(420°\text{C}-20°\text{C}\right)\right)\left(3.0\text{ m}\right)}{\pi \left(\text{6}\text{.5 }\Omega \right)}}\\ =0.86\text{ mm}\end{array}$

Therefore, the diameter of the wire is $\text{0}\text{.86 mm}$.

(d)

To determine

Find the current in the wire when it is turned on.

Answer to Problem 103P

The current in the wire when it is turned on is $\text{21 A}$.

Explanation of Solution

Write the equation for resistance.

$R=R_0\left(1+\alpha \Delta T\right)$ (VII)

Here, $R_0$ is the initial resistance.

Write the equation for current

$I=\frac{V}{R_0}$ (VIII)

Conclusion:

Substitute $0.00040°{\text{C}}^{-1}$ for $\alpha$ and $420°\text{C}-20°\text{C}$ for $\Delta T$ in equation VII.

$\begin{array}{c}R=R_0\left(1+\left(0.00040°{\text{C}}^{-1}\right)\left(420°\text{C}-20°\text{C}\right)\right)\\ R_0=\frac{R}{1.16}\end{array}$

Substitute the above value in equation VIII.

$\begin{array}{c}I=\frac{1.16V}{R}\\ =\frac{1.16V}{\raisebox{1ex}{$V^2$}\!\left/ \!\raisebox{-1ex}{$P$}\right.}\\ =\frac{1.16P}{V}\end{array}$

Substitute $\text{120 V}$ for $V$ and $2200\text{ W}$ for $P$ in above equaiton.

$\begin{array}{c}I=\frac{1.16\left(2200\text{ W}\right)}{\text{120 V}}\\ =21\text{ A}\end{array}$

Therefore, the current in the wire when it is turned on is $\text{21 A}$.