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Expert-verified Found in: Page 736 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # Nichrome wire is used in some radiative heaters. (a) Find the resistance needed if the average power output is to be $$1.00\;kW$$ utilizing $$120 - V$$$${\rm{AC}}$$. (b) What length of Nichrome wire, having a cross-sectional area of $$5.00{\rm{ }}m{m^2}$$, is needed if the operating temperature is $${500^o}C$$ ? (c) What power will it draw when first switched on?

1. The resistance needed if the average power output is $$1.00\;kW$$utilizing $$120 - V$$$${\rm{AC}}$$is, $$R = 14.4{\rm{ }}\Omega$$.
2. $$L = 60\;m$$ is length of Nichrome wire, having a cross-sectional area of $$5.00{\rm{ }}m{m^2}$$, is needed if the operating temperature is $${500^\circ }C$$.

3. The amount of power drawn by the Nichrome wire when first switched on is $${P_0} = 1.2\;kW$$.
See the step by step solution

## Step 1: Concept Introduction

The average power delivered by an $${\rm{AC}}$$ voltage source to a purely resistive load is,

$${P_{avg}} = {I_{rms}}\Delta {V_{rms}}$$

Because the potential difference across a resistor is given by $${V_{rms}} = {I_{rms}}R$$, we can express the average power delivered to a resistor as,

\begin{aligned}{c}P = {I^2}_{rms}R\\ = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{R}...(1)\end{aligned}

The energy delivered to a resistor by electrical transmission appears in the form of internal energy in the resistor.

The resistivity of a conductor varies approximately linearly with temperature according to the expression,

$$\rho = {\rho _0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)...(2)$$

Where $${\rho _0}$$ is the resistivity at some reference temperature $${T_0}$$ , and $$\alpha$$ is the temperature coefficient of resistivity.

The electrical resistance $$R$$ of a resistive circuit element depends on its geometric structure (its length $$L$$ and cross-sectional area $$A$$) and on the resistivity $$\rho$$ of the material of which it is made,

$$R = \rho \frac{L}{A}...(3)$$

The resistance of a conductor varies approximately linearly with temperature according to the expression,

$$R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)...(4)$$

Where $${R_0}$$ is the resistance at some reference temperature $${T_0}$$ , and $$\alpha$$ is the temperature coefficient of resistivity.

## Step 2: Information Provided

• The average power output of the heater is: $${P_{avg}} = \left( {1.00{\rm{ }}kW} \right)\left( {\frac{{1000{\rm{ }}W}}{{1{\rm{ }}kW}}} \right) = 1000{\rm{ }}W$$.
• The rms voltage across the heater is: $$\Delta {V_{rms}} = 120{\rm{ }}V$$.
• The cross-sectional area of the Nichrome wire is: $$A = \left( {500{\rm{ }}m{m^2}} \right)\left( {\frac{{1{\rm{ }}m}}{{{{10}^6}{\rm{ }}m{m^2}}}} \right) = 5.00 \times {10^{ - 6}}{\rm{ }}{m^2}$$.
• The operating temperature of the heater is: $$T = {500^o}C$$.
• The resistivity of Nichrome at temperature $${T_0} = {20.0^o}C$$ is: $${\rho _0} = 1.00 \times {10^{ - 6}}{\rm{ }}\Omega \cdot M$$.
• The temperature coefficient of resistivity for Nichrome is: $$\alpha = 4 \times {10^{ - 4}}^o{C^{ - 1}}$$.

## Step 3: Calculation for Resistance

a.

The average power output of the heater is found from Equation $$(1)$$,$${P_{avg}} = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{R}$$

Solving for $$R$$, it is obtained,

$${\rm{R = }}\frac{{{{\left( {{\rm{\Delta }}{{\rm{V}}_{{\rm{rms}}}}} \right)}^{\rm{2}}}}}{{{{\rm{P}}_{{\rm{avg}}}}}}$$

Entering the values for $$\Delta {V_{rms}}$$ and $${P_{avg}}$$, gives,

\begin{aligned}{}R = \frac{{{{\left( {120{\rm{ }}V} \right)}^2}}}{{1000{\rm{ }}W}}\\ = 14.4{\rm{ }}\Omega \end{aligned}

Therefore, the value for the resistance obtained is $$R = 14.4{\rm{ }}\Omega$$.

## Step 4: Calculation for Length

b.

The resistivity of Nichrome is expressed as a function of temperature from Equation $$(2)$$,

$$\rho = {\rho _0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)$$

Where $${\rm{\rho }}$$ is the resistivity of the Nichrome wire at the operating temperature $$T$$. Substitute numerical values,

\begin{aligned}{}\rho = \left( {1.00 \times {{10}^{ - 6}}{\rm{ }}\Omega \cdot m} \right)\left( {1 + \left( {4 \times {{10}^{ - 4o}}{C^{ - 1}})\left( {{{500}^o}C - {{20.0}^o}C} \right)} \right.} \right.\\ = 1.192 \times {10^{ - 6}}{\rm{ }}\Omega \cdot m\end{aligned}

The resistance of the Nichrome wire at the operating temperature $$T$$ is then found from Equation $$(3)$$,

$$R = \rho \frac{L}{A}$$

Solve for $$L$$,

$$L = \frac{{RA}}{\rho }$$

Entering the values for $$R$$, $$A$$, and $$\rho$$, it is obtained,

\begin{aligned}{}L = \frac{{(14.4{\rm{ }}\Omega )\left( {5.00 \times {{10}^{ - 6}}\;{m^2}} \right)}}{{1.192 \times {{10}^{ - 6}}{\rm{ }}\Omega \times m}}\\ = 60\;m\end{aligned}

Therefore, the value of the length is obtained as $$L = 60\;m$$.

## Step 5: Calculation for Power

c.

The resistance of the Nichrome wire is found as a function of temperature from Equation $$(4)$$ as,

$$R = {R_0}\left( {1 + \alpha \left( {T - {T_0}} \right)} \right)$$

Where $${R_0}$$ is the resistance of the wire at room temperature $${T_0} = {20.0^\circ }C$$ , which is the temperature of the wire when first switched on.

Rearrange and solve for $${R_0}$$as,

$${R_0} = \frac{R}{{1 + \alpha \left( {T - {T_0}} \right)}}$$

Substitute numerical values as,

\begin{aligned}{}{R_0} = \frac{{14.4{\rm{ }}\Omega }}{{1 + \left( {4 \times {{10}^{ - 4}}^\circ {C^{ - 1}}} \right)\left( {{{500}^\circ }C - {{20.0}^\circ }C} \right)}}\\ = 12.1{\rm{ }}\Omega \end{aligned}

The power drawn by the heater when it is first turned on is found from Equation $$(1)$$ as,

$${P_0} = \frac{{{{\left( {\Delta {V_{rms}}} \right)}^2}}}{{{R_0}}}$$

Entering the values for $$\Delta {V_{rms}}$$ and $${R_0}$$, it is obtained as,

\begin{aligned}{}{P_0} = \frac{{{{(120\;V)}^2}}}{{12.1{\rm{ }}\Omega }}\\ = 1.2 \times {10^3}\;W\\ = \left( {1.2 \times {{10}^3}\;W} \right)\left( {\frac{{1\;kW}}{{1000\;W}}} \right)\\ = 1.2\;kW\end{aligned}

Therefore, the value for the Power obtained is $${P_0} = 1.2\;kW$$. ### Want to see more solutions like these? 