19.9: Sample problems and solutions
- Page ID
- 19512
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A cylindrical wire as a current density that increases with radius as \(j(r) = ar\), where \(r\), is the radial distance from the center of the wire, and \(a\), is a constant. If the wire has a radius of \(R = 1.5\text{cm}\), what is the total current in the wire?
- Answer
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To determine the current through the entire cross section of the wire, we first divide the cross-section of the wire into infinitesimally small concentric rings of radius, \(r\), and width, \(dr\). The cross-sectional area of one ring is given by:
\[\begin{aligned} dA = 2\pi r dr\end{aligned}\]
so that the current through one ring is given by:
\[\begin{aligned} dI = j(r) dA = 2\pi a r^2 dr\end{aligned}\]
The current through the whole wire is then found by summing the currents through each ring:
\[\begin{aligned} I=\int dI = \int_0^R 2\pi a r^2 dr=\frac{2}{3}\pi aR^3\end{aligned}\]
A resistor is measured to have a resistance of \(R_1=103.4 \Omega\) at a temperature of \(T_1=30^{\circ}\text{C}\), and a resistance of \(R_2=106.8\Omega\) at a temperature of \(T_1=40^{\circ}\text{C}\). Using the values in Table 19.3.1, determine the material from which the resistor is made.
- Answer
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To determine the material of the resistor, we can find the temperature coefficient, \(\alpha\), since we are given measurements of resistance, \(R_1\) and \(R_2\), at two different temperatures, \(T_1\), and \(T_2\), respectively. The reference temperature is set to be \(T_0=20^{\circ}\text{Celsius}\), so that we can compare with Table 19.3.1.
We know that the resistance will vary with temperature, since the resistivity is temperature-dependent. The temperature dependence of resistivity is given by:
\[\begin{aligned} \rho(T)=\rho_0[1+\alpha(T-T_0)]\end{aligned}\]
If the resistor has length, \(L\), and cross-sectional area, \(A\), it will have resistance, \(R\), given by:
\[\begin{aligned} R(T)=\rho(T) \frac{L}{A}=\frac{\rho_0 L}{A}[1+\alpha(T-T_0)]=R_0[1+\alpha(T-T_0)]\end{aligned}\]
where \(R_0\) is the resistance at the reference temperature, \(T_0\). Since we are given the resistance at two different temperatures, we can determine both \(\alpha\) and \(R_0\), for a choice of \(T_0=20^{\circ}\text{C}\):
\[\begin{aligned} R_1&=R_0[1+\alpha(T_1-T_0)]\\[4pt] R_2&=R_0[1+\alpha(T_2-T_0)]\\[4pt] \therefore\frac{R_1}{R_2}&=\frac{1+\alpha(T_1-T_0)}{1+\alpha(T_2-T_0)}\\[4pt] R_1 [1+\alpha(T_2-T_0)]&=R_2 [1+\alpha(T_1-T_0)]\\[4pt] \alpha \left( R_1(T_2-T_0) - R_2(T_2-T_0) \right)&=R_2-R_1\\[4pt] \therefore \alpha &= \frac{R_2-R_1}{R_1(T_2-T_0) - R_2(T_1-T_0) }\\[4pt] &=\frac{(106.8\Omega) - (103.4\Omega)}{(103.4\Omega)((40^{\circ}\text{Celsius})-(20^{\circ}\text{Celsius})) - (106.8\Omega)((30^{\circ}\text{Celsius})-(20^{\circ}\text{Celsius})) }\\[4pt] &=0.0034\end{aligned}\]
Referring to Table 19.3.1, the material could likely be gold.