19.9: Sample problems and solutions
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Exercise \(\PageIndex{1}\)
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:
\[\begin{aligned} I=\int dI = \int_0^R 2\pi a r^2 dr=\frac{2}{3}\pi aR^3\end{aligned}\]
Exercise \(\PageIndex{2}\)
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.