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# 19.9: Sample problems and solutions

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?

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.

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)]\\ R_2&=R_0[1+\alpha(T_2-T_0)]\\ \therefore\frac{R_1}{R_2}&=\frac{1+\alpha(T_1-T_0)}{1+\alpha(T_2-T_0)}\\ R_1 [1+\alpha(T_2-T_0)]&=R_2 [1+\alpha(T_1-T_0)]\\ \alpha \left( R_1(T_2-T_0) - R_2(T_2-T_0) \right)&=R_2-R_1\\ \therefore \alpha &= \frac{R_2-R_1}{R_1(T_2-T_0) - R_2(T_1-T_0) }\\ &=\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})) }\\ &=0.0034\end{aligned} Referring to Table 19.3.1, the material could likely be gold.