# 19.9: Sample problems and solutions

- Page ID
- 19512

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**-
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**-
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.