Resistance and resistivity — practice questions

1. Convert diameter to radius and metres: Diameter = 0.5 mm = $0.5 \times 10^{-3} \text{ m}$ Radius ($r$) = Diameter / 2 = $0.25 \times 10^{-3} \text{ m}$

2. Calculate cross-sectional area ($A$): $A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 \text{ m}^2$ $A \approx 1.963 \times 10^{-7} \text{ m}^2$

3. Use the formula $R = \frac{\rho L}{A}$: $R = \frac{(1.68 \times 10^{-8} \Omega \text{m}) \times (2.5 \text{ m})}{1.963 \times 10^{-7} \text{ m}^2}$

4. Calculate the resistance: $R \approx 0.214 \Omega$

   The resistance of the copper wire is approximately $0.21 \Omega$ (to 2 significant figures).

This problem requires combining Ohm's Law with the resistivity formula.

1. Calculate the wire's resistance ($R$) using Ohm's Law: $R = V / I$ $R = 2.0 \text{ V} / 1.83 \text{ A} \approx 1.093 \text{ } \Omega$
2. Calculate the wire's cross-sectional area ($A$): First, find the radius in metres: Diameter = 0.80 mm = $0.80 \times 10^{-3} \text{ m}$ Radius ($r$) = Diameter / 2 = $0.40 \times 10^{-3} \text{ m}$ Now, calculate the area: $A = \pi r^2 = \pi (0.40 \times 10^{-3})^2 \approx 5.027 \times 10^{-7} \text{ m}^2$
3. Calculate the resistivity ($\rho$) using the resistivity formula: Rearrange $R = \frac{\rho L}{A}$ to solve for $\rho$: $\rho = \frac{RA}{L}$ Convert length to metres: $L = 50.0 \text{ cm} = 0.500 \text{ m}$ Substitute the values: $\rho = \frac{(1.093 \text{ } \Omega) \times (5.027 \times 10^{-7} \text{ m}^2)}{0.500 \text{ m}}$ $\rho \approx 1.099 \times 10^{-6} \text{ } \Omega \text{m}$
4. Final Answer: The input values have 2 or 3 significant figures. Rounding to 2 significant figures gives: $\rho = 1.1 \times 10^{-6} \text{ } \Omega \text{m}$