Question

The ratio of diameters of two copper wires of same length is 2 : 1. Compare their resistances.

Hint:

Remember the quick relationship: resistance is inversely proportional to the area (\(R \propto 1/A\)) and inversely proportional to the square of the diameter or radius (\(R \propto 1/d^2\)). If the diameter is doubled, the resistance becomes one-fourth.

Answer 1

Step 1: Understanding the Concept:
The electrical resistance (\(R\)) of a conductor depends on its intrinsic property (resistivity, \(\rho\)), its length (\(L\)), and its cross-sectional area (\(A\)). Since the problem specifies two copper wires of the same length, their resistivity and length are identical. Therefore, the difference in their resistance will solely depend on their cross-sectional areas, which are determined by their diameters.

Step 2: Key Formula or Approach:
The formula for the resistance of a wire is: \[ R = \rho \frac{L}{A} \] The cross-sectional area \(A\) of a wire with diameter \(d\) is circular, so its area is: \[ A = \pi r^2 = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4} \] Substituting the area into the resistance formula shows that resistance is inversely proportional to the square of the diameter: \[ R = \rho \frac{L}{(\pi d^2 / 4)} \implies R \propto \frac{1}{d^2} \]

Step 3: Detailed Explanation:
Let the two wires be designated as Wire 1 and Wire 2. We are given the following information: \begin{itemize} \item Same material (copper): \(\rho_1 = \rho_2 = \rho\) \item Same length: \(L_1 = L_2 = L\) \item Ratio of diameters: \(\frac{d_1}{d_2} = \frac{2}{1}\) \end{itemize} We want to find the ratio of their resistances, \(\frac{R_1}{R_2}\). Using the resistance formula for both wires: \[ \frac{R_1}{R_2} = \frac{\rho_1 \frac{L_1}{A_1}}{\rho_2 \frac{L_2}{A_2}} \] Since \(\rho\) and \(L\) are the same, they cancel out, leaving: \[ \frac{R_1}{R_2} = \frac{A_2}{A_1} \] Now, substitute the formula for area in terms of diameter: \[ \frac{R_1}{R_2} = \frac{\pi d_2^2 / 4}{\pi d_1^2 / 4} = \frac{d_2^2}{d_1^2} = \left(\frac{d_2}{d_1}\right)^2 \] We are given \(\frac{d_1}{d_2} = \frac{2}{1}\), which means \(\frac{d_2}{d_1} = \frac{1}{2}\). Substituting this value into our ratio: \[ \frac{R_1}{R_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Step 4: Final Answer:
The ratio of the resistances \(R_1 : R_2\) is 1 : 4. This means the thicker wire (\(d_1\)) has one-fourth the resistance of the thinner wire (\(d_2\)).