Problem 12 - College Physics

Question

If you triple the length of a cable and at the same time double its diameter, what will be its resistance if its original resistance was \(R\) ?

Step-by-Step Solution

Answer

The new resistance is \( \frac{3}{4} R \).

Step 1: Understanding Resistance Formula

The resistance of a wire is determined by the formula \( R = \rho \frac{L}{A} \), where \( R \) is the resistance, \( \rho \) is the resistivity, \( L \) is the length, and \( A \) is the cross-sectional area of the wire.

Step 2: Calculating Original Cross-Sectional Area

Let's denote the original length as \( L_0 \) and the original diameter as \( d_0 \). The cross-sectional area \( A_0 \) is \( \pi \left(\frac{d_0}{2}\right)^2 = \frac{\pi d_0^2}{4} \).

Step 3: Calculating New Length and Diameter

The new length \( L_1 = 3L_0 \) since the length is tripled, and the new diameter \( d_1 = 2d_0 \) since the diameter is doubled.

Step 4: Calculating New Cross-Sectional Area

The new cross-sectional area \( A_1 = \pi \left(\frac{d_1}{2}\right)^2 = \frac{\pi (2d_0)^2}{4} = \pi d_0^2 \).

Step 5: Deriving the New Resistance Formula

Substitute \( L_1 \) and \( A_1 \) into the resistance formula. We find \( R_1 = \rho \frac{L_1}{A_1} = \rho \frac{3L_0}{\pi d_0^2} \).

Step 6: Expressing New Resistance in Terms of Original Resistance

The original resistance \( R_0 = \rho \frac{L_0}{\frac{\pi d_0^2}{4}} = 4 \rho \frac{L_0}{\pi d_0^2} \). The new resistance \( R_1 = \frac{3}{4} \rho \frac{L_0}{\pi d_0^2} = \frac{3}{4} R_0 \).

Step 7: Simplifying the Calculation

Therefore, the new resistance \( R_1 = \frac{3}{4} R_0 \). This means the resistance decreases when both length is tripled, and the diameter is doubled.

Key Concepts

Resistivity

Resistivity is a fundamental property of materials which dictates how strongly a substance opposes the flow of electric current. It is symbolized by \( \rho \) and is measured in ohm-meters (\( \Omega \cdot m \)). This intrinsic property is constant for a given material and influences how a material's resistance changes with its geometry.

In the context of electrical wires, resistivity and resistance are directly related through the formula \( R = \rho \frac{L}{A} \), where \( R \) is resistance, \( L \) is length, and \( A \) is cross-sectional area. This formula shows that regardless of changes in length or diameter (which affect \( L \) and \( A \)), resistivity \( \rho \) remains constant.

Understanding resistivity helps in selecting materials for electrical applications, ensuring minimal energy loss across electrical components.

Cross-Sectional Area

When dealing with wires or cables, the cross-sectional area is a crucial factor that influences the resistance. It refers to the area of the wire's cross-section perpendicular to the flow of the current. Mathematically, this is calculated using the formula \( A = \pi \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter of the wire.

The cross-sectional area is inversely proportional to the resistance. This means a larger area results in lower resistance. In exercises where the dimensions of a wire change, such as doubling the diameter, recalculating the cross-sectional area is essential for determining the new resistance.

This principle is applied in designing cables with a lower resistance to conduct electricity more effectively.

Length-Diameter Relationship

The length and diameter of a wire have a significant relationship affecting its resistance. Changing one or both of these dimensions alters the overall resistance according to the formula \( R = \rho \frac{L}{A} \).

For example, if you triple the length of a wire, the resistance will generally increase, because resistance is directly proportional to length. Conversely, doubling the diameter decreases resistance because it increases the cross-sectional area. More area means more paths for electrons to travel, reducing overall opposition.

Therefore, in calculating resistance modifications, both changes in length and diameter must be carefully considered to predict resistance accurately. This relationship is crucial in electromechanical applications where different wire configurations affect system performance.

Physics Problem Solving

Solving physics problems involves breaking down exercises into clear, manageable steps. This approach applies well to solving resistance calculation problems.

First, identify the original geometric attributes such as length and diameter, then compute the initial cross-sectional area. Next, apply any given changes to these dimensions to find new ones. Afterward, calculate the new cross-sectional area and apply the resistance formula.

The key step lies in substituting these values back into the resistance formula \( R = \rho \frac{L}{A} \). Finally, compare the new and original resistance to understand how changes affect the electrical properties. Organizing steps systematically simplifies complex problems, making the solution more intuitive for students.