Question

Two cylindrical wires, 1 and \(2,\) made of the same material, have the same resistance. If the length of wire 2 is twice that of wire 1 , what is the ratio of their cross-sectional areas, \(A_{1}\) and \(A_{2} ?\) a) \(A_{1} / A_{2}=2\) c) \(\mathrm{A}_{1} / \mathrm{A}_{2}=0.5\) b) \(A_{1} / A_{2}=4\) d) \(A_{1} / A_{2}=0.25\)

Short answer

Answer: a) \(A_1 / A_2=2\)

Step-by-step solution

Recall the resistance formula for cylindrical wires

We need to recall the formula for the resistance \(R\) of a cylindrical wire, which is given by: \(R = \frac{\rho L}{A}\) where \(\rho\) is the resistivity of the material, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.

Set up an equation using the given information

We are given that both wires have the same resistance. Therefore, we can write: \(R_1 = R_2\) Using the formula for resistance, we obtain: \(\frac{\rho L_1}{A_1} = \frac{\rho L_2}{A_2}\)

Use the given relationship between the lengths of the wires

We are given that the length of wire 2 is twice that of wire 1 (\(L_2 = 2 L_1\)). We can substitute this into our equation: \(\frac{\rho L_1}{A_1} = \frac{\rho (2 L_1)}{A_2}\)

Solve for the ratio of cross-sectional areas

Now we need to solve the equation for the ratio \(A_1/A_2\): \(A_1 / A_2 = \frac{\rho L_1 A_2}{\rho L_1 A_1}\) By substituting the relationship between lengths, we get: \(A_1 / A_2 = \frac{2 L_1}{L_1}\) Simplifying this equation further, we find: \(A_1 / A_2 = 2\)

Choose the correct answer

Now we can compare our result with the given choices: a) \(A_1 / A_2=2\) b) \(A_{1} / A_{2}=4\) c) \(A_{1} / A_{2}=0.5\) d) \(A_{1} / A_{2}=0.25\) Our result agrees with option a, so the correct answer is: a) \(A_1 / A_2=2\)

Key concepts

Resistivity

Resistivity, represented by the symbol \( \rho \), is a fundamental property that quantifies how strongly a material opposes the flow of electric current. Imagine electrical current as water flowing through a pipe; resistivity would be the measure of how much the pipe hinders the water's flow.

For conductive materials, such as metals, the resistivity is low, meaning electricity can flow through them easily. On the other hand, insulators, like rubber, have high resistivity, hence they do not allow electric current to pass through as readily.

The unit of resistivity in the International System of Units (SI) is ohm-meter (\(\Omega\cdot m\)). The resistivity of a material is affected by temperature; in most materials, as the temperature increases, the resistivity also goes up, as increased atomic vibrations scatter more electrons and hinder their flow.

Understanding resistivity is crucial in designing electrical circuits and components, such as the wires in our exercise, because different applications require materials with appropriate levels of resistivity.

Cylindrical Wires

Cylindrical wires are a common form factor for conductive materials used to carry electric current. These wires are characterized by their circular cross-section and their length. In our daily lives, we encounter these types of wires in various applications, powering everything from household appliances to extensive power grids.

The resistance \( R \) of a cylindrical wire can be tailored for specific uses by altering its length \( L \) and the material from which it is made, as this affects its resistivity \( \rho \). Manufacturers can also adjust the thickness, or the cross-sectional area \( A \), to change the wire's resistance properties.

When solving problems involving cylindrical wires, we apply the formula \( R = \rho \cdot \frac{L}{A} \), which clearly shows the relationship between the wire’s physical dimensions, the material's resistivity, and the resulting electrical resistance.

Cross-sectional Area

The cross-sectional area \( A \) of a wire is the area of the 'slice' that is perpendicular to its length. This measurement plays a pivotal role in determining the wire's electrical resistance. Wires with larger cross-sectional areas have more space for electrons to move through, which reduces the resistance. Think of it as a highway: more lanes allow more cars to move without congestion. Similarly, a large cross-sectional area allows more electric current to flow with less resistance.

For cylindrical wires, the cross-sectional area is calculated using the formula for the area of a circle \( A = \pi r^2 \), where \( r \) is the radius of the wire's cross-section. When comparing two wires of the same material and with identical resistances but different lengths, the one with the greater length will require a proportionally larger cross-sectional area to maintain the same level of resistance, as described in the step-by-step solution of the problem above.

In practical applications, engineers select wire gauges with suitable cross-sectional areas to ensure sufficient current capacity and to avoid overheating, which can arise from excessive resistance.