Question

Which of the following wires has the largest current flowing through it? a) a 1 -m-long copper wire of diameter \(1 \mathrm{~mm}\) connected to a \(10-V\) battery b) a \(0.5-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery c) a 2 -m-long copper wire of diameter \(2 \mathrm{~mm}\) connected to a \(20-V\) battery d) a \(1-\mathrm{m}\) -long copper wire of diameter \(0.5 \mathrm{~mm}\) connected to a 5 -V battery e) All of the wires have the same current flowing through them.

Short answer

Answer: The wire with the largest current flowing through it is wire (c).

Step-by-step solution

Calculate the cross-sectional area (A)

To calculate the area (A), we will use the formula for the area of a circle: A = π\((\frac{d}{2})^2\), where d is the diameter of the wire. For each wire: a) \(A_a = \pi(\frac{1\,\text{mm}}{2})^2 = \pi (\frac{0.001\,\text{m}}{2})^2\) b) \(A_b = \pi(\frac{0.5\,\text{mm}}{2})^2 = \pi (\frac{0.0005\,\text{m}}{2})^2\) c) \(A_c = \pi(\frac{2\,\text{mm}}{2})^2 = \pi (\frac{0.002\,\text{m}}{2})^2\) d) \(A_d = \pi(\frac{0.5\,\text{mm}}{2})^2 = \pi (\frac{0.0005\,\text{m}}{2})^2\)

Calculate the resistance (R)

To calculate the resistance (R), we will use the formula \(R = \frac{\rho L}{A}\), where \(\rho\) is the resistivity of copper, which is approximately \(1.7\times10^{-8}\,\Omega\cdot\mathrm{m}\). For each wire: a) \(R_a = \frac{1.7\times10^{-8}\,\Omega\cdot\mathrm{m} \cdot 1\,\text{m}}{A_a}\) b) \(R_b = \frac{1.7\times10^{-8}\,\Omega\cdot\mathrm{m} \cdot 0.5\,\text{m}}{A_b}\) c) \(R_c = \frac{1.7\times10^{-8}\,\Omega\cdot\mathrm{m} \cdot 2\,\text{m}}{A_c}\) d) \(R_d = \frac{1.7\times10^{-8}\,\Omega\cdot\mathrm{m} \cdot 1\,\text{m}}{A_d}\)

Calculate the current (I)

To calculate the current (I), we will use Ohm's law: \( I = \frac{V}{R}\), where V is the voltage of each battery. For each wire: a) \(I_a = \frac{10\,\text{V}}{R_a}\) b) \(I_b = \frac{5\,\text{V}}{R_b}\) c) \(I_c = \frac{20\,\text{V}}{R_c}\) d) \(I_d = \frac{5\,\text{V}}{R_d}\) Now, we will compare the currents to determine which wire has the largest current. Calculate the values and compare them. The wire with the largest current is wire (c) which is connected to a 20-V battery with a diameter of 2 mm and a length of 2 meters.

Key concepts

Electrical Current

Understanding electrical current is akin to looking at the flow of water through a pipe, where the water represents the moving charge in a conductor. In electrical terms, the current is the rate at which electronic charge flows through a conductor. It's measured in amperes (A), with an ampere defined as one coulomb of charge passing through a point in a circuit per second.

When speaking of a wire in a circuit, like in our exercise, the higher the voltage (V) applied to it, the higher the driving force for the charge carriers, and subsequently, the higher the current, assuming resistance (R) remains constant. This is summarized by the equation from Ohm's Law, which states that current (I) equals voltage (V) divided by resistance (R), expressed as \(I = \frac{V}{R}\).

The wire connected to a higher voltage source will typically carry a larger current, again assuming resistance is not a limiting factor. In the given exercise, wire (c) has the highest voltage of 20 V which influences the current flowing through it, potentially making it the wire with the largest current.

Electrical Resistance

Resistance is essentially the opposition to the flow of current within a material. Think of it as a rock in our imaginary water pipe, it makes it more difficult for water to flow past. In electrical circuits, the resistance of a conductor is influenced by several factors including the material's intrinsic resistance, the length of the conductor, and its cross-sectional area.

Resistivity, \(\rho\), is a fundamental property of the material a wire is made of and varies from substance to substance. Longer wires have more resistance because the charges encounter more obstacles as they travel, just like a longer river has more twists and turns. However, larger cross-sectional areas allow more room for the charge to move, just as a wider river allows for more water to flow easily, thus reducing resistance.

The formula for resistance (R) is \(R = \frac{\rho L}{A}\), where L is the length and A is the cross-sectional area. For the problem at hand, the wire's material and length were key factors in determining its resistance, and in turn, the amount of current it could carry.

Cross-Sectional Area

Imagine a slice through our hypothetical pipe. The size of that slice is what we refer to as the cross-sectional area (A) in the context of a conductor in an electrical circuit. This area is vitally important as it directly affects the amount of resistance in the wire and consequently, the current that can flow through it.

The cross-sectional area is proportional to the diameter of the wire squared. In mathematical terms for a circular wire, this is expressed as \(A = \pi\left(\frac{d}{2}\right)^2\), where \(\pi\) is the constant Pi and d is the diameter of the wire. A larger cross-sectional area, like a wide chute for water, allows greater ease for the flow of electric charge, thus decreasing resistance.

In our exercise, comparing wires with different diameters will show that those with a greater diameter (and therefore larger area for the same length) will generally have less resistance, allowing more current to flow in response to a given voltage.

Resistivity

The final piece to the puzzle in understanding why certain wires carry more current than others is resistivity. Resistivity is an inherent property of materials that quantifies how strongly a material opposes the flow of electric current. It has the symbol \(\rho\) and is measured in ohm-meters (\(\Omega\cdot m\)).

Every material's resistivity is different: copper, for example, has a low resistivity which makes it an excellent conductor for electrical current, while rubber has a high resistivity and is used as an insulator. In calculations, we consider resistivity to determine the resistance of a wire given its length and area using the formula \(R = \frac{\rho L}{A}\).

In the context of the original exercise, since all wires are made of copper, they will have the same resistivity; however, varying lengths and diameters will lead to different resistances and thus, different currents when connected to their respective batteries. The resistivity is the factor that ties together the length and cross-sectional area of the wire to result in a measure of resistance to electric current flow.