Question

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

Short answer

The largest current is flowing through wire c) a 2-m-long copper wire of diameter 2 mm connected to a 20V battery, with a current of 1869.2 A.

Step-by-step solution

Use the formula for resistance

Recall that the resistance of a wire, denoted by \(R\), can be calculated using the formula: \(R = \frac{\rho L}{A}\), where \(\rho\) is the resistivity of the material, \(L\) is the length of the wire, and \(A\) is the cross-sectional area.

Calculate the cross-sectional area for each wire

To find the area \(A\) for the wires, use the formula for the area of a circle (\(A = \pi r^2\), where \(r\) is the radius of the wire) since the wires are given in terms of their diameter: a) \(A_1 = \pi \cdot (0.5 \mathrm{~mm})^2 = 0.7854 \mathrm{~mm}^2\) b) \(A_2 = \pi \cdot (0.25 \mathrm{~mm})^2 = 0.1963 \mathrm{~mm}^2\) c) \(A_3 = \pi \cdot (1 \mathrm{~mm})^2 = 3.1416 \mathrm{~mm}^2\) d) \(A_4 = \pi \cdot (0.25 \mathrm{~mm})^2 = 0.1963 \mathrm{~mm}^2\)

Calculate the resistance for each wire using the formula

We know that all the wires are made of copper, so the resistivity, \(\rho\), is constant ( \(\rho_{Cu}=1.68\times 10^{-8} \Omega \cdot \mathrm{m}\)). Now, let's calculate the resistance \(R\) for each wire: a) \(R_1 = \frac{(1.68\times 10^{-8} \Omega \cdot \mathrm{m})(1\mathrm{~m})}{0.7854\times 10^{-6}\mathrm{~m}^2} = 0.0214 \Omega\) b) \(R_2 = \frac{(1.68\times 10^{-8} \Omega \cdot \mathrm{m})(0.5\mathrm{~m})}{0.1963\times 10^{-6}\mathrm{~m}^2} = 0.0428 \Omega\) c) \(R_3 = \frac{(1.68\times 10^{-8} \Omega \cdot \mathrm{m})(2\mathrm{~m})}{3.1416\times 10^{-6}\mathrm{~m}^2} = 0.0107 \Omega\) d) \(R_4= R_2 = 0.0428 \Omega\)

Find the current flowing through each wire using Ohm's Law

Ohm's Law states that: \(I = \frac{V}{R}\), where \(I\) is the current, \(V\) is the voltage, and \(R\) is the resistance. Calculate the current \(I\) for each wire: a) \(I_1 = \frac{10 \mathrm{V}}{0.0214 \Omega} = 467.3 \mathrm{A}\) b) \(I_2 = \frac{5 \mathrm{V}}{0.0428 \Omega} = 116.8 \mathrm{A}\) c) \(I_3 = \frac{20 \mathrm{V}}{0.0107 \Omega} = 1869.2 \mathrm{A}\) d) \(I_4 = \frac{5 \mathrm{V}}{0.0428 \Omega} = 116.8 \mathrm{A}\)

Compare the currents and determine the largest

Comparing the calculated current values, we can clearly see that the largest current is flowing through wire c) (\(I_3 = 1869.2 \mathrm{A}\)). So, the answer is c) a 2-m-long copper wire of diameter 2 mm connected to a \(20 \mathrm{V}\) battery.