Topic: Intro to Current. For copper wire of the same length, which American Wire Gauge (AWG) will carry the most current before excessive heating (i.e., has the lowest resistance)?

• A. 22 AWG
• B. 12 AWG
• C. 4 AWG
• D. 18 AWG

Answer: (C)

Verified step by step guidance

1. Understand that the resistance of a wire depends on its material, length, and cross-sectional area. Since all wires are copper and have the same length, the key factor is the cross-sectional area, which varies with the American Wire Gauge (AWG) number.
2. Recall that in the AWG system, a smaller gauge number corresponds to a thicker wire, meaning a larger cross-sectional area. Conversely, a larger gauge number means a thinner wire with a smaller cross-sectional area.
3. Use the formula for resistance of a wire: \(R = \rho \frac{L}{A}\), where \(R\) is resistance, \(\rho\) is resistivity of copper, \(L\) is length, and \(A\) is cross-sectional area. Since \(\rho\) and \(L\) are constant, resistance is inversely proportional to \(A\).
4. Recognize that a wire with lower resistance can carry more current before excessive heating occurs, because power dissipated as heat is \(P = I^2 R\). Lower resistance means less heat for the same current.
5. Conclude that the wire with the lowest AWG number (thickest wire) will have the lowest resistance and thus can carry the most current before excessive heating. Therefore, 4 AWG carries more current than 12, 18, or 22 AWG.