Wire Gauge worksheet

1. Nearest standard gauge: 9 AWG

1. Run the AWG formula backwards to get the exact gauge number n, which may fall between two sizes

   $n$$=36-39\times \frac{\text{ln}23.6220472441}{\text{ln}92}$$=8.726492$

2. Round to the nearest standard gauge between 0000 and 40, since wire is sold in whole sizes

   $\text{Nearest:}\text{9 AWG}$

3. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times 3^2$$=7.0685834706\text{\hspace{0.17em}mm²}$

4. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{7.0685834706}$$=2.43896108\text{\hspace{0.17em}Ω/km}$

2. Nearest standard gauge: 9 AWG

1. Run the AWG formula backwards to get the exact gauge number n, which may fall between two sizes

   $n$$=36-39\times \frac{\text{ln}23.6220472441}{\text{ln}92}$$=8.726492$

2. Round to the nearest standard gauge between 0000 and 40, since wire is sold in whole sizes

   $\text{Nearest:}\text{9 AWG}$

3. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times 3^2$$=7.0685834706\text{\hspace{0.17em}mm²}$

4. Resistance of 1 km of aluminum at 20 °C. Divide the metal's resistivity ρ (2.82e-8 Ω·m, or 0.0282 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.0282\times \mathrm{1,000}}{7.0685834706}$$=3.98948391\text{\hspace{0.17em}Ω/km}$

3. Diameter: 1.63 mm

1. Use the AWG formula for the diameter d. Each gauge step shrinks the wire by the same ratio, starting from 0.127 mm at 36 AWG. Here n = 14 for 14 AWG

   $d$$=0.127\times {92}^{\left(\frac{36-14}{39}\right)}$$=1.6277266338\text{\hspace{0.17em}mm}$

2. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {1.6277266338}^2$$=2.0809077171\text{\hspace{0.17em}mm²}$

3. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{2.0809077171}$$=8.28484601\text{\hspace{0.17em}Ω/km}$

4. Diameter: 1.02 mm

1. Use the AWG formula for the diameter d. Each gauge step shrinks the wire by the same ratio, starting from 0.127 mm at 36 AWG. Here n = 18 for 18 AWG

   $d$$=0.127\times {92}^{\left(\frac{36-18}{39}\right)}$$=1.0236873428\text{\hspace{0.17em}mm}$

2. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {1.0236873428}^2$$=0.8230468337\text{\hspace{0.17em}mm²}$

3. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{0.8230468337}$$=20.94656014\text{\hspace{0.17em}Ω/km}$

5. Nearest standard gauge: 18 AWG

1. Run the AWG formula backwards to get the exact gauge number n, which may fall between two sizes

   $n$$=36-39\times \frac{\text{ln}7.874015748}{\text{ln}92}$$=18.201919$

2. Round to the nearest standard gauge between 0000 and 40, since wire is sold in whole sizes

   $\text{Nearest:}\text{18 AWG}$

3. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times 1^2$$=0.7853981634\text{\hspace{0.17em}mm²}$

4. Resistance of 1 km of aluminum at 20 °C. Divide the metal's resistivity ρ (2.82e-8 Ω·m, or 0.0282 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.0282\times \mathrm{1,000}}{0.7853981634}$$=35.90535516\text{\hspace{0.17em}Ω/km}$

6. Nearest standard gauge: 12 AWG

1. Run the AWG formula backwards to get the exact gauge number n, which may fall between two sizes

   $n$$=36-39\times \frac{\text{ln}15.7480314961}{\text{ln}92}$$=12.22359$

2. Round to the nearest standard gauge between 0000 and 40, since wire is sold in whole sizes

   $\text{Nearest:}\text{12 AWG}$

3. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times 2^2$$=3.1415926536\text{\hspace{0.17em}mm²}$

4. Resistance of 1 km of aluminum at 20 °C. Divide the metal's resistivity ρ (2.82e-8 Ω·m, or 0.0282 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.0282\times \mathrm{1,000}}{3.1415926536}$$=8.97633879\text{\hspace{0.17em}Ω/km}$

7. Nearest standard gauge: 10 AWG

1. Find the diameter of a round wire with that area, by working the circle-area formula backwards

   $d$$=\sqrt{\frac{4\times 5}{\pi }}$$=2.523132522\text{\hspace{0.17em}mm}$

2. Run the AWG formula backwards to get the exact gauge number n, which may fall between two sizes

   $n$$=36-39\times \frac{\text{ln}19.8671852128}{\text{ln}92}$$=10.219558$

3. Round to the nearest standard gauge between 0000 and 40, since wire is sold in whole sizes

   $\text{Nearest:}\text{10 AWG}$

4. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {2.523132522}^2$$=5\text{\hspace{0.17em}mm²}$

5. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{5}$$=3.448\text{\hspace{0.17em}Ω/km}$

8. Diameter: 0.25 mm

1. Use the AWG formula for the diameter d. Each gauge step shrinks the wire by the same ratio, starting from 0.127 mm at 36 AWG. Here n = 30 for 30 AWG

   $d$$=0.127\times {92}^{\left(\frac{36-30}{39}\right)}$$=0.254639003\text{\hspace{0.17em}mm}$

2. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {0.254639003}^2$$=0.0509260195\text{\hspace{0.17em}mm²}$

3. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{0.0509260195}$$=338.5302873\text{\hspace{0.17em}Ω/km}$

9. Diameter: 1.45 mm

1. Use the AWG formula for the diameter d. Each gauge step shrinks the wire by the same ratio, starting from 0.127 mm at 36 AWG. Here n = 15 for 15 AWG

   $d$$=0.127\times {92}^{\left(\frac{36-15}{39}\right)}$$=1.4495324284\text{\hspace{0.17em}mm}$

2. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {1.4495324284}^2$$=1.6502348437\text{\hspace{0.17em}mm²}$

3. Resistance of 1 km of copper at 20 °C. Divide the metal's resistivity ρ (1.724e-8 Ω·m, or 0.01724 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.01724\times \mathrm{1,000}}{1.6502348437}$$=10.44699793\text{\hspace{0.17em}Ω/km}$

10. Diameter: 0.32 mm

1. Use the AWG formula for the diameter d. Each gauge step shrinks the wire by the same ratio, starting from 0.127 mm at 36 AWG. Here n = 28 for 28 AWG

   $d$$=0.127\times {92}^{\left(\frac{36-28}{39}\right)}$$=0.3210938543\text{\hspace{0.17em}mm}$

2. Find the cross-sectional area A, the size of the cut end, from the diameter

   $A$$=\frac{\pi }{4}\times {0.3210938543}^2$$=0.0809755428\text{\hspace{0.17em}mm²}$

3. Resistance of 1 km of aluminum at 20 °C. Divide the metal's resistivity ρ (2.82e-8 Ω·m, or 0.0282 Ω·mm²/m) by the area, because a thicker wire resists less

   $R$$=\frac{0.0282\times \mathrm{1,000}}{0.0809755428}$$=348.25330005\text{\hspace{0.17em}Ω/km}$