Voltage Drop worksheet

1. Voltage drop: 5.55 V

1. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{3.2}$$=0.0053875\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $130\text{\hspace{0.17em}ft}\times 0.3048$$=39.624\text{\hspace{0.17em}m}$

3. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 39.624\times 0.0053875\times 13$$=5.550332\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{5.550332}{90}\times 100$$=\mathrm{6.167\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $90-5.550332$$=84.449668\text{\hspace{0.17em}V}$

2. Voltage drop: 0.18 V

1. Find the wire's thickness (diameter d) from its gauge number n = -2, using the standard AWG formula

   $d$$=0.127\times {92}^{\left(\frac{36-\left(\mathrm{-2}\right)}{39}\right)}$$=10.404902\text{\hspace{0.17em}mm}$

2. Work out the area of the round wire's cut end (A). A bigger area means less resistance

   $A$$=\frac{\pi }{4}\times {10.404902}^2$$=85.028773\text{\hspace{0.17em}mm²}$

3. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{85.028773}$$=0.00020275\text{\hspace{0.17em}Ω/m}$

4. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $115\text{\hspace{0.17em}ft}\times 0.3048$$=35.052\text{\hspace{0.17em}m}$

5. Three-phase drop between two lines: multiply √3 × length × resistance per meter × current

   $\mathrm{Vd}$$=\sqrt{3}\times 35.052\times 0.00020275\times 15$$=0.184644\text{\hspace{0.17em}V}$

6. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{0.184644}{110}\times 100$$=\mathrm{0.1679\%}$

7. Take the drop away from the source voltage to get what is left at the load

   $110-0.184644$$=109.815356\text{\hspace{0.17em}V}$

3. Voltage drop: 5.46 V

1. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{2}$$=0.00862\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $100\text{\hspace{0.17em}ft}\times 0.3048$$=30.48\text{\hspace{0.17em}m}$

3. Three-phase drop between two lines: multiply √3 × length × resistance per meter × current

   $\mathrm{Vd}$$=\sqrt{3}\times 30.48\times 0.00862\times 12$$=5.460898\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{5.460898}{140}\times 100$$=\mathrm{3.9006\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $140-5.460898$$=134.539102\text{\hspace{0.17em}V}$

4. Voltage drop: 6.88 V

1. Resistance per meter of aluminum at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 2.82e-8 Ω·m or 0.0282 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.0282}{3.6}$$=0.00783333\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $90\text{\hspace{0.17em}ft}\times 0.3048$$=27.432\text{\hspace{0.17em}m}$

3. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 27.432\times 0.00783333\times 16$$=6.876288\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{6.876288}{120}\times 100$$=\mathrm{5.7302\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $120-6.876288$$=113.123712\text{\hspace{0.17em}V}$

5. Voltage drop: 8.59 V

1. Find the wire's thickness (diameter d) from its gauge number n = 14, using the standard AWG formula

   $d$$=0.127\times {92}^{\left(\frac{36-14}{39}\right)}$$=1.627727\text{\hspace{0.17em}mm}$

2. Work out the area of the round wire's cut end (A). A bigger area means less resistance

   $A$$=\frac{\pi }{4}\times {1.627727}^2$$=2.080908\text{\hspace{0.17em}mm²}$

3. Resistance per meter of aluminum at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 2.82e-8 Ω·m or 0.0282 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.0282}{2.080908}$$=0.01355178\text{\hspace{0.17em}Ω/m}$

4. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $80\text{\hspace{0.17em}ft}\times 0.3048$$=24.384\text{\hspace{0.17em}m}$

5. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 24.384\times 0.01355178\times 13$$=8.591611\text{\hspace{0.17em}V}$

6. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{8.591611}{110}\times 100$$=\mathrm{7.8106\%}$

7. Take the drop away from the source voltage to get what is left at the load

   $110-8.591611$$=101.408389\text{\hspace{0.17em}V}$

6. Voltage drop: 61.94 V

1. Find the wire's thickness (diameter d) from its gauge number n = 25, using the standard AWG formula

   $d$$=0.127\times {92}^{\left(\frac{36-25}{39}\right)}$$=0.454666\text{\hspace{0.17em}mm}$

2. Work out the area of the round wire's cut end (A). A bigger area means less resistance

   $A$$=\frac{\pi }{4}\times {0.454666}^2$$=0.162359\text{\hspace{0.17em}mm²}$

3. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{0.162359}$$=0.10618476\text{\hspace{0.17em}Ω/m}$

4. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $85\text{\hspace{0.17em}ft}\times 0.3048$$=25.908\text{\hspace{0.17em}m}$

5. Three-phase drop between two lines: multiply √3 × length × resistance per meter × current

   $\mathrm{Vd}$$=\sqrt{3}\times 25.908\times 0.10618476\times 13$$=61.944116\text{\hspace{0.17em}V}$

6. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{61.944116}{105}\times 100$$=\mathrm{58.9944\%}$

7. Take the drop away from the source voltage to get what is left at the load

   $105-61.944116$$=43.055884\text{\hspace{0.17em}V}$

7. Voltage drop: 4.1 V

1. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{2.8}$$=0.00615714\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $90\text{\hspace{0.17em}ft}\times 0.3048$$=27.432\text{\hspace{0.17em}m}$

3. Three-phase drop between two lines: multiply √3 × length × resistance per meter × current

   $\mathrm{Vd}$$=\sqrt{3}\times 27.432\times 0.00615714\times 14$$=4.095674\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{4.095674}{134}\times 100$$=\mathrm{3.0565\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $134-4.095674$$=129.904326\text{\hspace{0.17em}V}$

8. Voltage drop: 6.57 V

1. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{2.4}$$=0.00718333\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $125\text{\hspace{0.17em}ft}\times 0.3048$$=38.1\text{\hspace{0.17em}m}$

3. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 38.1\times 0.00718333\times 12$$=6.56844\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{6.56844}{145}\times 100$$=\mathrm{4.53\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $145-6.56844$$=138.43156\text{\hspace{0.17em}V}$

9. Voltage drop: 8.99 V

1. Resistance per meter of copper at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 1.724e-8 Ω·m or 0.01724 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.01724}{1.8}$$=0.00957778\text{\hspace{0.17em}Ω/m}$

2. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $110\text{\hspace{0.17em}ft}\times 0.3048$$=33.528\text{\hspace{0.17em}m}$

3. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 33.528\times 0.00957778\times 14$$=8.991465\text{\hspace{0.17em}V}$

4. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{8.991465}{125}\times 100$$=\mathrm{7.1932\%}$

5. Take the drop away from the source voltage to get what is left at the load

   $125-8.991465$$=116.008535\text{\hspace{0.17em}V}$

10. Voltage drop: 5.93 V

1. Find the wire's thickness (diameter d) from its gauge number n = 11, using the standard AWG formula

   $d$$=0.127\times {92}^{\left(\frac{36-11}{39}\right)}$$=2.304847\text{\hspace{0.17em}mm}$

2. Work out the area of the round wire's cut end (A). A bigger area means less resistance

   $A$$=\frac{\pi }{4}\times {2.304847}^2$$=4.172286\text{\hspace{0.17em}mm²}$

3. Resistance per meter of aluminum at 20 °C: divide the metal's resistivity ρ (how strongly it resists current, 2.82e-8 Ω·m or 0.0282 Ω·mm²/m) by the area

   $\mathrm{R/m}$$=\frac{0.0282}{4.172286}$$=0.00675889\text{\hspace{0.17em}Ω/m}$

4. Change the length to meters to match the resistance per meter (1 ft = 0.3048 m)

   $120\text{\hspace{0.17em}ft}\times 0.3048$$=36.576\text{\hspace{0.17em}m}$

5. The current flows out and back, so the drop covers twice the one-way length: 2 × length × resistance per meter × current

   $\mathrm{Vd}$$=2\times 36.576\times 0.00675889\times 12$$=5.933112\text{\hspace{0.17em}V}$

6. Divide the drop by the source voltage to see what share of it is lost, as a percent

   $\frac{5.933112}{130}\times 100$$=\mathrm{4.5639\%}$

7. Take the drop away from the source voltage to get what is left at the load

   $130-5.933112$$=124.066888\text{\hspace{0.17em}V}$