Electrical Code Calculations Lv2 - NEC 2017
Calculate the resistance of 150 feet of 14 AWG THWN solid uncoated copper conductor. Calculate the answer to four decimal places.
14 AWG solid uncoated copper = 3.07 Ω/kft R = (DC Res. × L) ÷ 1,000 = (3.07 × 150) ÷ 1,000 = 460.05 ÷ 1,000 = 0.4605 Ω
Calculate the resistance of 175 feet of 14 AWG stranded uncoated copper conductor. Calculate the answer to four decimal places.
14 AWG stranded uncoated copper = 3.14 Ω/kft R = (DC Res. × L) ÷ 1,000 = (3.14 × 175) ÷ 1,000 = 549.5 ÷ 1,000 = 0.5495 Ω
A 10 AWG copper stranded conductor is somewhat larger in diameter than a solid conductor of the same conductor material and size. How much greater is the resistance in ohms per 1,000 feet (ohms/kFT) of an uncoated 10 AWG stranded conductor than that of a 10 AWG solid copper conductor?
0.03 ohms/kFT Chapter 9 Table 8
An 8 AWG stranded conductor is somewhat larger in diameter than a solid conductor of the same size. How much larger is the diameter of an 8 AWG stranded conductor than that of an 8 AWG solid conductor?
0.018" Chapter 9 Table 8
Calculate the resistance of 300 feet of 10 AWG solid aluminum conductor. Calculate the answer to two decimal places.
10 AWG solid aluminum = 2.0 Ω/kft R = (DC Res. × L) ÷ 1,000 = (2.0 × 300) ÷ 1,000 = 600 ÷ 1,000 = 0.60 Ω
What is the voltage drop on a single-phase, 240-volt, branch circuit supplying a 48-ampere load a distance of 88 feet using 4 AWG RHW aluminum conductors? Calculate your answer to two decimal places.
4 AWG aluminum = 0.508 ohms/kFT Vd = (DC res. × I × 2L) ÷ 1,000 = (0.508 × 48 × 2 × 88) ÷ 1,000 = 4,290 ÷ 1,000 = 4.29 volts
What is the voltage drop on a single-phase branch circuit supplying a 42-ampere load a distance of 125 feet with 6 AWG TW uncoated copper conductors? Calculate your answer to two decimal places.
6 AWG uncoated copper = 0.491 ohms/kFT Vd = (DC Res. × I × 2L) ÷ (1,000 × 1.05) = (0.491 × 42 × 2 × 125) ÷ (1,000 × 1.05) = 5,155.5 ÷ 1,050 = 4.91 volts
What is the voltage at the load with a 240-volt supply for a single-phase branch circuit supplying a 40-ampere load at a distance of 72 feet with 8 AWG THW stranded, uncoated copper conductors? Calculate your answer to two decimal places.
Vd = (DC Res. × I × 2L) ÷ 1,000= (0.778 × 40 × 2 × 72) ÷ 1,000= 4,480 ÷ 1,000= 4.48 volts V load = V supply − Vd= 240 − 4.48= 235.52 volts at load
What is the voltage drop on a 3-phase, 480-volt, branch circuit supplying a 120-ampere load a distance of 65 feet with 1 AWG THW uncoated copper conductors? Calculate your answer to two decimal places.
1 AWG uncoated copper = 0.154 ohms/kFT Vd = (DC Res. × I × 1.73L) ÷ 1,000 = (0.154 × 120 × 1.73 × 65) ÷ 1,000 = 2,080 ÷ 1,000 = 2.08 volts
Calculate the resistance of 190 feet of 12 AWG TW solid uncoated copper conductor. Calculate your answer to four decimal places.
12 AWG solid uncoated copper = 1.93 ohms/kFT R = (DC Res. × L) ÷ (1,000 × 1.05) = (1.93 × 190) ÷ (1,000 × 1.05) = 366.7 ÷ 1,050 = 0.3490 ohms
What is the voltage drop on a 480-volt branch circuit supplying 96 amperes to a 3-phase load 200 feet from supply with 3 AWG THHN uncoated copper conductors? Calculate your answer to two decimal places.
3 AWG uncoated copper = 0.245 ohms/kFT Vd = (DC Res. × I × 1.73L × 1.05) ÷ 1,000 = (0.245 × 96 × 1.73 × 200 × 1.05) ÷ 1,000 = 8,540 ÷ 1,000 = 8.54 volts
What is the voltage drop on a single-phase, 460-volt, branch circuit supplying a 215-ampere load a distance of 118 feet, with 4/0 AWG THWN copper uncoated conductors? Calculate your answer to three decimal places.
4/0 AWG uncoated copper = 0.0608 ohms/kFT Vd = (DC Resist. × I × 2L ) ÷ 1,000 = (0.0608 × 215 × 2 × 118) ÷ 1,000 = 3,084.992 ÷ 1,000 = 3.085 volts
Calculate the resistance of 250 feet of 6 AWG THHN aluminum conductor. Calculate your answer to four decimal places.
6 AWG aluminum = 0.808 ohms/kFT R = (DC Res. × L × 1.05) ÷ 1,000 = (0.808 × 250 × 1.05) ÷ 1,000 = 212.1 ÷ 1,000 = 0.2121 ohms
What is the voltage drop of a single-phase branch circuit supplying a 28-ampere load a distance of 58 feet with 8 AWG THHN stranded aluminum conductors? Calculate your answer to two decimal places.
8 AWG stranded aluminum = 1.28 ohms/kFT Vd = (DC Res. × I × 2L × 1.05) ÷ 1,000 = (1.28 × 28 × 2 × 58 × 1.05) ÷ 1,000 = 4,365.312 ÷ 1,000 = 4.37 volts
Calculate the resistance of 200 feet of 8 AWG coated solid copper conductor. Calculate the answer to four decimal places.
Chapter 9, Table 8, DC resistance8 AWG solid coated copper = 0.786 Ω/kft R = (DC Res. × L) ÷ 1,000 = (0.786 × 200) ÷ 1,000 = 157.2 ÷ 1,000 = 0.1572 Ω
Which of the following does not specifically change the direct current resistance of a particular conductor?
Conductor insulation Chapter 9 Table 8
Calculate the resistance of 500 feet of 250,000 circular mils uncoated copper. Calculate the answer to five decimal places.
250 kcmil uncoated copper = 0.0515 Ω/kft R = (DC Res. × L) ÷ 1,000 = (0.0515 × 500) ÷ 1,000 = 25.75 ÷ 1,000 = 0.02575 Ω
What is the maximum length of a single-phase, 120-volt branch circuit supplying a 10-ampere load with 14 AWG solid copper conductors permitted without exceeding a 3% voltage drop?
Vdmax = supply voltage × 3% = 120 × 0.03 = 3.6 Vcmils 14 AWG copper = 4,110 cmilsL = (cmils × Vd) ÷ (2 × k × I ) = (4,110 × 3.6) ÷ (2 × 12.9 × 10) = 14,796 ÷ 258 = 57.35 ft
What size copper conductors are needed to supply a 3-phase, 208-volt, 200-ampere load at a distance of 250 feet and not exceed a 3% voltage drop?
Vdmax = supply voltage × 3% = 208 × 0.03 = 6.24 Vcmils = (k × L × I × 1.73) ÷ Vd = (12.9 × 200 × 250 × 1.73) ÷ 6.24 = 1,115,850 ÷ 6.24 = 178,822 cmils
A 3-phase service is 240 volts at the service equipment. Presently, there is a 5-volt drop to the downstream distribution panel. What size copper branch-circuit conductors are needed to supply a 38-ampere, 3-phase branch-circuit load at a distance of 110 feet and not exceed a total 5% voltage drop on the feeder plus the branch circuit?
Vdmax = supply voltage × 5%= 240 × 0.05= 12 V Vd (BC) = Vdmax − Vd (feeder)= 12 − 5= 7 V (branch circuit)cmils = (k × I × 1.73 × L ) ÷ Vd (BC)= (12.9 × 38 × 1.73 × 110) ÷ 7= 93,285.06 ÷ 7= 13,326 cmils
What is the maximum current for a single-phase 230-volt branch circuit supplying a load at a distance of 140 feet with 4 AWG copper conductors and holding the voltage drop to 2%?
Vdmax = supply voltage × 2% = 230 × 0.02 = 4.6 Vcmils 4 AWG copper = 41,740 cmilsI max = (cmils × Vd) ÷ (k × 2L ) = (41,740 × 4.6) ÷ (12.9 × 2 × 140) =192,004 ÷ 3,612 = 53.16 A
What minimum size copper wire is needed for a 230-volt, single-phase branch circuit supplying a 65-ampere load a distance of 250 feet to limit the voltage drop within the recommended Code limit?
Vd = supply voltage × 3%= 230 × 0.03= 6.9 voltcmils = (k × I × 2L ) ÷ Vd= (12.9 × 65 × 2 × 250) ÷ 6.9= 419,250 ÷ 6.9= 60,761 cmils Chapter 9, Table 8, Area, cmilsRead (more than) 60,761 = 66,360 cmils66,360 = 2 AWG copper
What minimum size aluminum wire is needed for a 440-volt, 3-phase branch circuit supplying a 60-ampere load a distance of 325 feet and holding the voltage drop within a reasonable limit of efficiency?
Vd = supply voltage × 3%= 440 × 0.03= 13.2 volt cmils = (K × L × I × 1.73) ÷ Vd= (21.2 × 60 × 325 × 1.73) ÷ 13.2= 715,182 ÷ 13.2= 54,180 cmils
What is the maximum length of a single-phase, 110-volt branch circuit supplying a 40-ampere load with 6 AWG aluminum conductors permitted without exceeding a 3% voltage drop?
Vdmax = supply voltage × 3% = 110 × 0.03 = 3.3 Vcmils 6 AWG aluminum = 26,240 cmilsL = (cmils × Vd) ÷ (2 × k × I ) = (26,240 × 3.3) ÷ (2 × 21.2 × 40) = 86,592 ÷ 1,696 = 51.06 ft. max. length (one way)
