Lesson 5: Calculating Voltage Drop in Feeders and Branch Circuits Flashcards
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?
Use k = 12.9.
a.
2/0 AWG
b.
3/0 AWG
c.
4/0 AWG
d.
250 kcmil
c. 4/0 AWG
Note:
Vdmax = supply voltage × 3%
= 208 × 0.03
= 6.24 V
cmils = (k × L × I × 1.73) ÷ Vd
= (12.9 × 200 × 250 × 1.73) ÷ 6.24
= 1,115,850 ÷ 6.24
= 178,822 cmils
Chapter 9, Table 8, Area, cmils
Read (more than) 178,822 = 211,600 cmils
211,600 cmils = 4/0 AWG copper
How much greater is the resistance in ohms per 1,000 feet (Ω/kFT) of an uncoated 10 AWG stranded copper conductor than that of a 10 AWG solid copper conductor?
a.
0.01 Ω/kFT
b.
0.02 Ω/kFT
c.
0.03 Ω/kFT
d.
0.04 Ω/kFT
c. 0.03 Ω/kFT
What is the voltage drop on a single-phase, 240-volt branch circuit supplying a 100-ampere load at a distance of 105 feet with 1 AWG THWN aluminum conductors? (Round the FINAL answer to two decimal places.)
5.32 V
Note:
Chapter 9, Table 8, DC resistance
1 AWG THWN aluminum = 0.253 Ω/kFT
Vd = (DC Resist. × I × 2L) ÷ 1,000
= (100 × 2 × 105 × 0.253) ÷ 1,000
= 5,313 ÷ 1,000
= 5.313 V
How much larger is the diameter of an 8 AWG stranded conductor than that of an 8 AWG solid conductor?
a.
0.014”
b.
0.018”
c.
0.021”
d.
0.024”
b.
0.018”
What is the voltage drop on a single-phase, 460-volt branch circuit supplying a 215-ampere load at a distance of 118 feet, with 4/0 AWG THWN copper uncoated conductors at maximum operating temperature? (Round the FINAL answer to three decimal places.)
3.097 V
Note:
Chapter 9, Table 8, DC resistance
4/0 AWG uncoated copper = 0.0608 Ω/kFT
Vd = (DC Resist. × I × 2L) ÷ 1,000
= (0.0608 × 215 × 2 × 118) ÷ 1,000
= 3,084.992 ÷ 1,000
= 3.085 V
What is the voltage drop on a single-phase, 208-volt branch circuit supplying a 130-ampere load at a distance of 185 feet with 2/0 AWG THWN aluminum conductors? (Round the FINAL answer to two decimal places.)
7.66 V
Note:
Chapter 9, Table 8, DC resistance
2/ 0 AWG THWN aluminum = 0.159 Ω/kFT
Vd = (DC Resist. × I × 2L) ÷ 1,000
= (0.159 × 130 × 2 × 185) ÷ 1,000
= 7,647.9 ÷ 1,000
= 7.65 V
What is the resistance of 500 feet of 250,000 circular mils uncoated copper? (Round the FINAL answer to five decimal places.)
0.02575 ohms
Note:
Chapter 9, Table 8, DC resistance
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 minimum size copper wire is needed for a 120-volt, single-phase branch circuit supplying a 30-ampere load at a distance of 205 feet to limit the voltage drop within the recommended Code limit?
Use k = 12.9.
3 AWG
Note:
210.19(A), Informational Note No. 3
Voltage drop limit = 3%
Vd = supply voltage × 3%
= 120 × 0.03
= 3.6 V
cmils = (k × I × 2L) ÷ Vd
= (12.9 × 30 × 2 × 205) ÷ 3.6
= 158,670 ÷ 3.6
= 44,075 cmils
Chapter 9, Table 8, Area, cmils
Read (more than) 44,075 = 52,620 cmils
52,620 cmils = 3 AWG copper
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 voltage drop of 3.6 volts? (Round the FINAL answer to two decimal places.)
58.63 ft
Note: Since K is not given, use Chapter 9, Table 8 values.
Vd=IxR
=Ix (2xLxDC resist.)/1000
L =(Vdx1000)/Ix2xDC resist.
=(3.6x1000)/10x2x3.07 =3600/61.4 =58.63192
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 4.6 volts? (Round the FINAL answer to two decimal places.)
Use k = 12.9.
53.15 A
Note:
Chapter 9, Table 8, circular mils 4 AWG copper = 41,740 circular mils
Imax = (circular mils × Vd) ÷ (k × 2L)
= (41,740 × 4.6) ÷ (12.9 × 2 × 140)
= 192,004 ÷ 3,612
= 53.16 A
What is the voltage drop on a single-phase, 208-volt branch circuit supplying a 54-ampere load at a distance of 98 feet using 6 AWG RHW uncoated copper conductors at maximum operating temperature? (Round the FINAL answer to two decimal places.)
5.21 V
Note:
Chapter 9, Table 8, DC resistance
6 AWG copper = 0.491 Ω/kFT
Vd = (DC res. × I × 2L) ÷ 1,000
= (0.491 × 54 × 2 × 98) ÷ 1,000
= 5,197 ÷ 1,000
= 5.20 V
What is the resistance of 200 feet of 8 AWG coated solid copper conductor? (Round the FINAL answer to four decimal places.)
0.1572 ohms
Note:
Chapter 9, Table 8, DC resistance
8 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
The correct answer is: 0.1572
What is the voltage drop on a 3-phase, 230-volt branch circuit supplying a 125-ampere load at a distance of 175 feet with 4/0 AWG THWN copper uncoated conductors? (Round the FINAL answer to two decimal places.)
2.3 V
Note:
Chapter 9, Table 8, DC resistance
4/0 AWG uncoated copper = 0.0608 Ω/kFT
Vd = (DC Resist. × I × 1.73L) ÷ 1,000
= (0.0608 × 125 × 1.73 × 175) ÷ 1,000
= 2,300.9 ÷ 1,000
= 2.30 V
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 voltage drop of 3.3 volts? (Round the FINAL answer to two decimal places.)
Use k = 21.2.
51.05
Note:
Chapter 9, Table 8, cmils 6 AWG aluminum = 26,240 cmils
L = (cmils × Vd) ÷ (2 × k × I)
= (26,240 × 3.3) ÷ (2 × 21.2 × 40)
= 86,592 ÷ 1,696
= 51.06 ft. max. length (one way)
What is the voltage drop on a 3-phase, 480-volt feeder supplying a 400-ampere load at a distance of 95 feet, with 500 kcmil THWN copper uncoated conductors? (Round the FINAL answer to two decimal places.)
1.7 V
Note:
Chapter 9, Table 8, DC resistance
500 kcmil uncoated copper = 0.0258 Ω/kFT
Vd = (DC Resist. × I × 1.73L) ÷ 1,000
= (0.0258 × 400 × 1.73 × 95) ÷ 1,000
= 1,696.092 ÷ 1,000
= 1.70 V
The correct answer is: 1.7
What minimum size copper wire is needed for a 230-volt, single-phase branch circuit supplying a 65-ampere load at a distance of 250 feet to limit the voltage drop within the recommended Code limit?
Use k = 12.9.
2 AWG
210.19, Informational Note
Voltage drop limit = 3%
Vd = supply voltage × 3%
= 230 × 0.03
= 6.9 V
cmils = (k × I × 2L ) ÷ Vd
= (12.9 × 65 × 2 × 250) ÷ 6.9
= 419,250 ÷ 6.9
= 60,761 cmils
Chapter 9, Table 8, Area, cmils
Read (more than) 60,761 = 66,360 cmils
66,360 cmils = 2 AWG copper
What is the voltage drop on a single-phase, 240-volt branch circuit supplying a 48-ampere load at a distance of 88 feet using 4 AWG RHW aluminum conductors at maximum operating temperature? (Round the FINAL answer to two decimal places.)
4.30 V
Note:
Chapter 9, Table 8, DC resistance
4 AWG aluminum = 0.508 Ω/kFT
Vd = (DC res. × I × 2L) ÷ 1,000
= (0.508 × 48 × 2 × 88) ÷ 1,000
= 4,290 ÷ 1,000
= 4.29 V
The correct answer is: 4.29
Which of the following does not specifically change the DC resistance of a particular conductor?
a.
Coating
b.
Insulation
c.
Material
d.
Stranding
b. insulation
What is the resistance of 300 feet of 10 AWG solid aluminum conductor? (Round the FINAL answer to two decimal places.)
.6 ohms
Note:
Chapter 9, Table 8, DC resistance
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
The correct answer is: 0.6
What is the voltage drop on a 3-phase, 480-volt branch circuit supplying a 120-ampere load at a distance of 65 feet with 1 AWG THW uncoated copper conductors at maximum operating temperature? (Round the FINAL answer to two decimal places.
2.08 V
Note:
Chapter 9, Table 8, DC resistance
1 AWG uncoated copper = 0.154 Ω/kFT
Vd = (DC Res. × I × 1.73L) ÷ 1,000
= (0.154 × 120 × 1.73 × 65) ÷ 1,000
= 2,080 ÷ 1,000
= 2.08 V
What minimum size aluminum wire is needed for a 440-volt, 3-phase branch circuit supplying a 60-ampere load at a distance of 325 feet and holding the voltage drop to 13.2 volts?
Use k = 21.2
2 AWG
Note:
cmils = (k × L × I × 1.73) ÷ Vd
= (21.2 × 60 × 325 × 1.73) ÷ 13.2
= 715,182 ÷ 13.2
= 54,180 cmils
Chapter 9, Table 8, Area, cmils
Read (more than) 54,180 = 66,360 cmils
66,360 cmils = 2 AWG aluminum
What is the resistance of 175 feet of 14 AWG stranded uncoated copper conductor? (Round the FINAL answer to four decimal places.)
.5495 ohms
Note:
Chapter 9, Table 8, DC resistance
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 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?
Use k = 12.9.
8 AWG
Note:
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
Chapter 9, Table 8, Area, cmils
Read (more than) 13,326 = 16,510 cmils
16,510 cmils = 8 AWG copper
The resistance of a conductor will change according to the ambient temperature.
true or false
True