# BUS Ele Tech Lib Electrical Formulas

```Short-Circuit Current Calculations
Basic Point-to-Point Calculation Procedure
Step 1.
Determine the transformer full load amps (F.L.A.) from
At some distance from the terminals, depending upon wire size, the L-N fault
current is lower than the L-L fault current. The 1.5 multiplier is an approximation
and will theoretically vary from 1.33 to 1.67. These figures are based on change in
turns ratio between primary and secondary, infinite source available, zero feet from
terminals of transformer, and 1.2 x %X and 1.5 x %R for L-N vs. L-L resistance and
reactance values. Begin L-N calculations at transformer secondary terminals, then
proceed point-to-point.
Step 5.
Calculate "M" (multiplier) or take from Table 2.
either the nameplate, the following formulas or Table 1:
100
Multiplier =
*% Z transformer
Step 2.
Step 6.
Find the transformer multiplier. See Notes 1 and 2
* Note 1. Get %Z from nameplate or Table 1. Transformer impedance (Z) helps to
determine what the short circuit current will be at the transformer secondary.
Transformer impedance is determined as follows: The transformer secondary is short
circuited. Voltage is increased on the primary until full load current flows in the
secondary. This applied voltage divided by the rated primary voltage (times 100) is the
impedance of the transformer.
Example: For a 480 Volt rated primary, if 9.6 volts causes secondary full load current to
flow through the shorted secondary, the transformer impedance is 9.6/480 = .02 = 2%Z.
* Note 2. In addition, UL 1561 listed transformers 25kVA and larger have a ± 10%
impedance tolerance. Short circuit amps can be affected by this tolerance. Therefore, for
high end worst case, multiply %Z by .9. For low end of worst case, multiply %Z by 1.1.
Transformers constructed to ANSI standards have a ±7.5% impedance tolerance
(two-winding construction).
Step 3. Determine by formula or Table 1 the transformer letthrough short-circuit current. See Notes 3 and 4.
1
1 +f
M=
Calculate the available short circuit symmetrical RMS
current at the point of fault. Add motor contribution, if
applicable.
I S.C. sym. RMS = IS.C. x M
Step 6A. Motor short circuit contribution, if significant, may be
added at all fault locations throughout the system. A
practical estimate of motor short circuit contribution is to
multiply the total motor current in amps by 4. Values of 4
to 6 are commonly accepted.
Calculation of Short-Circuit Currents When Primary
Available Short-Circuit Current is Known
Use the following procedure to calculate the level of fault current at the secondary
of a second, downstream transformer in a system when the level of fault current at
the transformer primary is known.
MAIN
TRANSFORMER
IS.C. = TransformerF.L.A. x Multiplier
Note 3. Utility voltages may vary ±10% for power and ±5.8% for 120 Volt lighting
services. Therefore, for highest short circuit conditions, multiply values as calculated in
step 3 by 1.1 or 1.058 respectively. To find the lower end worst case, multiply results in
step 3 by .9 or .942 respectively.
Note 4. Motor short circuit contribution, if significant, may be added at all fault locations
throughout the system. A practical estimate of motor short circuit contribution is to
multiply the total motor current in amps by 4. Values of 4 to 6 are commonly accepted.
Step 4. Calculate the "f" factor.
3Ø Faults
1Ø Line-to-Line (L-L) Faults
See Note 5 & Table 3
1Ø Line-to-Neutral (L-N) Faults
See Note 5 & Table 3
f=
Step A.
2 x L x I L-L
f=
C x n x EL-L
2 x L x I L-N†
f=
C x n x EL-N
† Note 5. The L-N fault current is higher than the L-L fault current at the secondary
terminals of a single-phase center-tapped transformer. The short-circuit current available
(I) for this case in Step 4 should be adjusted at the transformer terminals as follows: At
L-N center tapped transformer terminals, IL-N = 1.5 x IL-L at Transformer Terminals.
IS.C. primary
1.732 x L x I 3Ø
C x n x E L-L
Where:
L = length (feet) of conductor to the fault.
C = constant from Table 4 of “C” values for conductors and
Table 5 of “C” values for busway.
n = Number of conductors per phase (adjusts C value for
parallel runs)
I = Available short-circuit current in amperes at beginning
of circuit.
E = Voltage of circuit.
IS.C. primary
H.V. UTILITY
CONNECTION
Step B.
IS.C. secondary
IS.C. secondary
Calculate the "f" factor (IS.C. primary known)
3Ø Transformer
(I S.C. primary and
I S.C. secondary are
3Ø fault values)
f=
1Ø Transformer
(I S.C. primary and
I S.C. secondary are
1Ø fault values:
I S.C. secondary is L-L)
f=
I S.C. primary x V primary x 1.73 (%Z)
100,000 x kVA
transformer
I S.C. primary x V primary x (%Z)
100,000 x kVA
transformer
Calculate "M" (multiplier).
1
1 +f
Step C. Calculate the short-circuit current at the secondary of the
transformer. (See Note under Step 3 of "Basic Point-toPoint Calculation Procedure".)
M=
I S.C. secondary =
Vprimary
Vsecondary
x M x I S.C. primary
237
Short-Circuit Current Calculations
Three-Phase Short Circuits
System A
One-Line Diagram
Available Utility
Infinite Assumption
1500 KVA Transformer
480V, 3Ø, 3.5%Z,
3.45% X, 0.56%R
1
Fault X1
Fault X3
Step 1. If.l. = 1500 X 1000 = 1804.3A
480 X 1.732
Step 4. f = 1.732 X 50 X 55,137 = 0.4484
22,185 X 1 X 480
Step 2. Multipler
100
= 31.746
3.5 X 0.9†
Step 5. M =
1
= 0.6904
1 + 0.4483
I =1804A
f.l.
Step 3. Is.c. = 1804.3 X 31.746 = 57,279A
Is.c. motor contribution** = 4 X 1804.3 = 7217A
Itotal s.c. sym RMS = 57,279 + 7217 = 64,496A
25’ - 500kcml Cu
3 Single Conductors
6 Per Phase
Magnetic Conduit
Step 6. Is.c. sym RMS = 55,137 X 0.6904 = 38,067A
Is.c. motor contribution** = 4 X 1804.3 = 7217A
Itotal s.c. sym RMS (X3) = 38,067 + 7217 = 45,284A
Fault X2
2
Step 4. f = 1.732 X 25 X 57,279 = 0.0388
22,185 X 6 X 480
2000A Switch
KRP-C 2000SP Fuse
Step 5. M =
400A Switch
LPS-RK-400SP Fuse
1
= 0.9626
1 + 0.0388
Step 6. Is.c. sym RMS = 57,279 X 0.9626 = 55,137A
Is.c. motor contribution** = 4 X 1804.3 = 7217A
Itotal s.c. sym RMS = 55,137 + 7217 = 62,354A
50’ - 500 kcmil Cu
3 Single Conductors
Magnetic Conduit
3
Motor Contribution*
M
*See note 4 on page 240.
**Assumes 100% motor load. If 50% of this load was from motors. Is.c. motor contrib. = 4 X 1804 X 0.5 = 3,608A
† See note 2 on page 240
System B
One-Line Diagram
Available Utility
Infinite Assumption
1000 KVA Transformer
480V, 3Ø, 3.5%Z,
Fault X1
Fault X3
Step 1. Is.c. = 1000 X 1000 = 1202.8A
480 X 1.732
Step 4. f = 1.732 X 20 X 36,761 = 0.1161
2 X 11,424 X 480
Step 2. Multipler =
If.I.=1203A
1
30’ - 500kcml Cu
3 Single Conductors
4 Per Phase
PVC Conduit
100
= 31.746
3.5 X 0.9†
Step 5. M =
=
1
= 0.8960
1 + 0.1161
Step 3. Is.c. = 1202.8 X 31.746 = 38,184A
Step 6. Is.c. sym RMS = 36,761 X 0.8960 = 32,937A
Fault X2
Fault X4
Step 4. f = 1.732 X 30 X 38,184 = 0.0387
26,706 X 4 X 480
Step A. f = 32,937 X 480 X 1.732 X (1.2 X 0.9) = 1.3144
100,000 X 225
2
1600A Switch
KRP-C 1500SP Fuse
=
Step 5. M =
400A Switch
LPS-RK-350SP Fuse
1
= 0.9627
1 + 0.0387
Step 6. Is.c. sym RMS = 38,184 X 0.9627 = 36,761A
20’ - 2/0 Cu
3 Single Conductors
2 Per Phase
PVC Conduit
Step B. M =
1
= 0.4321
1 + 1.3144
Step C. Is.c. sym RMS = 480 X 0.4321 X 32,937 = 32,842A
208
3
225 KVA Transformer
208V, 3Ø
1.2%Z
This example assumes no motor contribution.
4
238
Short-Circuit Current Calculations
Single-Phase Short Circuits
Short circuit calculations on a single-phase center tapped transformer system
require a slightly different procedure than 3Ø faults on 3Ø systems.
A
B
C
1. It is necessary that the proper impedance be used to represent the primary system.
For 3Ø fault calculations, a single primary conductor impedance is used from the
source to the transformer connection. This is compensated for in the 3Ø short circuit
formula by multiplying the single conductor or single-phase impedance by 1.73.
Primary
Secondary
However, for single-phase faults, a primary conductor impedance is considered from
the source to the transformer and back to the source. This is compensated in the
calculations by multiplying the 3Ø primary source impedance by two.
2. The impedance of the center-tapped transformer must be adjusted for the
half-winding (generally line-to-neutral) fault condition.
Short
Circuit
The diagram at the right illustrates that during line-to-neutral faults, the full primary
winding is involved but, only the half-winding on the secondary is involved.
Therefore, the actual transformer reactance and resistance of the half-winding
condition is different than the actual transformer reactance and resistance of the full
winding condition. Thus, adjustment to the %X and %R must be made when
considering line-to-neutral faults. The adjustment multipliers generally used for this
condition are as follows:
• 1.5 times full winding %R on full winding basis.
• 1.2 times full winding %X on full winding basis.
Note: %R and %X multipliers given in “Impedance Data for Single Phase
Transformers” Table may be used, however, calculations must be adjusted to
indicate transformer kVA/2.
3. The impedance of the cable and two-pole switches on the system must be
considered “both-ways” since the current flows to the fault and then returns to the
source. For instance, if a line-to-line fault occurs 50 feet from a transformer, then
100 feet of cable impedance must be included in the calculation.
Primary
Secondary
Short Circuit
L2
N
L1
The calculations on the following pages illustrate 1Ø fault calculations on a
single-phase transformer system. Both line-to-line and line-to-neutral faults are
considered.
L1
Note in these examples:
a. The multiplier of 2 for some electrical components to account for the single-phase
fault current flow,
b. The half-winding transformer %X and %R multipliers for the line-to-neutral fault
situation, and
N
50 Feet
Short Circuit
L2
239
Short-Circuit Current Calculations
Single-Phase Short Circuits
System A
One-Line Diagram
Available Utility
Infinite Assumption
75KVA, 1Ø Transformer.
1.22%X, 0.68%R
1.40%Z
120/240V
Line-to-Line (L-L) Fault
Line-to-Neutral (L-N) Fault
Fault X1
Fault X1
Step 1. If.l. = 75 X 1000 = 312.5A
240
Step 1. If.l. = 75 X 1000 = 312.5A
240
Step 2. Multipler =
100
= 79.37
1.4 X 0.9†
Step 3. Is.c. (L-L) = 312.5 X 79.37 = 24,802A
25’ - 500kcml Cu
Magnetic Conduit
3 Single Conductors
Fault X2
Step 5. M =
100
= 79.37
1.4 X 0.9†
Step 3*. Is.c. (L-N) = 24,802 X 1.5 = 37,202A
Fault X2
Step 4. f = 2 X 25 X 24,802 = 0.2329
22,185 X 1 X 240
2
Step 2. Multipler =
1
= 0.8111
1 + 0.2329
Step 4. f = 2 X 25 X 37,202 = 0.6987
22,185 X 1 X 120
Step 5. M =
1
= 0.5887
1 + 0.6987
Step 6. Is.c. (L-L) (X2) = 24,802 X 0.8111 = 20,116
400A Switch
LPN-RK-400SP Fuse
Step 6*. Is.c. (L-N) (X2) = 37,202 X 0.5887 = 21,900A
50’ - 3 AWG Cu
Magnetic Conduit
3 Single Conductors
Fault X3
Fault X3
Step 4. f = 2 X 50 X 20,116 = 1.7557
4774 X 1 X 240
Step 4. f = 2 X 50 X 21,900** = 3.8323
4774 X 1 X 120
3
Step 5. M =
240
1
= 0.3629
1 + 1.7557
Step 5. M =
1
= 0.2073
1 + 3.823
Step 6. Is.c. (L-L) (X3) = 20,116 X 0.3629 = 7,300A
Step 6*. Is.c. (L-N) (X3) = 21,900 X 0.2073 = 4,540A
† In addition, UL 1561 listed transformers 25kVA
and larger have a ± 10% impedance tolerance.
Short circuit amps can be affected by this
tolerance. Therefore, for high end worst case,
multiply %Z by 0.9. For low end of worst case,
multiply %Z by 1.1. Transformers constructed to
ANSI standards have a ±7.5% impedance
tolerance (two-winding construction).
* Note 5. The L-N fault current is higher than the L-L
fault current at the secondary terminals of a singlephase center-tapped transformer. The short-circuit
current available (I) for this case in Step 4 should be
adjusted at the transformer terminals as follows: At L-N
center tapped transformer terminals, IL-N = 1.5 x IL-L at
Transformer Terminals.
**Assumes the neutral conductor and the line conductor
are the same size.
Short-Circuit Current Calculations
Impedance & Reactance Data
Transformers
Table 1. Short-Circuit Currents Available from
Various Size Transformers
(Based upon actual field nameplate data or from utility transformer worst case
impedance)
Voltage
and
Phase
Full
%
Short
Impedance††
Circuit
kVA
Amps
(Nameplate)
Amps†
25
104
1.5
12175
37.5
156
1.5
18018
120/240
50
208
1.5
23706
1 ph.*
75
313
1.5
34639
100
417
1.6
42472
167
696
1.6
66644
45
125
1.0
13879
75
208
1.0
23132
112.5
312
1.11
31259
150
416
1.07
43237
120/208
225
625
1.12
61960
3 ph.**
300
833
1.11
83357
500
1388
1.24
124364
750
2082
3.50
66091
1000
2776
3.50
88121
1500
4164
3.50
132181
2000
5552
4.00
154211
2500
6940
4.00
192764
75
90
1.00
10035
112.5
135
1.00
15053
150
181
1.20
16726
225
271
1.20
25088
300
361
1.20
33451
277/480
500
602
1.30
51463
3 ph.**
750
903
3.50
28672
1000
1204
3.50
38230
1500
1806
3.50
57345
2000
2408
4.00
66902
2500
3011
4.00
83628
* Single-phase values are L-N values at transformer terminals. These figures
are based on change in turns ratio between primary and secondary, 100,000
KVA primary, zero feet from terminals of transformer, 1.2 (%X) and 1.5 (%R)
multipliers for L-N vs. L-L reactance and resistance values and transformer
X/R ratio = 3.
** Three-phase short-circuit currents based on “infinite” primary.
††
UL listed transformers 25 KVA or greater have a ±10% impedance toler ance. Short-circuit amps shown in Table 1 reflect –10% condition. Transformers constructed to ANSI standards have a ±7.5% impedance tolerance
(two-winding construction).
† Fluctuations in system voltage will affect the available short-circuit current.
Impedance Data for Single-Phase Transformers
Normal Range
Impedance Multipliers**
Suggested
X/R Ratio
of Percent
For Line-to-Neutral
kVA
for
Impedance (%Z)*
Faults
Calculation
for %X
for %R
1Ø
0.75
1.1
1.2–6.0
0.6
25.0
0.6
0.75
37.5
1.4
1.2–6.5
0.75
1.6
1.2–6.4
0.6
50.0
0.75
1.8
1.2–6.6
0.6
75.0
100.0
2.0
1.3–5.7
0.6
0.75
0.75
167.0
2.5
1.4–6.1
1.0
0.75
3.6
1.9–6.8
1.0
250.0
4.7
2.4–6.0
1.0
0.75
333.0
1.0
0.75
5.5
2.2–5.4
500.0
* National standards do not specify %Z for single-phase transformers. Consult
manufacturer for values to use in calculation.
** Based on rated current of the winding (one–half nameplate kVA divided by
secondary line-to-neutral voltage).
Note: UL Listed transformers 25 kVA and greater have a ± 10% tolerance on
their impedance nameplate.
This table has been reprinted from IEEE Std 242-1986 (R1991), IEEE
Recommended Practice for Protection and Coordination of Industrial and
Commercial Power Systems, Copyright© 1986 by the Institute of Electrical and
Electronics Engineers, Inc. with the permission of the IEEE Standards
Department.
.
Im p ed ance Data for Sing le-Phase and Three-Phase Transform ersSupplement †
kVA
1Ø
10
15
%Z
3Ø
—
1.2
1.3
—
75
1.11
1.07
150
225
1.12
300
1.11
—
1.9
333
500
1.24
500
—
2.1
†These represent actual transformer
installations.
Suggested
X/R Ratio for Calculation
1.1
1.1
1.5
1.5
1.5
1.5
4.7
1.5
5.5
nameplate ratings taken from field
Note: UL Listed transformers 25kVA and greater have a ±10% tolerance on
their impedance nameplate.
For example, a 10% increase in system voltage will result in a 10% greater
available short-circuit currents than as shown in Table 1.
241
Short-Circuit Current Calculations
Conductors & Busways "C" Values
Table 4. “C” Values for Conductors
Copper
AWG Three Single Conductors
Conduit
or
kcmil Steel
15kV
5kV
600V
389
14
617
12
981
10
1551
8
1557
2389
2406
2425
6
3696
3751
3806
4
4577
4674
4774
3
5574
5736
5907
2
1
7293
7029
6759
7973
1/0
8925
8544
9390
10062
2/0 10755
11022
3/0 12844
11804
12543
13606
4/0 15082
13644
14925
250 16483
14769
16293
300 18177
15678
17385
350 19704
16366
18235
400 20566
17492
19172
500 22185
17962
600 22965
20567
18889
21387
750 24137
19923
22539
1,000 25278
Aluminum
237
14
376
12
599
10
950
8
951
1472
1476
1481
6
2319
2333
4
2346
2904
2928
2952
3
3626
3670
2
3713
4498
4575
4645
1
5493
5670
5777
1/0
6733
6968
2/0
7187
8163
8467
8826
3/0
9700
10167
4/0 10741
10849
11460
250 12122
12193
13009
300 13910
14280
13288
350 15484
14188
15355
400 16671
16828
15657
500 18756
16484
18428
600 20093
17686
19685
750 21766
1,000 23478
21235
19006
Nonmagnetic
5kV
600V
389
617
982
1555
1559
2418
2430
3789
3826
4745
4811
5926
6044
7307
7493
9034
9317
10878
11424
13048
13923
16673
15351
17121
18594
20868
18975
20526
22737
21786
24297
23277
26706
25204
28033
26453
29735
28083
31491
2407
3753
4679
5809
7109
8590
10319
12360
14347
15866
17409
18672
19731
21330
22097
23408
24887
Three-Conductor Cable
Conduit
Steel
5kV
600V
389
617
982
1557
1559
2431
2425
3812
3830
4785
4820
5930
5989
7365
7454
9086
9210
11045
11245
13333
13656
16392
15890
17851
18311
20617
20052
21914
22646
23372
24253
25449
26980
27975
28752
30024
31051
32689
33864
951
1479
2342
2945
3702
4632
5766
7153
8851
10749
12343
14183
15858
17321
19503
21718
23702
26109
1476
2333
2929
3673
4580
5646
6986
8627
10387
11847
13492
14955
16234
18315
19635
21437
23482
237
376
599
952
1482
2351
2963
3734
4686
5852
7327
9077
11185
12797
14917
16795
18462
21395
23633
26432
29865
237
376
599
952
1482
2350
2961
3730
4678
5838
7301
9110
11174
12862
14923
16813
18506
21391
23451
25976
28779
15kV
951
1480
2347
2955
3719
4664
5820
7271
8981
11022
12636
14698
16490
18064
20607
23196
25790
29049
2415
3779
4726
5828
7189
8708
10500
12613
14813
16466
18319
19821
21042
23126
24897
26933
29320
Nonmagnetic
5kV
600V
389
617
982
1558
1560
2433
2428
3838
3823
4803
4833
6023
6087
7507
7579
9373
9473
11529
11703
14119
14410
17020
17483
19352
19779
22525
21938
24126
24904
26044
26916
28712
30096
31258
32154
33315
34605
35749
37197
2421
3798
4762
5958
7364
9053
11053
13462
16013
18001
20163
21982
23518
25916
27766
29735
31959
1478
2339
2941
3693
4618
5717
7109
8751
10642
12115
13973
15541
16921
19314
21349
23750
26608
237
376
599
952
1482
2353
2966
3740
4699
5876
7373
9243
11409
13236
15495
17635
19588
23018
25708
29036
32938
952
1481
2350
2959
3725
4682
5852
7329
9164
11277
13106
15300
17352
19244
22381
25244
28262
31920
1479
2344
2949
3709
4646
5771
7202
8977
10969
12661
14659
16501
18154
20978
23295
25976
29135
15kV
15kV
Note: These values are equal to one over the impedance per foot and based upon resistance and reactance values found in IEEE Std 241-1990 (Gray Book), IEEE Recommended Practice for Electric Power
Systems in Commerical Buildings & IEEE Std 242-1986 (Buff Book), IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems. Where resistance and reac tance values differ or are not available, the Buff Book values have been used. The values for reactance in determining the C Value at 5 KV & 15 KV are from the Gray Book only (Values for 14-10 AWG at 5 kV
and 14-8 AWG at 15 kV are not available and values for 3 AWG have been approximated).
Table 5. “C” Values for Busway
Ampacity
Busway
Plug-In
Feeder
Copper
Aluminum
Copper
225
28700
23000
18700
400
38900
34700
23900
600
41000
38300
36500
800
46100
57500
49300
1000
69400
89300
62900
1200
94300
97100
76900
1350
119000
104200
90100
1600
129900
120500
101000
2000
142900
135100
134200
2500
143800
156300
180500
3000
144900
175400
204100
4000
—
—
277800
Note: These values are equal to one over the
impedance in a survey of industry.
242
High Impedance
Aluminum
Copper
—
12000
—
21300
—
31300
—
44100
15600
56200
16100
69900
84000
17500
90900
19200
125000
20400
166700
21700
188700
23800
256400
—
impedance per foot for
Voltage Drop Calculations
Ratings of Conductors and Tables to Determine Volt Loss
With larger loads on new installations, it is extremely important to consider volt
loss, otherwise some very unsatisfactory problems are likely to be
encountered.
The actual conductor used must also meet the other sizing requirements such a
full-load current, ambient temperature, number in a raceway, etc.
How to Figure Volt Loss
Multiply distance (length in feet of one wire) by the current (expressed in amps) by the
figure shown in table for the kind of current and the size of wire to be used, by one over
the number of conductors per phase.
Then, put a decimal point in front of the last 6 digits–you have the volt loss to be
Example – 6 AWG copper wire, one per phase, in 180 feet of steel
conduit–3 phase, 40 amp load at 80% power factor.
Multiply feet by amperes: 180 x 40 = 7200
Multiply this number by number from table for 6 AWG wire threephase at 80% power factor: 7200 x 745† = 5364000
1
1 x 5364000 = 5364000
Multiply by
x 5364000 =
#/phase
1
Place decimal point 6 places to left:
This gives volt loss to be expected: 5.364V
(For a 240V circuit the % voltage drop is 5.364 x 100 or 2.23%).
240
Table A and B take into consideration reactance on AC circuits as
well as resistance of the wire.
Remember on short runs to check to see that the size and type
of wire indicated has sufficient ampacity.
expected on that circuit.
How to Select Size of Wire
Multiply distance (length in feet of one wire) by the current (expressed in
amps), by one over the number of conductors per phase.
Divide that figure into the permissible volt loss multiplied by 1,000,000.
Example – Copper in 180 feet of steel conduit–3 phase, 40 amp
Ioad at 80% power factor–maximum volt loss permitted from local
code equals 5.5 volts.
Multiply feet by amperes by 1
1
180 x 40 x = 7200.
#/phase
1
Divide permissible volt loss multiplied by 1,000,000 by this
5.5 x 1,000,000 = 764.
number:
7200
Look under the column in Table A and B applying to the type of current and
power factor for the value nearest, but not above your result – you have the
size of wire needed.
† Value from Table A
Select number from Table A, three-phase at 80% power factor, that is nearest
but not greater than 764. This number is 745 which indicates the size of wire
needed: 6 AWG.
Line-to-Neutral
For line to neutral voltage drop on a 3 phase system, divide the three phase
value by 1.73. For line to neutral voltage drop on a single phase system,
divide single phase value by 2.
Open Wiring
The volt loss for open wiring installations depends on the separation between
conductors. The volt loss is approximately equal to that for conductors in
non-magnetic conduit.
Installation in Conduit, Cable or Raceway
NEC® Tables 310.15(B)(16) through 310.15(B)(19) give allowable ampacities
(current-carrying capacities) for not more than three current carrying
conductors in a conduit, cable, or raceway. Where the number of current
carrying conductors exceeds three the allowable ampacity of each conductor
must be reduced as shown in the following tables:
Installation in Conduit, Cable or Raceway per 310.15(B)(2)(a)
The Number of
Conductors In One
Conduit, Raceway
Or Cable
4 to 6
7 to 9
10 to 20
21 to 30
31 to 40
41 and over
Percentage of Values
In Tables 310.16 And
310.18
80%
70%
50%
45%
40%
35%
Conditions Causing Higher Volt Loss
The voltage loss is increased when a conductor is operated at a higher
temperature because the resistance increases.
Room Temperature Affects Ratings
The ampacities (carrying capacities) of conductors are based on a room
temperature of either 30°C or 40ºC. For derating based upon 30ºC ambient, if
room temperature is higher, the ampacities are reduced by using the following
multipliers; (for 0-2000 volt, insulated conductors not more than 3 conductors
in raceway or direct buried, Table 310.15(B)(2)(a)). For room temperatures
based upon a 40ºC ambient, see Table 310.15(B)(2)(b).
Room Temperature Affects Ratings
Room
Temperature
°C
31-35
36-40
41-45
46-50
51-55
56-60
61-70
71-80
TW
°F
87-95
96-104
105-113
114-122
123-131
132-140
141-158
159-176
Ampacity Multiplier
THHN, XHHW*
THW, THWN
(75°C Wire)
(60°C Wire)
.94
.91
.88
.82
.82
.71
.75
.58
.67
.41
.58
–
.33
–
–
–
(90°C Wire)
.96
.91
.87
.82
.76
.71
.58
.41
243
Voltage Drop Calculations
Table A — Copper Conductors — Ratings & Volt Loss†
Conduit
Steel
Conduit
NonMagnetic
Conduit
Covered
Cables or
Installation
in Fibre or
Other
NonMagnetic
Conduit,
Etc.)
Wire
Size
14
12
10
8
6
4
3
2
1
0
00
000
0000
250
300
350
400
500
600
750
1000
14
12
10
8
6
4
3
2
1
0
00
000
0000
250
300
350
400
500
600
750
1000
Ampacity
Type
T, TW
(60°C
Wire)
20*
25*
30
40
55
70
85
95
110
125
145
165
195
215
240
260
280
320
335
400
455
20*
25*
30
40
55
70
85
95
110
125
145
165
195
215
240
260
280
320
335
400
455
Type
RH,
THWN,
RHW,
THW
(75°C
Wire)
20*
25*
35*
50
65
85
100
115
130
150
175
200
230
255
285
310
335
380
420
475
545
20*
25*
35*
50
65
85
100
115
130
150
175
200
230
255
285
310
335
380
420
475
545
Type
RHH,
THHN,
XHHW
(90°C
Wire)
25*
30*
40*
55
75
95
110
130
150
170
195
225
260
290
320
350
380
430
475
535
615
25*
30*
40*
55
75
95
110
130
150
170
195
225
260
290
320
350
380
430
475
535
615
Direct
Current
6140
3860
2420
1528
982
616
490
388
308
244
193
153
122
103
86
73
64
52
43
34
26
6140
3464
2420
1528
982
616
470
388
308
244
193
153
122
103
86
73
64
52
43
34
26
Volt Loss (See explanation prior page.)
Three-Phase
(60 Cycle, Lagging Power Factor.)
60%
70%
90%
80%
100%
Single-Phase
(60 Cycle, Lagging Power Factor.)
70%
90%
80%
100%
60%
5369
3464
2078
1350
848
536
433
346
277
207
173
136
109
93
77
67
60
50
43
36
31
5369
3464
2078
1350
848
536
433
329
259
207
173
133
107
90
76
65
57
46
39
32
25
6200
4000
2400
1560
980
620
500
400
320
240
200
158
126
108
90
78
70
58
50
42
36
6200
4000
2400
1560
980
620
500
380
300
240
200
154
124
104
88
76
66
54
46
38
30
3836
2508
1540
1040
690
468
394
331
283
232
206
178
157
148
135
126
120
111
106
102
95
3812
2486
1520
1019
669
448
375
300
253
214
188
159
140
128
118
109
103
94
90
83
77
4887
3169
1918
1264
812
528
434
354
292
228
196
162
136
123
108
98
91
81
75
68
62
4876
3158
1908
1255
802
519
425
330
268
220
188
151
127
112
99
89
81
71
65
58
51
4371
2841
1728
1148
745
491
407
336
280
223
194
163
140
128
115
106
99
90
84
78
72
4355
2827
1714
1134
731
479
395
310
255
212
183
150
128
114
103
94
87
77
72
65
59
3848
2508
1532
1026
673
450
376
312
264
213
188
160
139
129
117
109
103
94
89
84
78
3830
2491
1516
1010
657
435
361
286
238
199
174
145
125
113
104
95
89
80
76
70
63
3322
2172
1334
900
597
405
341
286
245
200
178
154
136
128
117
109
104
96
92
88
82
3301
2153
1316
882
579
388
324
259
219
185
163
138
121
110
102
94
89
82
77
72
66
5643
3659
2214
1460
937
610
501
409
337
263
227
187
157
142
125
113
105
94
86
79
72
5630
3647
2203
1449
926
599
490
381
310
254
217
175
147
129
114
103
94
82
75
67
59
5047
3281
1995
1326
860
568
470
388
324
258
224
188
162
148
133
122
114
104
97
91
84
5029
3264
1980
1310
845
553
456
358
295
244
211
173
148
132
119
108
100
90
83
76
68
4444
2897
1769
1184
777
519
434
361
305
246
217
184
161
149
135
126
118
109
103
97
90
4422
2877
1751
1166
758
502
417
330
275
230
201
167
145
131
120
110
103
93
87
80
73
* The overcurrent protection for conductor types marked with an (*) shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper; or 15
amperes for 12 AWG and 25 amperes for 10 AWG aluminum and copper-clad aluminum after any correction factors for ambient temperature and number of conductors have
been applied.
† Figures are L-L for both single-phase and three-phase. Three-phase figures are average for the three-phase.
244
Voltage Drop Calculations
Table B — Aluminum Conductors — Ratings & Volt Loss†
Conduit
Steel
Conduit
NonMagnetic
Conduit
Covered
Cables or
Installation
in Fibre or
Other
NonMagnetic
Conduit,
Etc.)
Wire
Size
12
10
8
6
4
3
2
1
0
00
000
0000
250
300
350
400
500
600
750
1000
12
10
8
6
4
3
2
1
0
00
000
0000
250
300
350
400
500
600
750
1000
Ampacity
Type
T, TW
(60°C
Wire)
20*
25
30
40
55
65
75
85
100
115
130
150
170
190
210
225
260
285
320
375
20*
25
30
40
55
65
75
85
100
115
130
150
170
190
210
225
260
285
320
375
Type
RH,
THWN,
RHW,
THW
(75°C
Wire)
20*
30*
40
50
65
75
90
100
120
135
155
180
205
230
250
270
310
340
385
445
20*
30*
40
50
65
75
90
100
120
135
155
180
205
230
250
270
310
340
385
445
Type
RHH,
THHN,
XHHW
(90°C
Wire)
25*
35*
45
60
75
85
100
115
135
150
175
205
230
255
280
305
350
385
435
500
25*
35*
45
60
75
85
100
115
135
150
175
205
230
255
280
305
350
385
435
500
Direct
Current
6360
4000
2520
1616
1016
796
638
506
402
318
259
200
169
141
121
106
85
71
56
42
6360
4000
2520
1616
1016
796
638
506
402
318
252
200
169
141
121
106
85
71
56
42
Volt Loss (See explanation two pages prior.)
Three-Phase
(60 Cycle, Lagging Power Factor.)
100%
90%
80%
70%
60%
Single-Phase
(60 Cycle, Lagging Power Factor.)
100%
90%
80%
70%
60%
5542
3464
2251
1402
883
692
554
433
346
277
225
173
148
124
109
95
77
65
53
43
5542
3464
2251
1402
883
692
554
433
346
277
225
173
147
122
105
93
74
62
50
39
6400
4000
2600
1620
1020
800
640
500
400
320
260
200
172
144
126
110
90
76
62
50
6400
4000
2600
1620
1020
800
640
500
400
320
260
200
170
142
122
108
86
72
58
46
3948
2500
1663
1074
708
574
475
391
328
278
239
201
186
168
155
144
130
122
114
103
3926
2480
1643
1053
668
555
456
373
310
260
223
185
167
150
137
128
114
105
95
86
5039
3165
2075
1310
840
668
541
432
353
290
241
194
173
150
135
122
106
95
84
73
5029
3155
2065
1301
831
659
532
424
344
281
234
186
163
141
125
114
96
85
73
63
4504
2836
1868
1188
769
615
502
405
334
277
234
191
173
152
139
127
112
102
92
82
4490
2823
1855
1175
756
603
490
394
322
266
223
181
160
140
125
116
100
90
79
70
3963
2502
1656
1061
692
557
458
373
310
260
221
184
168
150
138
127
113
105
96
87
3946
2486
1640
1045
677
543
443
360
296
247
209
171
153
136
123
114
100
91
82
73
3419
2165
1441
930
613
497
411
338
284
241
207
174
161
145
134
125
113
106
98
89
3400
2147
1423
912
596
480
394
323
268
225
193
160
145
130
118
111
98
91
82
75
5819
3654
2396
1513
970
771
625
499
407
335
279
224
200
174
156
141
122
110
97
85
5807
3643
2385
1502
959
760
615
490
398
325
270
215
188
163
144
132
111
98
85
73
5201
3275
2158
1372
888
710
580
468
386
320
270
221
200
176
160
146
129
118
107
95
5184
3260
2142
1357
873
696
566
455
372
307
258
209
185
162
145
134
115
104
92
81
4577
2889
1912
1225
799
644
529
431
358
301
256
212
194
173
159
146
131
121
111
100
4557
2871
1894
1206
782
627
512
415
342
285
241
198
177
157
142
132
115
106
94
85
* The overcurrent protection for conductor types marked with an (*) shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper; or 15
amperes for 12 AWG and 25 amperes for 10 AWG aluminum and copper-clad aluminum after any correction factors for ambient temperature and number of conductors have been
applied.
† Figures are L-L for both single-phase and three-phase. Three-phase figures are average for the three-phase.
245
Glossary
Common Electrical Terminology
Ohm
The unit of measure for electric resistance. An ohm is the amount of resistance that will
allow one amp to flow under a pressure of one volt.
Semiconductor Fuses
Fuses used to protect solid-state devices. See “High Speed Fuses.”
Short-Circuit
The relationship between voltage, current, and resistance, expressed by the equation E
= IR, where E is the voltage in volts, I is the current in amps, and R is the resistance in
ohms.
Can be classified as an overcurrent which exceeds the normal full load current of a
circuit by a factor many times (tens, hundreds or thousands greater). Also characteristic
of this type of overcurrent is that it leaves the normal current carrying path of the circuit
– it takes a “short cut” around the load and back to the source.
One Time Fuses
Short-Circuit Current Rating
Overcurrent
Single-Phasing
Ohm’s Law
Generic term used to describe a Class H nonrenewable cartridge fuse, with a single
element.
A condition which exists on an electrical circuit when the normal load current is
exceeded. Overcurrents take on two separate characteristics – overloads and shortcircuits.
Can be classified as an overcurrent which exceeds the normal full load current of a
circuit. Also characteristic of this type of overcurrent is that it does not leave the normal
current carrying path of the circuit – that is, it flows from the source, through the
conductors, through the load, back through the conductors, to the source again.
Peak Let-Through Current, lp
The instantaneous value of peak current let-through by a current-limiting fuse, when it
operates in its current-limiting range.
Renewable Fuse (600V & below)
A fuse in which the element, typically a zinc link, may be replaced after the fuse has
opened, and then reused. Renewable fuses are made to Class H standards.
An electrical load which is characteristic of not having any significant inrush current.
When a resistive load is energized, the current rises instantly to its steady-state value,
without first rising to a higher value.
RMS Current
The RMS (root-mean-square) value of any periodic current is equal to the value of the
direct current which, flowing through a resistance, produces the same heating effect in
the resistance as the periodic current does.
The maximum short-circuit current an electrical component can sustain without the
occurrence of excessive damage when protected with an overcurrent protective device.
That condition which occurs when one phase of a three-phase system opens, either in a
low voltage (secondary) or high voltage (primary) distribution system. Primary or
secondary single-phasing can be caused by any number of events. This condition
results in unbalanced currents in polyphase motors and unless protective measures are
taken, causes overheating and failure.
Threshold Current
The symmetrical RMS available current at the threshold of the current-limiting range,
where the fuse becomes current-limiting when tested to the industry standard. This
value can be read off of a peak let-through chart where the fuse curve intersects the
A - B line. A threshold ratio is the relationship of the threshold current to the fuse’s
continuous current rating.
Time-Delay Fuse
A fuse with a built-in delay that allows temporary and harmless inrush currents to pass
without opening, but is so designed to open on sustained overloads and short-circuits.
Voltage Rating
The maximum open circuit voltage in which a fuse can be used, yet safely interrupt an
overcurrent. Exceeding the voltage rating of a fuse impairs its ability to clear an overload
or short-circuit safely.
Withstand Rating
The maximum current that an unprotected electrical component can sustain for a
specified period of time without the occurrence of extensive damage.
Electrical Formulas
Single-Phase
To Find
Amperes when kVA is known
Amperes when horsepower is known
Amperes when kilowatts are known
Kilowatts
Kilovolt-Amperes
Horsepower
Watts
I = Amperes
HP = Horsepower
Two-Phase
Three-Phase
kVA ≈ 1000
E
HP ≈ 746
E ≈ % eff. ≈ pf
kW ≈ 1000
E ≈ pf
I ≈ E ≈ pf
1000
I≈ E
1000
I ≈ E % eff. ≈ pf
746
E ≈ I ≈ pf
kVA ≈ 1000
kVA ≈ 1000
E≈ 2
E ≈ 1.73
HP ≈ 746
HP ≈ 746
E ≈ 2 ≈ % eff. ≈ pf
E ≈ 1.73 ≈ % eff. ≈ pf
kW ≈ 1000
kW ≈ 1000
E ≈ 2 pf
E ≈ 1.73 ≈ pf
I ≈ E ≈ 2 ≈ pf
I ≈ E ≈ 1.73 ≈ pf
1000
1000
I≈ E≈ 2
I ≈ E ≈ 1.73
1000
1000
I ≈ E ≈ 2 ≈ % eff. ≈ pf I ≈ E ≈ 1.73 ≈ % eff. ≈ pf
746
746
I ≈ E ≈ 2 ≈ pf
I ≈ E ≈ 1.73 ≈ pf
Energy Efficiency = Load Horsepower ≈ 746
Load Input kVA ≈ 1000
Power
Consumed = W or kW = cosθ
Power Factor = pf =
kVA
VA
Apparent Power
E = Volts
% eff. = Percent Efficiency
kW = Kilowatts
pf = Power Factor
Direct Current
Not Applicable
HP ≈ 746
E ≈ % eff.
kW ≈ 1000
E
I≈ E
1000
Not Applicable
I ≈ E ≈ % eff.
746
E≈ I
kVA = Kilovolt-Amperes
263
```