Lesson 8: Calculation of Available Fault Current (2023)

When calculating the multiplier for transformer available fault current, the transformer impedance (%Z) is multiplied by ? to determine the worst-case condition to achieve the highest available fault current.

0.9 Note: The marked impedance values on a transformer may vary ± 10% from the actual values determined by UL product standard. To establish a worst-case condition requires multiplying % Z by 0.9.

Which of the following is/are a major source(s) of fault current contribution?

A.) On-site generators C.) Utility source Both a. and c.

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 7 is ? than the available fault current at position 6.

Lower

Various methods have been developed to calculate the available fault current. All are based on ? .

Ohm's Law

The first step to determine the fault current at any point in the system is to draw a(n) ? showing all the sources of the available fault current.

One-line diagram

Which of the following is not a critical point where short-circuit calculations should be completed in a system?

Push button device on 120V control circuit

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 Use the single-phase formula for this calculation. The IAFC at the beginning of the run is at panel LPA. The IAFC from panel LPA must be multiplied by 0.5 (use 100% of the IAFC at the transformer, 50% everywhere else for L-N or L-G) per Figure 4 Table procedure found in the Reference. Use 50% of the MDP IAFC to calculate the "f" factor and 50% of the MDP IAFC for the total IAFC.) The available fault current at position 9 (florescent luminaire) is ? . Note: Round to three decimal points for all calculations. The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

1067 A f = (1.732 × 30 × 15570 × 0.5) ÷ (617 × 1 × 120) = 6.309 M = 1 ÷ (1 + 6.309) = 0.137 Iᴀғᴄ = 15570 A × 0.5* × 0.137 = 1067 A Note: Use 0.5(50%) for L - N available fault current calculations per the Bolted Fault Approximation section in the Reference.

Per NEC ? , the overcurrent protective devices, the total impedance, the component short-circuit current ratings, and other characteristics of the circuit to be protected shall be selected and coordinated to permit the circuit protective devices to clear a fault without extensive damage to the electrical components of the circuit.

110.10

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 6 (AC-1) is ? . (Note that AC-1 and AC-2 are 3-phase, L-L-L loads. They do not require a neutral conductor to operate.) Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

11589 A f = (1.732 × 35 × 39417) ÷ (4774 × 1 × 208) = 2.407 M = 1 ÷ (1 + 2.407) = 0.294 Iᴀғᴄ = 39417 A × 0.294 = 11589.598 or 11589 A

What is the "C" value for one 4/0 AWG/phase (600-volt, copper, three conductors) in a steel conduit used in a 3-phase circuit?

15,082

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 3 (LPA) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

15570 A f = (1.732 × 60 × 39417) ÷ (12844 × 1 × 208) = 1.534 M = 1 ÷ (1 + 1.534) = 0.395 Iᴀғᴄ = 39417 A × 0.395 = 15569.715 or 15570 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 10 (combination motor controller) is ?. Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

16449 A f = (1.732 × 4 × 25385) ÷ (1557 × 1 × 208) = 0.543 M = 1 ÷ (1 + 0.534) = 0.648 Iᴀғᴄ = 25385 A × 0.648 = 16449 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 5 (LPA) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

16595 A f = (1.732 × 20 × 39417) ÷ (4774 × 1 × 208) = 0.552 M = 1 ÷ (1 + 1.375) = 0.421 Iᴀғᴄ = 39417 A × 0.421 = 16595.557 or 16595 A

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 7 (AC-2) is ? . (Select the answer closest to the calculated result.)

20504 A f = (1.732 × 35 × 64276) ÷ (380 × 1 × 480) = 2.133 M = 1 ÷ (1 + 1.2.133) = 0.139 Iᴀғᴄ = 64276 A × 0.139 = 20504 A

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 6 (AC-1) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

23782 A f = (1.732 × 35 × 64276) ÷ (4774 × 1 × 480) = 1.700 M = 1 ÷ (1 + 1.700) = 0.370 Iᴀғᴄ = 64276 A × 0.370 = 23782 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 4 (LPA) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

25385 A f = (1.732 × 15 × 39417) ÷ (8925 × 1 × 208) = 0.552 M = 1 ÷ (1 + 0.552) = 0.644 Iᴀғᴄ = 39417 A × 0.644 = 25384.548 or 25384.548 A

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 3 (LPA) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

30852 A f = (1.732 × 60 × 64276) ÷ (12844 × 1 × 480) = 1.083 M = 1 ÷ (1 + 1.083) = 0.480 Iᴀғᴄ = 64276 A × 0.480 = 30852 A

For this question, use the following criteria for the various point-to-point calculation step results in calculating the answers: • Transformer full load current: round to a whole number, such as 601 • Square root of three: 1.732 • f: three significant digits right of the decimal point, such as 1.036 • Transformer multiplier: three significant digits right of the decimal point • M: three significant digits right of the decimal point • Final calculated available fault current in amperes: round to whole number, such as 31,564 Determine the available fault current with motor contribution on the 480/277-volt secondary of a 500-kilovolt-ampere transformer with 2% impedance. Assume motor load is 50% of the transformer rating. (Use a 0.9 multiplier for %Z and assume the primary of the transformer has an infinite available fault current. Select the answer closest to the calculated result.)

34591 A Note: Iᴀғᴄ without motor contribution: 33412.48 A Motor contribution: 1202 A Iᴀғᴄ with motor contribution: 34591.15 A Iғʟᴀ = (500 × 1000) ÷ (480 × 1.732) = 601 A M = 100 ÷ (0.9 × 2) = 55.556 Iᴀғᴄ = 601 A × 55.556 = 33389.15 A (without motor contribution) Motor contribution: = (motor load 50% of transformer): 0.5 × 4 × 601 = 1202 A Iᴀғᴄ = 33389.15 + 1202 = 34591.15 A (with motor contribution)

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 5 (LPB) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

38459 A f = (1.732 × 20 × 64276) ÷ (4774 × 1 × 480) = 0.972 M = 1 ÷ (1 + 0.972) = 0.507 Iᴀғᴄ = 64276 A × 0.507 = 38459 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 2 (MDP) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

39,417 A f = (1.732 × 20 × 46264) ÷ (22185 × 2 × 208) = 0.174 M = 1 ÷ (1 + 0.174) = 0.852 Iᴀғᴄ = 46264 A × 0.852 = 39416.926 or 39417 A Note: The conductor run used in this problem has two conductors per phase. The formula used in Step 4 has a variable "n," which represents the conductors per phase. Often students will forget about using the value for variable "n" when a conductor run has multiple conductors per phase. Multiple conductors per phase provide parallel paths for the fault current to flow, which, in effect, reduces the impedance of that portion of the electrical system. Parallel conductor fault current paths must be considered for accuracy in the equations.

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 6 (AC-1) is ? . (Note that AC-1 and AC-2 are 3-phase, L-L-L loads. They do not require a neutral conductor to operate.) Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

9815 A f = (1.732 × 35 × 39417) ÷ (3086 × 1 × 208) = 3.019 M = 1 ÷ (1 + 3.019) = 0.249 Iᴀғᴄ = 39417 A × 0.249 = 9814.833 or 9815 A

For this question, use the following criteria for the various point-to-point calculation step results in calculating the answers: • Transformer full load current: round to a whole number, such as 601 • Square root of three: 1.732 • f: three significant digits right of the decimal point, such as 1.036 • Transformer multiplier: three significant digits right of the decimal point • M: three significant digits right of the decimal point • Final calculated available fault current in amperes: round to whole number, such as 31,564. Calculate the available fault current at the transformer secondary terminals. The transformer is rated at 1,000 kilovolt-amperes, 3% impedance, 480/277-volt secondary, 3-phase. Use a 0.9 multiplier for %Z transformer impedance (−10% tolerance). Do not include motor contribution. Select the answer closest to the calculated result.

44556 A Iғʟᴀ = (1000 × 1000) ÷ (480 × 1.732) = 1203 A M = 100 ÷ (0.9 × 1.732) = 37.037 Iᴀғᴄ = 1203 A × 37.037 = 44556 A Note: The FC² Calculator result is 44458 amperes. Using different rounding or significant decimals can result in slightly different outcomes.

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 4 (LPC) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

46214 A f = (1.732 × 15 × 64276) ÷ (8925 × 1 × 480) = 0.390 M = 1 ÷ (1 + 0.390) = 0.719 Iᴀғᴄ = 64276 A × 0.719 = 46214 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 1 (transformer secondary terminals) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

46264 A Iғʟᴀ = (300 × 1000) ÷ (208 × 1.732) = 832.741 A M = 100 ÷ (0.9 × 2) = 55.556 Iᴀғᴄ = 832.741 A × 55.556 = 46263.759 A or 46264 A

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at the end of a 480-volt, 3-phase, 50-foot, 800-ampere, copper feeder bus with an available fault current of 60,000 amperes at the beginning of the run is ? . (Select the answer closest to the calculated result.)

49200 A f = (1.732 × 50 × 60000) ÷ (49300 × 1 × 480) = 0.220 M = 1 ÷ (1 + 0.220) = 0.820 Iᴀғᴄ = 60000 A × 0.820 = 49200 A

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 2 (MDP) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

64276 A f = (1.732 × 15 × 66815) ÷ (22965 × 4 × 480) = 0.220 M = 1 ÷ (1 + 0.039) = 0.962 Iᴀғᴄ = 66815 A × 0.962 = 64276 A

For this question, use the following criteria for the various point-to-point calculation step results in calculating the answers: • Transformer full load current: round to a whole number, such as 601 • Square root of three: 1.732 • f: three significant digits right of the decimal point, such as 1.036 • Transformer multiplier: three significant digits right of the decimal point • M: three significant digits right of the decimal point • Final calculated available fault current in amperes: round to whole number, such as 31,564 Calculate the available fault current at the transformer secondary terminals of a transformer rated 750 kilovolt-amperes, 1.5% impedance, 480/277-volt secondary, 3-phase. Use a 0.9 multiplier for %Z transformer impedance (−10% tolerance). Do not include motor contribution. Select the answer closest to the calculated results.

66815 A I = 750000 ÷ (480 × 1.732) = 902 A M = 100 ÷ (0.9 × 1.5) = 74.074 Iᴀғᴄ = 902 A × 74.074 = 66815 A Note: The FC² Calculator result is 66822 amperes.

Follow the steps outlined in the short-circuit calculation procedure. For these calculations, do not consider adding motor contribution to the available fault current. For the calculation step using transformer %Z, use the multiplier of 0.9 times the transformer impedance. The available fault current at position 1 (transformer secondary terminals) is ? . Note: The values calculated for this question will be used for additional questions. Select the answer closest to the calculated result.

66815 A Iғʟᴀ = (750 × 1000) ÷ (480 × 1.732) = 902 A M = 100 ÷ (0.9 × 1.5) = 74.074 Iᴀғᴄ = 902 A × 74.074 = 66815 A

For this question, use the following criteria for the various point-to-point calculation step results in calculating the answers: • Transformer full load current: round to a whole number, such as 601 • Square root of three: 1.732 • f: three significant digits right of the decimal point, such as 1.036 • Transformer multiplier: three significant digits right of the decimal point • M: three significant digits right of the decimal point • Final calculated available fault current in amperes: round to whole number, such as 31,564 Calculate the available fault current at the transformer secondary terminals of a transformer rated 1,000 kilovolt-amperes, 1.5% impedance, 480/277-volt secondary, 3-phase. Use a 0.9 multiplier for %Z transformer impedance (−10% tolerance). Do not include motor contribution. Select the answer closest to the calculated results.

89111 A Iғʟᴀ = (1000 × 1000) ÷ (480 × 1.732) = 1203 A M = 100 ÷ (0.9 × 1.5) = 74.074 Iᴀғᴄ = 1203 A × 74.074 = 89111 A Note: The FC² Calculator result is 89096 amperes. Using different rounding or significant decimals can result in slightly different outcomes.

Which of the following are reasons why short-circuit studies involve calculating a bolted 3-phase fault condition?

A.) Three phases bolted together create a near-zero impedance connection. B.) A "worst-case" (highest current) condition that results in maximum 3-phase thermal and mechanical stress in the system is established. C.) A "worst-case" condition is typically what is needed to ensure proper equipment ratings, such as OCPD interrupting rating and equipment short-circuit current ratings (SCCR). All of the above

Refer to the figures and follow the steps outlined in the short-circuit calculation procedure. Do not consider adding motor contribution to the available fault current. Calculation of Available Fault Currents - Figure 10 Calculation of Available Fault Currents - Figure 11 The available fault current at position 5 is ? than the available fault current at position 6 due to the shorter conductor run supplying LPB compared to the length of conductor supplying AC-1.

Higher Note: Longer conductors of the same size will have more impedance and allow less fault current to flow.

Which of the following is not a reason why short-circuit calculations are necessary for electrical systems?

They are needed to determine the standard ampere rating of the OCPD.

When a short-circuit condition occurs, running motors may contribute four to six times their normal full load into the fault, adding to the magnitude of the available fault current.

True

When doing an arc-flash incident energy hazard analysis, the available bolted fault current is necessary for determining the hazard. It is recommended that the calculation be as accurate as possible.

True

For this question, use the following criteria for the various point-to-point calculation step results in calculating the answers: • Transformer full load current: round to a whole number, such as 601 • Square root of three: 1.732 • f: three significant digits right of the decimal point, such as 1.036 • Transformer multiplier: three significant digits right of the decimal point • M: three significant digits right of the decimal point • Final calculated available fault current in amperes: round to whole number, such as 31,564 Comparing the answers in the previous three questions, the lower the impedance for a transformer, the greater the available fault current at the transformer secondary terminals. The higher the kVA of a transformer, the greater the available fault current at the transformer secondary terminals.

True Note: The replacement of an existing transformer with one with a different kVA or impedance can impact compliance with requirements in 110.9 for the interrupting rating of OCPDs. The interrupting rating of existing OCPDs may no longer be adequate for the higher available fault current after the transformer is replaced. This also includes short-circuit current ratings of service equipment. In addition, if the available fault current is increased at the service equipment, the available fault current will increase at points downstream. This may result in downstream OCPDs having inadequate interrupting ratings and equipment having inadequate SCCRs. Although 110.24 does not require marking the equipment downstream of the service with the new value, inadequate OCPD interrupting ratings and equipment SCCR does not comply with NEC or OSHA regulations. These are safety hazards for the Electrical Worker and others as well as fire hazards for facilities.

To comply with 110.9 of the NEC, the overcurrent protective device selected must have an interrupting rating equal to or greater than the available fault current.

True Note: NEC 240.86 permits the available fault current at the lineside terminals of a circuit breaker to be greater than a circuit breaker's interrupting rating. In this case, a circuit breaker with an inadequate interrupting rating must have a series combination rating with supply side fuses or circuit breaker.

Short-circuit calculations are performed without overcurrent protective devices in the system. Calculations are done as though these devices are replaced with copper bars to determine the available fault current.

True Note: The point is that when doing a short-circuit calculation, the OCPDs are not considered in the calculations. Only after the calculations are determined are the OCPDs considered. For instance, does each OCPD have an interrupting rating equal to or greater than the available fault current, or do the OCPDs protect the circuit devices or equipment for the available fault current?