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Short-circuit duty by IEEE C37.010

On the ANSI / IEEE set, the duty of a circuit breaker follows IEEE C37.010-2016, and the breaker's capability follows IEEE C37.04-2018. The method is the E/X method with adjustment for the alternating-current (AC) and direct-current (DC) decrements (IEEE C37.010-2016 §5.2.3). A breaker is rated on a symmetrical basis for a DC time constant of 45 ms. A fault with a higher X/R ratio carries more DC component at contact parting, and its symmetrical current is raised by a duty factor before it is compared with the rating.

Read first: Short-circuit currents by IEC 60909-0

1The E/X method

The method divides a driving voltage $E$ by the reactance $X$ of the network at the fault. $E$ is the highest typical operating voltage at the breaker's location (IEEE C37.010-2016 §5.1.2). The design takes its upper limit, the maximum system voltage of the level, for the maximum case: for example 242 kV at 230 kV, or 1.052 pu (per unit). The minimum case takes 1.0 pu without the plant's infeed.

I_{3} = \frac{E}{X _{1}}

(1)

IEEE C37.010-2016 §5.2.2

I_{1} = \frac{3 E}{2 X _{1} + X _{0}}

(2)

IEEE C37.010-2016 §5.2.2

The standard allows E/Z in place of E/X (§5.2.1). The design uses E/Z, since the neutral grounding resistor (NGR) of a medium-voltage (MV) section is a resistance that E/X would leave out. The networks are those of the IEC study (Short-circuit currents by IEC 60909-0): the grid behind its lines at the high-voltage (HV) bus, the transformers, the MV sections with their NGRs and the plant. The grid feeder follows from its stated fault levels at 1.0 pu, and the transformers enter at their rated impedance. None of the IEC 60909 corrections ( $c$ , $K_{T}$ , $K_{S}$ ) applies. A generator enters by its subtransient reactance $X_{d}^{''}$ (Table 20), with an X/R ratio of 80 (Table 23).

2The X/R ratio

The X/R ratio of the fault is calculated from two separate reductions of the network: one of the reactances with the resistances set to zero, and one of the resistances with the reactances set to zero (§5.2.3). A three-phase fault takes the positive-sequence values. A line-to-ground fault combines the sequences as the fault current does:

\frac{X}{R} = \frac{X _{1}}{R _{1}}, \frac{X}{R} = \frac{2 X _{1} + X _{0}}{2 R _{1} + R _{0}}

(3)

IEEE C37.010-2016 §5.2.3

On a MV section grounded through an NGR, the resistance sets the line-to-ground ratio, 0.048 in Example 1.

3The duty factor

The DC component decays with the time constant $τ$ of the fault. At contact parting, the total current is the symmetrical current raised by the remaining DC component. The breaker's capability at the same instant follows from its rated time constant of 45 ms. The duty factor $M F$ is the ratio of the two, and at least 1:

τ = \frac{X / R}{2 π f} M F = max (1, \frac{1 + 2 e ^{- 2 t / τ}}{S})

(4)

IEEE C37.010-2016 §5.2.2, Eq. 39 to 42

S = 1 + 2 e^{- 2 t /45 ms}

(5)

IEEE C37.04-2018 §5.6.2.2, Eq. 1

Up to an X/R ratio of 17 at 60 Hz (14 at 50 Hz), the fault's time constant is within the rated one, and E/X is compared with the rating directly (§5.1.4, §5.2.2). Above it, the network's current is multiplied by $M F$ . The converter plant's current has no DC component, and is added without the factor. §5.2.2 allows the X/R adjustment to be left out when E/X is at most 80 % of the breaker's symmetrical interrupting capability, 70 % for a line-to-ground fault. The design calculates the factor in both cases.

The contact parting time $t$ is half a cycle of relay time plus the breaker's minimum opening time (§5.1.2). It is 1.5 cycles for a breaker of 2 cycles rated interrupting time and 2 cycles for one of 3 cycles (Table 17), and 3 cycles for one of 5 cycles (§5.3.1). The design does not choose the breaker's interrupting time. It takes the larger factor at the rated interrupting times IEEE C37.04-2018 lists for the level's rated maximum voltage (Table 12, Table 16). The factor rises with the X/R ratio and with the contact parting time (Figure 1).

Ratio X/R28.0 · Contact parting 3 cycles: 1.1178 · Contact parting 2 cycles: 1.1171 · Contact parting 1.5 cycles: 1.1037

Figure 1.The duty factor is 1 up to the rated time constant and rises with the X/R ratio above it, at 60 Hz. A longer contact parting time gives a higher factor, since the rated capability falls faster than the fault's DC component. The design service calculated the values by Eq. (4) and Eq. (5).

The design's factor takes no AC decrement, as for a fault fed from remote sources. For a fault fed from nearby generators, the AC decrement lowers the current, so the factor is at or above the standard's curves for local sources. The design also checks that the DC time constant is at most 120 ms, the limit of the derating method of IEEE C37.010-2016. Above it, the breaker's capability is the manufacturer's.

where
$E$	driving voltage, the maximum system voltage line to neutral	kV
$X_{1}$ , $R_{1}$ , $X_{0}$ , $R_{0}$	positive- and zero-sequence reactance and resistance at the fault, from the separate reductions	Ω
$I_{3}$ , $I_{1}$	symmetrical three-phase and line-to-ground currents	kA
$τ$	DC time constant of the fault	ms
$f$	system frequency	Hz
$t$	contact parting time	ms
$S$	rated asymmetrical capability of the breaker per unit of its rated short-circuit current, at $t$
$M F$	duty factor on the symmetrical current

4Closing and latching and short-time current

The rated closing and latching current is a peak current, a multiple of the rated short-circuit current $I_{sc}$ :

I_{CL, rated} = 2.6 I_{sc}

(6)

IEEE C37.04-2018 §5.6.2.3

The multiple is 2.6 at 60 Hz and 2.5 at 50 Hz, for a time constant up to 45 ms. The design takes the first-cycle peak of the fault current as the same multiple of the symmetrical current, or $2 (1 + e^{- π / (X / R)})$ times it where that is higher (IEEE C37.010-2016 Eq. 43). The converter current enters as its own peak, $2$ times. A breaker's 2.7 rating for a time constant up to 133 ms is not taken. The short-time current is the symmetrical current, against $I_{sc}$ for the rated duration $Y$ : 2 s below 100 kV and 1 s at 100 kV and above (IEEE C37.04-2018 Table 12, Table 16).

Example 1Breaker duty of a 230 kV project of the ANSI / IEEE set

Given a 230 kV / 34.5 kV project of the ANSI / IEEE set at 60 Hz, for a 250 MW plant, with the default grid, line, transformers and NGRs.

Quantity	230 kV bus	34.5 kV section
Maximum system voltage, $E$	242 kV, 1.052 pu	36.5 kV, 1.058 pu
Positive-sequence impedance $Z_{1}$	0.501 Ω + j6.412 Ω	0.033 Ω + j1.030 Ω
Rated interrupting times listed	2 cycles and 3 cycles	3 cycles and 5 cycles
Rated short-circuit current $I_{sc}$ , duration $Y$	50 kA, 1 s	31.5 kA, 2 s

Symmetrical currents by Eq. (1) and Eq. (2), E/Z, with the plant's infeed in the three-phase current: 22.52 kA and 20.41 kA at the HV bus, 23.10 kA and 1.06 kA on a MV section.

X/R ratios by Eq. (3): 12.8 three-phase and 14.9 line to ground at the HV bus, 30.8 and 0.048 on a MV section. The larger HV ratio is at or below 17, and the HV factors are 1.000 three-phase and 1.000 line to ground. The MV three-phase ratio is above 17.

Duty factor of the MV three-phase fault by Eq. (4) and Eq. (5). The contact parting time of 3 cycles (50.0 ms), for a breaker of 5 cycles rated interrupting time, gives the larger factor:

τ = \frac{30.8}{2 π \times 60} = 81.8 ms, M F = \frac{1.260}{1.103} = 1.143

Interrupting current of the MV section, the network's current times the factor plus the plant's 2.64 kA:

(23.10 - 2.64) \times 1.143 + 2.64 = 26.02 kA

Line-to-ground interrupting current, the symmetrical current times its factor: 20.41 kA × 1.000 = 20.41 kA at the HV bus, and 1.06 kA × 1.000 = 1.06 kA on a MV section.

Closing and latching current of the MV section by Eq. 43, with the factor 2.691 at an X/R ratio of 30.8: 58.79 kA against 81.9 kA.

Check	HV calculated	HV rating	HV result	MV calculated	MV rating	MV result
Interrupting, three-phase	22.52 kA	50 kA	Pass	26.02 kA	31.5 kA	Pass
Interrupting, line to ground	20.41 kA	50 kA	Pass	1.06 kA	31.5 kA	Pass
Closing and latching	57.60 kA	130 kA	Pass	58.79 kA	81.9 kA	Pass
Short-time, for $Y$	22.52 kA	50 kA	Pass	23.10 kA	31.5 kA	Pass
DC time constant	39.5 ms	120 ms	Pass	81.8 ms	120 ms	Pass

Result The breaker ratings of the example cover the calculated duty. The MV three-phase duty of 26.02 kA is the network's E/Z current of 20.46 kA times 1.143, plus the plant's 2.64 kA. The factor follows from the X/R ratio of 30.8.

Sources

[1]IEEE C37.010-2016, IEEE Application Guide for AC High-Voltage Circuit Breakers. §5.2.2; §5.2.3; §5.2.2, Eq. 39 to 42.

[2]IEEE C37.04-2018, IEEE Standard for Ratings and Requirements for AC High-Voltage Circuit Breakers. §5.6.2.2, Eq. 1; §5.6.2.3.

CitationSubstationLayout Reference, Short-circuit duty by IEEE C37.010, revision 1, 8 Nov 2026, https://substation.solarlayout.app/library/reference/short-circuit-duty-ansi

Short-circuit duty by IEEE C37.010

1The E/X methodCopy link to The E/X method

2The X/R ratioCopy link to The X/R ratio

3The duty factorCopy link to The duty factor

4Closing and latching and short-time currentCopy link to Closing and latching and short-time current

1The E/X method

2The X/R ratio

3The duty factor

4Closing and latching and short-time current