Asymmetry in fault currents due to the inductive nature of power systems results in higher-than-normal transient conditions that may exceed the interrupting rating of devices or cause saturation of relaying CTs (and thus device misoperation).

Asymmetrical fault currents are AC fault currents with a DC offset that quickly decays to zero. The result is higher than normal fault currents that may exceed the interrupting rating of devices or cause saturation of relaying CTs on protective devices. Asymmetry occurs to some degree in all three-phase faults since all phase currents cannot be 0 at the same time. Conversely, line-to-ground are mostly symmetric, since insulator failure or flashoiver usually occurs near voltage peaks (zero-current crossing for naturally reactive systems)1. Here we’ll talk about asymmetric currents out on the distribution system, far from a generator, so we won’t be taking into account the generator subtransient, transient, and steady state impedances.

Asymmetry occurs to some degree in all three-phase faults since all three-phase currents cannot be 0 simultaneously.

The asymmetry, with respect to the zero axis of the sinusoidal current, arises from the need for currents and voltages to satisfy two rules:

1. Current must lag voltage by the system phase angle (90° for purely reactive systems)
2. Current cannot change instantaneously across an inductance (and both large conductors and large machines are inductive)

Asymmetrical current can be broken into both a purely symmetrical AC component and a purely aperiodic DC component. The AC component is symmetrical about the y-axis. The DC component is exponentially decaying with a time constant proportional to X/R. The initial magnitude of the DC current depends on the time of occurence due to the closing phase angle determined by the offset from the zero-crossing of the voltage waveform and the system phase angle, $\theta=\tan^{-1}(X/R)$. The asymmetrical waveform is the sum of the DC and AC components. In a power system, it will develop from zero as shown.

## Case: Inductive System, $X>>R$, $\theta \approx 90°$

### Conditions for Maximum DC Offset

The worst case is for a fault occuring near a zero-crossing of the voltage where the closing angle $\alpha=0$. At this point the current is at a peak, and so an opposing DC current of the same magnitude will be induced. The asymmetric current will then increase until a point when the current sine wave is lagging the voltage by 90°.

$I_{ac}$, the RMS value of the symmetrical waveform, depends on the type of fault, the pre-fault voltage and equivalent impedance. For three phase faults, $I_{ac} = V/(Z_1+Z_F)$.

$I_{dc}$ is the starting value (and maximum) of the DC component. Having no DC voltage to maintain it, the DC component decays with a time constant $\tau=\frac{X/R}{\omega} =\frac{L}{R}$. Usually the system X/R value is given, where a larger value will cause a slower decay. IEEE/IEC standards use X/R $=$ 17 ($\tau =$ 45ms or ~2.7 cycles), which is an expected value near a distribution breaker, but X/R > 50 is not uncommon for a large generator2.

Since current cannot change instantaneously in a power system, the range of the maximum DC offset is

In general, for a asymmetric sinsoidal wave with a DC offset of $A_0$ and a purely symmetric peak of $A_1$, the RMS value is equal to3

thus the maximum RMS value that the asymmetric fault current can obtain is

Since in reality the DC component will decay during the first half cycle before the asymmetric current reaches its peak,

(this approximation is also shown in the limit of the graph of asymmetric curve factors below)

### Conditions for No DC Offset

Conversely if the fault occurs at a voltage peak (zero current crossing), no offset will develop since the closing angle is equal to the system angle.

In fact the full asymmetrical waveform based on the closing ($\alpha$) and system ($\theta$) angles is4

With $\alpha=\theta\approx 90°$,

## Equation & Curve of Asymmetric Factor

The Asymmetry Factor, $\kappa$, can be used to calculate the total asymmetric current from just the symmetric current and the X/R value of the system.

And taking $t=\text{half cycle}=1/120s$ since the worst case peak value of the asymmetric current would occur a half cycle after the fault occurence,

The equation as a function of X/R is shown in the interactive graph below.

1. Zocholl, Stanley E., Analyzing and Applying Current Transformers

2. Yuen, Moon H. “Short Circuit ABC–Learn It in an Hour, Use It Anywhere, Memorize No Formula.”

3. de Metz-Noblat, Benoit, Frédéric Dumas, and Georges Thomasset. “Calculation of short-circuit currents.” Cahier technique 158