Symmetrical components is a subject that makes most engineers wince in pain and frustration when they first hear these words.

Before the days of computer-assisted fault current calculations, all fault calculations were done by hand.

In the modern world, short circuit calculations are completed with the help of computer simulated calculations, however, the NCEES® Electrical Power PE Exam still tests on the basic understanding of symmetrical components for unbalanced faults, and for good reason.

It’s important to understand how each type of unbalanced fault has an effect on the actual available short circuit current at any point in an electrical system.

Below is a realistic NCEES® like practice problem demonstrating how to calculate the short circuit current for an unbalanced single line to ground fault using symmetrical components.

**What’s in this article? – Single Line to Ground Fault Symmetrical Components
**Click below to jump to any section.

**1. NCEES® Electrical Power PE Exam Practice Problem – Single Line to Ground Fault**

2. Example Problem Demonstration Video

3. Single Line to Ground Fault Single Phase Equivalent Circuit

4. Solve for the Per Unit Current in the Single Line to Ground Fault Circuit

5. Calculate the Base Current

6. Calculate the Current in Amps

7. Calculate the Single Line to Ground Fault Current

8. Need More Practice?

9.Thoughts? Comments? Questions?

**1. NCEES® Electrical Power PE Exam Practice Problem – Single Line to Ground Fault**

Please note: this is a practice problem created by the authors at Electrical PE Review, INC to simulate a realistic PE exam practice problem, it is not an official practice problem from NCEES®.

**Symmetrical Components Single Line to Ground Fault Example Problem:**

A single line to ground fault occurs somewhere along the 50 mile transmission line connecting the utility substation bus #1 to transmission bus A. Using the transmission line diagram, solve for the fault current magnitude in amps at this point if the zero sequence impedance is 0.241pu, and the positive and negative sequence impedances are both equal to five times the zero sequence impedance. Use the generator power rating and the system voltage at the fault as base values.

**(Transmission Line Diagram)**

**2. Example Problem Demonstration Video**

Follow along step by step by watching the above practice problem worked out live with the symmetrical component single line to ground fault video:

**3. Single Line to Ground Fault Single Phase Equivalent Circuit**

The first step is to draw the single-phase equivalent circuit for the single line to ground fault using symmetrical components. Then plug in the values given in the problem. Each type of fault has its own unique symmetrical component circuit so be sure to have a reference handy.

A single line to ground fault has all three sequence networks – zero, positive, and negative – connected in series:

**(Single Line to Ground Fault Symmetrical Components Single Phase Equivalent Circuit)**

We use 1pu<0 as the per unit value of the phase voltage in the equivalent circuit. If you’re unsure why, or rusty with the per unit system, the following article may help explain why:

Electrical PE Review – Per Unit Example, Tips, and Tricks

**4. Solve for the Per Unit Current in the Single Line to Ground Fault Circuit**

From circuit analysis, we can determine that Ia0, Ia1, and Ia2 are equal to each other since they are in series:

Let’s use ohm’s law to solve for the current by adding all three series impedances together:

The per unit current in the single line to ground fault single-phase equivalent circuit is 0.377pu.

**5. Calculate the Base Current**

We know the value of the current in per unit, but the question specifically asks for the current in amps.

To convert from per unit to amps, we will need to first calculate the base current.

We will use the power rating of the generator for base power and the system voltage for base voltage, as instructed by the problem statement.

Don’t forget, since this is really just the formula for the magnitude of three phase power, we’ll need to use the line voltage value for the base voltage value, and don’t forget the square root of three.

Confused as to why we used this formula? The following article may help to explain why:

Electrical PE Review – The Three Most Common Mistakes to Avoid for Power Formulas

The base current is equal to 1,255.1 amps.

**6. Calculate the Current in Amps**

Now that we know both the per unit and base values of the current, it’s time to calculate it in amps using the per unit system:

The current in the single line to ground fault single-phase equivalent circuit is 473.2 Amps.

**7. Calculate the Single Line to Ground Fault Current**

For a single line to ground fault, the current flowing through the fault will only be from the phase that is faulted. Assuming the fault occurred on the A phase, this means the fault current will equal the A line current at the time of the fault.

Or we can say that I_F = I_A:

**(Single Line to Ground Fault on A Phase – Three Phase Diagram)**

Next, since this is a three phase transmission system, the line current in the system will equal the phase current in the single phase equivalent circuit (think of wye : wye relationships where I_phase = I_Line).

That means we will need to use the symmetrical components formula that calculates the A phase current from the sequence currents:

Since we already demonstrated that the positive (1), negative (2), and zero (0) sequence currents are all equal in a single line to ground fault, we can solve for the A line current using the value we calculated for Iao in the previous step:

Since:

Then:

The A line current in the transmission system during the single line to ground fault is 1,420A(rounded to the nearest amp).

Since we demonstrated that the fault current will equal the line current for a single line to ground fault, we can solve for the fault current:

The fault current is equal to 1,420 amps (rounded to the nearest amp) in the single line fault for the transmission system in the given problem.

That wasn’t so bad, was it?

Okay, maybe it was a little challenging if this was your first time working this type of problem. It get’s easier with practice, I promise.

**8. Need More Practice?**

Make sure to watch the video embedded at the top of this article.

The video demonstration takes the single line to ground fault example one step further by demonstrating how to prove that the voltage on line that is faulted is equal to zero – (assuming it is a zero impedance fault) – by calculating the A phase voltage from the sum of the positive (1), negative (2), and zero (0) sequence voltages.

Want to try it on your own?

First, you’ll need to calculate the positive (1), negative (2), and zero (0) sequence voltages.

If you are already a student of ours, you can go directly to the chapter on symmetrical components that offers video demonstrations of each type of fault by clicking the following link (you must log in):

Electrical PE Review – Chapter 4.2 Symmetrical Components

You can also work the practice problem quiz that features 10 sample PE exam problems with full worked out solutions, similar to this article:

Electrical PE Review – Chapter 4.2 Quiz! Symmetrical Components

**9. Thoughts? Comments? Questions?**

Have you ever come across a tough symmetrical components question that you couldn’t answer?

Is this the first time you’ve seen a full-length explanation on solving a single line to ground fault?

Have any lingering questions after working your way to the bottom of this article?

**Join the discussion by leaving your comment below.**

We respond to all comments in a timely manner.

Justin Obenchain says

Excellent article Zach. This was very helpful!

Zach Stone, P.E. says

Glad you enjoyed it, Justin.