Gridcraft Power Lab · Faults & Phasors

Network inputs

Try a scenario

Presets replace the inputs with a reproducible case.

The study network

BREAKER CLOSED

Single-source equivalent · no pre-fault load · fault downstream of the breaker.

Phase A fault current—Primary RMS

Phase A relay current—Ideal CT secondary RMS

Residual current · 3I₀—Primary RMS

Current phasors

A · primary

ABCRMS · auto-scaled

Bus voltage phasors

kV · L–G

ABCSeparate voltage scale

Phase quantities

Phase	Primary current · A ∠°	Relay current · A ∠°	Bus voltage · kV ∠°

Sequence quantities

Sequence	Current · A ∠°	Bus voltage · kV ∠°

Angles reference pre-fault phase-A voltage at 0°. A dash means the angle of a zero-magnitude phasor is undefined.

FOLLOW THE NUMBERS

How this result is calculated

Lesson 1 · keep beside the problem

ABC FAULT / COMPLEX NUMBERS / CTs

Work through the default case

Select Reset to match these numbers. This worked example stays fixed; “How this result is calculated” follows your current inputs.

Convert line voltage.
E₁ = 13,800 / √3
= 7,967.4 ∠ 0° V
A per-phase impedance requires phase-to-neutral voltage.
Divide by complex impedance.
Iₐ = 7,967.4 / (0.2 + j2)
Multiply the numerator and denominator by the conjugate, 0.2 − j2.
Iₐ = 7,967.4(0.2 − j2) / (0.2² + 2²)
= 394.43 − j3,944.27 A
= 3,963.95 ∠ −84.29° A
The negative angle means current lags the voltage for this inductive impedance.
Convert through the CT.
CT ratio = 600 / 5 = 120
Irelay = 3,963.95 / 120
= 33.03 ∠ −84.29° A
This is an ideal-CT prediction. CT saturation and actual relay input limits are not simulated.

Predict, then test

Select Weaker source. All sequence impedances double; the bolted ABC current halves.
Reset, then change only CT primary from 600 A to 1,200 A. Primary current stays the same; secondary current halves.
Select A–G fault. Check I₀ = I₁ = I₂, Ia = 3I₁, and Ib = Ic = 0 for this unloaded case.
Open the breaker. The downstream bus loses its only source in this model.

Read the code lesson and assumptions

First predict. Then change one thing.

01 / PHYSICS

Voltage drives current through impedance.

Start with Reset. Convert 13.8 kV line-to-line to 7.967 kV phase-to-neutral. For the bolted ABC case, divide that phasor by 0.2 + j2 Ω. The current lags because the impedance is inductive.

02 / EXPERIMENT

Make the source weaker.

Predict the result before selecting Weaker source. The preset doubles all sequence impedances. For the bolted fault, the primary current halves. Changing only the CT rating changes the relay quantity, not the primary fault.

03 / EXPLAIN

Compare ABC with A–G.

For an A–G fault with no pre-fault load, I₀ = I₁ = I₂, and Iₐ = 3I₁. The B and C currents are zero. Their bus voltages need not be zero. Increase Z₀ and explain why the ground-fault current falls.

How the code is organized

power-math.mjs contains the complex-number operations, sequence-network equations and phase reconstruction. power-lab.mjs reads the form, calls the solver and draws the results. Keeping those separate lets us check the engineering without relying on the display.

// Phase-A-to-ground fault; all quantities are primary.
const I1 = E1 / (Z1 + Z2 + Z0 + 3 * Rf);
const Ia = 3 * I1;
const relayCurrent = Ia / (CTprimary / CTsecondary);

This excerpt is mathematical pseudocode. JavaScript needs the explicit complex-number functions in the solver; ordinary division cannot divide complex objects.

Your first coding exercise

Find the solve function. Trace the input voltage to E, then the sequence current to phases, then the CT ratio to secondary. Before changing the solver, reproduce the default current using a separate hand calculation.

Model assumptions, validation and references

This is an educational single-bus, single-source Thevenin model, with balanced positive-sequence pre-fault voltage and ABC rotation. It calculates steady RMS fundamental-frequency phasors. No load, shunt capacitance, motor/IBR contribution, CT saturation, DC offset, transformer vector-group conversion, protection operation or clearing-time simulation is included. The CTs are ideal and located on the only source contribution path.

All sequence impedances are entered at the bus voltage base in primary ohms. Include grounding/neutral contributions in Z₀; do not add them twice. For ABC, Rf is the same per-phase fault resistance. For A–G, it is the A-to-ground fault resistance. Opening the breaker isolates this otherwise unsupplied bus: all bus voltages and currents are zero in this model.

This is not a multi-bus study or a substitute for a reviewed engineering model. Calculation checks cover analytic reference cases, fault-boundary conditions, sequence reconstruction, CT scaling and invalid inputs.

See the fault. Understand the quantities.