Power Systems Lab · 18 modules · all computed in your browser

Bus fault lab (13.8 kV)

The original Power Lab: one source, ABC/A-G, CT secondary.

Lab 01 · Faults and phasors

See the fault. Understand the quantities.

One bus, one source, a Thevenin equivalent. Change the network. Predict the result. Check the calculation.

IA primary RMS3964 A

IA relay secondary33.03 A

3I0 primary0 A

Phase	Primary A	Relay A
A	3964 ∠ −84.3°	33.0 ∠ −84.3°
B	3964 ∠ +155.7°	33.0 ∠ +155.7°
C	3964 ∠ +35.7°	33.0 ∠ +35.7°

Sequence	Current A	Voltage kV
0	0	0
1	3964 ∠ −84.3°	0
2	0	0

Angles reference pre-fault phase-A voltage at 0°. A zero-magnitude phasor has no angle. 13.8 kV line-to-line is 7.967 kV phase-to-neutral.

01 Physics. Voltage drives current through impedance. For the bolted ABC case, divide the phase-to-neutral phasor by 0.2 + j2 Ω. The current lags because the impedance is inductive.

02 Experiment. Weaker source doubles every sequence impedance. For the bolted fault the primary current halves. Changing only the CT rating changes the relay quantity, not the primary fault.

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

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 or IBR contribution, CT saturation, DC offset, transformer vector-group conversion, protection operation or clearing time is included. The CTs are ideal and on the only source path. Sequence impedances are primary ohms at the bus voltage. Include grounding in Z₀; do not add it 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. This is not a multi-bus study or a substitute for a reviewed engineering model.

Learn it

A bolted or resistive fault on a line is solved by connecting the sequence networks at the fault point. The relay then sees the result on the secondary side of its CTs, scaled by the CT ratio.

Change the sequence impedances and Rf to see which fault type is most severe, and how far the secondary current is from the relay's pickup.

Quiz · 0/2

1. Raising Rf from 0 to 5 Ω on an SLG fault mostly…
2. With a 1200:5 CT, 6000 A primary is how much secondary current?

Loading module…