Monthly Archives: January 2013

Jamestech Electrical Contracting and Projects

Can we tender for your project?

Jamestech has many years of contracting experience  and having Electrical Contracting and Engineering Licenses inPapua New Guinea and Australia. This gives Jamestech the ability to be a complete EPCM Contractor for your projects.

Having the Company Director as a licensed Electrician and the Endorsee of the Jamestech Electrical Contractors licenses, puts us well ahead of other Large Contracting Companies having statutory requirements as our priority on all projects.

With our large database of Electrical workers also gives Jamestech the ability to complete any sized project with challanged timeframes.

We consider our knowledge of Electrical Standards, Regulations and Acts to be superior to many of our competitors and have compliance to these regulations as our primary focus.

Jamestech Construction team has been busy completing the Electrical Installations for the Stage 1, 2 & 3 of Xstrata’s Ernest Henry Mine Village Expansion. Jamestech was sucessful in winning the tender for the project.

The final stages of the expansion shall be completed early 2013.

Jamestech has been a contact partner with Xstrata since 2002 and we look forward to tendering on and assisting Xstrata with all future projects.

For your next project, expansion or upgrade, Jamestech would welcome the opportunity to provide a quote for you to consider.

” Jamestech prioritises Electrical Safety “

Should you require any further information, please do not hesitate to contact any of the Jamestech staff on:
Main Telephone: 1300 732881 (freecall within Australia)
Brisbane Office: +61-7-55493471
Townsville Office: +61-7-47267709
Papua New Guinea: +675-7004 2655
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Darren Caulfield

Fault Currents & Short Circuit Impedance, in Electricians terms

Short Circuits and Fault Levels

An overview, in electrician’s terms, with some examples


The basic electrical theorem (Ohm’s Law) says the amount of current that will flow through a short circuit depends on two variable values,
1. the system voltage and
2. the connected total impedance of the current flow path from the source to the point of the fault.

The typical system voltages are very familiar to all of us. The connected total impedance of the shortcircuit current flow path needs a little clarification, however.

This impedance normally includes the feeder conductors’ resistance and reactance, any transformers’ impedances (going from the point of fault back to the energy source), and any other equipment connected in the path of current flow.

Fig. 1 is a very simplified one-line, with the following: a power source, transformer, and an overcurrent protective device (OCPD) having a specific short-circuit current interrupting rating and a load with a fault.

Let’s talk about the power source first. In many short-circuit current calculation examples, you’ll see references like “Assume the power source has infinite capacity” or “The source has an infinite bus.” What does this mean, and why is it important to our sample calculation?

All that is being said in this case, is: “The source voltage has no or an extremely low internal impedance”. Since the source has been assumed to have infinitesimal impedance of its own, the corresponding shortcircuit current will be at its worst case.

Now let’s look at the transformer. The impedance determining the amount of short-circuit current on its secondary, is made up of two separate impedances: Its own impedance plus that of the secondary conductors which run to the point of the fault.

The transformers own impedance is the amount of its own opposition to the flow of short-circuit current through it. Now, all transformers have impedance, and it’s generally expressed as a voltage percentage.

This is the percentage of normal rated primary voltage, that must be applied to the transformer to cause full-load rated current to flow in the short-circuited secondary. (You remember that from TAFE) For instance, if a 11kV to 415V transformer has an impedance of 5%, this means that 5% of 11kV, or ≈550V, applied to its primary will cause rated load current flow in its secondary. If 5% of primary voltage will cause such current, then 100% of primary voltage will cause 20 times (100 divided by 5) full-load-rated secondary current to flow through a solid short circuit on its secondary terminals.

Obviously, then, the lower the impedance of a transformer of a given kVA rating, The higher the amount of short-circuit current it can deliver.

For the full document please open and print the full document:

A basic Outline of Fault current.pdf

Want to know more?
James technologies can assist.

Main Telephone, 1300 732881 (freecall within Australia)
Brisbane Office, +61-7-55493471
Townsville Office, +61-7-47267709

Ken Cryer
Engineering Manager