Electricity Transmission Considerations

All transmission cables; whether they be huge overhead transmission lines or small leads from the wall to your appliance; have resistance. Current through this resistance results in power loss due to heat. This has two effects:

1] Some of the energy transmitted never makes it to the end appliance/device

2] It is the heat dissipated in the cable and the cables ability to shed this heat that limits the amount of current a cable can handle.

Power ‘Lost’ to a cable is dependent on the resistance of the cable times the current squared (P = R * I * I). (Current has the unit Amps but gets the symbol ‘I’ in equations.)

If we transmit power at a higher Voltage we reduce the current inversely (P = V * I) thus doubling the Voltage, halves the current and therefore quarters the power lost in the (same) transmission line (or permits a much smaller diameter and therefore less expensive transmission line).

This is why the national grid transmits power at many thousands of Volts.

Cable or wire comes in a number of standard sizes (a ‘cable’ is just a big wire). The amount of current that a wire is rated to carry is affected by a number of things:

• The transmission material, I’ll assume copper here as most small wire in the UK is copper.
• The insulation material
• The ambient temperature
• Whether the cable is bunched or on its own.
• The airflow (a cable outside in the breeze can shed heat more readily than one buried).

As a result of all these factors it’s impossible to give a definitive list of the maximum current a wire can carry but the following is a guide of typical values for maximum current carrying capability of a cable:

This is the maximum current the cable can carry, it does not consider the Voltage drop or power loss.

For our applications we need to consider a grid down scenario and how we intend to generate, store and transmit relatively small amounts of power.

For a small scale system we will probably want to use 12V. This is simple and provided the parts are all physically close this works well.

If we want to transmit 1A around the house for low Voltage lighting at 12V.

The table above suggests that 0.5 mm^2 would be more than capable of carrying this current.

Cable length from battery to lights, 20m (each way) 40m total.

Resistivity of copper is 17.2 n.Ohm.m (1.72*10-8 Ohm.m)

Our desired load is 12W (1A at 12V, Power = Voltage * Current)

The resistance of the load is 12 Ohms (Resistance = Voltage / Current)

The resistance of the cable is R = ρ l / A (Resistance = Resistivity * length / Area)

= 1.72*10-8 * 40 / 0.5*10-6

= 1.376 Ohms

Given this additional line resistance only 0.9A flows resulting in 10.7V and 9.6W at our load and 1.1W lost as heat in the cable. We actually only get 80.5% of the power we wanted to the load.

So while the cable is capable of handling this current without melting, for a run this long it’s probably too thin for our needs.

If instead we used a 1.5mm^2 cable the resistance drops to 0.46 Ohms and we get 11.6V and 11.1W of power to the load (92.8% of what we wanted) with only 0.4W lost in the cable.

If instead of using a 12V source we changed the wiring such that we could transmit at 24V we halve the current to 0.5A, even with our 0.5mm^2 cable we get 23.3V and 11.3W of the power at the load (94.5% of what we wanted).

For higher power applications we quickly see the limitations of operating at 12V especially over long cable runs.

A typical fridge is about 100W when operating. If we used 1mm^2 cable and fed it power from 20m away at 230V 99.9% of the power gets to the fridge.

If we supplied the fridge at 12V using the same 1mm^2 cable we would get only 65% transmission efficiency (and the fridge probably wouldn’t work), even if we used whopping 70mm^2 cable we would still only get 99.3% transmission efficiency.

This is why high power applications are fed at High Voltage and why all but the smallest off grid systems run at 24V or 48V rather than 12V.

Alternating current is relatively easily stepped up and down Voltage using a transformer. While the transformers at either end have their own losses these are one off losses and not dependent on the cable run. Additionally the transformer is rated including its own internal Voltage loss. It steps the Voltage a little higher to compensate for the Voltage loss internal to the transformer, thus the output Voltage is the Voltage you wanted. Unfortunately Direct Current is not as simple to step up to a higher Voltage but that’s a topic for another article.

As a side note I did the calculations for this article on a spreadsheet, something that will not be available to me in a grid down situation. While I’m pretty good with mental and long hand arithmetic I expect when I’m tired, stressed and working by torchlight I’m going to make mistakes. Mistakes I can’t afford to base my emergency power wiring on. Fortunately I have a solar powered calculator, something to consider adding to your preps (and yes the calculator does work if I shine a bright torch at the solar cells even if the display is a bit dim).