Voltage Drop calculation

[scubapete]

I understand how to do this for a single point to point cable, e.g. 2m cable from 12V transformer to 20W bulb.

How do I start calculating it for daisy-chained cables? Is it simply determining the load on each cable section and calculating the voltage drop over each section and then summing these to the terminating points? Or is it more arcane?

Pointers to books would be appreciated as I would like to buy some good guides/references and I'm not afraid of the technical stuff.

 

[securespark]

Imagine a cable with 4 lights attached to it via junction boxes.

The Supply is at point A, points B, C, D & E are the lights.

The lights are (for argument's sake) 500W each.

So, the cable between A/B carries 2kW, B/C carries 1.5kW, C/D carries 1kW, and D/E carries 500W.

Edit. Sorry, John's just pointed out that your post concerns SELV. With 12V, it is very important to make sure the cables can carry the load. Because SELV currents are vary high, as John says, it is better to have 1 load per cable, rather than the set-up I described above.

Comparison:

60W LV lamp @ 230V = 0.26A

50W SELV lamp @12V = 4.17A


Now imagine 4 50W SELV lamps fed via one long cable...

 

[JohnD]

And incidentally, this is why, for 12v lamps driven by a transformer, it is better to run each lamp on its own radial, like spiders legs, and preferably cut all the radials to the same length.

 

[scubapete]

Given this circuit:

Assume distance from transformer to chocbox 2 is negligible
All other cables are 1m
All bulbs are 40W

 

[scubapete]

Are these calcs correct?


Max volt drop is 4%, therefore a minimum of 11.52V supplied to each bulb
From tables, 1.5mm2 2-core has a volt drop rating of 29mV/A/m

Bulb A:
i) 2 -> 1:
I = 40W / 12V = 3.33A
Volt-drop = 29 * 3.33 * 1m / 1000 = 0.097V
ii) 1 -> A:
I = 40 / (12-0.097) = 3.36A
Volt-drop = 29 * 3.36 * 1 / 1000 = 0.097V

Total volt-drop = 0.097 + 0.097 = 0.194V = 1.6% of 12V

Bulb B:
i) 2 -> 3:
I = 80 / 12V = 6.67A
Volt-drop = 29 * 6.67 * 1m / 1000 = 0.19V
ii) 3 -> B:
I = 40 / (12-0.19) = 3.39A
Volt-drop = 29 * 3.39 * 1 / 1000 = 0.098V

Total volt-drop = 0.19 + 0.098 = 0.288V = 2.4% of 12V

Bulb C:
i) 2 -> 3 (same as Bulb B):
I = 80 / 12V = 6.67A
Volt-drop = 29 * 6.67 * 1m / 1000 = 0.19V
ii) 3 -> 4:
I = 40 / (12-0.19) = 3.39A
Volt-drop = 29 * 3.39 * 1 / 1000 = 0.098V
iii) 4 -> C:
I = 40 / (12-0.19-0.098) = 3.42A
Volt-drop = 29 * 3.42 * 1 / 1000 = 0.099V

Total volt-drop = 0.19 + 0.098 + 0.099 = 0.387V = 3.2% of 12V