Theoretical Problem

[Silly Sausage]

I found this somewhere else.
What do you reckon?
WARNING - Don't waste too much time on it, just something to exercise the brain with!
Note, the lamps are constant power.

I had to calculate something today, and I got completely stumped. What bugs me is that I cannot work out how to do this, but I'm sure that about 20 years ago I would have known. In the end I just added in enough of a safety factor that it didn't matter, but I'm curious about how you'd go about solving it...
I have a long pair of wires with lamps on them, and the resistance is significant. So, each lamp gets a lower voltage as you get further from the EMF. This drop is not linear, because each section of cable carries less current the further you go, because it's supplying fewer lamps.
I need to find the PD at the last lamp.
If each lamp is a constant current, then this is trivial.
For example, 100V supply with 10 lamps each drawing 1A. Each wire section is 1 Ohm. Lamp 1 gets 90V, because the first section carries 10A. Lamp 2 gets 90V-9*1 = 81V as the second section carries 9A. And so on.
The trouble is, my lamps are constant power. So, the current increases as the voltage drops. Lamp 1 would draw 1.11A at 90V, but it doesn't get 90V, because all the lamps are drawing more current as they are getting a lower voltage. Every lamp current depends on the voltage that all the others are getting.
I reckon this can be solved, something to do with differential equations or series. Anyone know?

 

[gnuuser]

i think it would be dependant on whether they were wired in parallel or series
in parallel the resistance would decrease for every lamp thats attached and more current would flow but the voltage would decrease with every lamp added.
if this is the case then the formula for a series/parallel resister array would be used.

 

[Silly Sausage]

The lamps are in parallel and the volt drop of the cables is significant, therein lies the problem!
I haven't had a serious think yet, might have a go on my 15 mile hike at the weekend.

 

[gnuuser]

method 3 in the link and the formulas are below
hope this helps
How to Calculate Series and Parallel Resistance (with Cheat Sheets)

 

[Darkwood]

Come on Archy ...spill it out - your Xmas lights didn't work and you couldn't fix them!

 

[Marvo]

If the lamps are constant power ie they're all consuming the same number of watts then it's easy to work out the total circuit current using I=P/V x the number of lamps (where P is the wattage of one lamp).

Next draw out the entire circuit as a ladder network of resistors where the cross ones are the lamps and the side resistances are the wires (refer to manufacturers specs). Now take each individual lamp as its own circuit so for lamp 1 you have one resistor each side with the full load current, work out the two volt drops to give you your load voltage. This load voltage is now the starting point for the same calculation for lamp 2 and so on.

This is how we'd do it in Africa....low tech is the name of the game

In the UK you'd probably get someone to write a damn app for your Crapple device

 

[HandySparks]

I don't think that some of you are seeing the true difficulty of this one from Archy. The thing that makes it difficult is the constant power loads.

I thought that I'd cracked it by looking at the power drawn through each section, starting at the source and using a quadratic equation to work out the voltage drop in each section, based on the total load supplied by that section. Then do the same calc for the next section, using the voltage at the end of the first section as the start of the next.

Then I realised the flaw in that approach; each section of cable is itself a load (power is dissipated in it), but you can't know its value until you know the current in each section, which is dependent on the voltage applied to each of the constant power loads.

I guess the easy way is some sort of iterative calculation on a spreadsheet.

 

[Silly Sausage]

Exactly HS!

It's really more of a heavy duty Maths problem. Non trivial anyway!
I just thought I'd throw it in for a bit of 'fun'. Maybe I need to get out more.
Interesting problem though but.

 

[happysteve]

Alright, here's my attempt.

I have not solved the equations. What I have done is provide equations for the "worst" case (all the constant power loads at the end), a "middle" case (half the loads at the end of the lossy cables, half at the source), and a trivial "best" case (all the loads at the source). These are for current, from which you can easily derive voltage drops. To solve the equations for a continuous system will involve differential equations.

Workings:

View attachment const_power_loads.pdf

For those on tapatalk or similar (can't view PDFs):

Page 1 (JPG)
Page 2 (JPG)
Page 3 (JPG)
Page 4 (JPG)
Page 5 (JPG)

I'd be interested in comments. I haven't checked this rigorously, beyond putting in some numbers for the "worst" case and checking that the powers, currents and voltages are all sane. Anyway - happy reading

 

[gnuuser]

ok this is the info I found on it
regardless of series or parallel circuit power consumption is additive.

so this may help solve it
When calculating the power dissipation of resistive components, use any one of the three power equations to derive the answer from values of voltage, current, and/or resistance pertaining to each component:

This is easily managed by adding another row to our familiar table of voltages, currents, and resistances:

Power for any particular table column can be found by the appropriate Ohm's Law equation (appropriate based on what figures are present for E, I, and R in that column).
An interesting rule for total power versus individual power is that it is additive for any configuration of circuit: series, parallel, series/parallel, or otherwise. Power is a measure of rate of work, and since power dissipated must equal the total power applied by the source(s) (as per the Law of Conservation of Energy in physics), circuit configuration has no effect on the mathematics.

• REVIEW:
• Power is additive in any configuration of resistive circuit: P[SUB]Total[/SUB] = P[SUB]1[/SUB] + P[SUB]2[/SUB] + . . . P[SUB]n[/SUB]

 

[tazz]

Reading this, I can presume the following diagram to show the resistance.......correct me if im wrong, but if agreed, we could use this to work this out...

 

[happysteve]

Reading this, I can presume the following diagram to show the resistance.......correct me if im wrong, but if agreed, we could use this to work this out...
View attachment 23490

Yep, that's it.

Unless you have some very bizarre wiring, R1=R2, R3=R4 etc. Also, I would argue that you could combine them in the top branch of the circuit to make the analysis easier.

if all your lamps are equally spaced, you could generalise: R1=R2=R3=R4... Then the cable resistance of each cell would be 2R.

L1, L2 etc all draw the same power, but because the voltage across each of them will be different, their resistances would all be different. So R(L1) does not equal R(L2) etc (or if you intended that L1, L2 etc are resistances, then L1 does not equal L2 etc).

i think in my post above I got as far as I could without getting into differentials. The way forward from that point, I reckon, would be to take the "middle" case and start adding a small series resistance to the front of it ("delta R") to see what effect that has on the remainder of the circuit. That will give you a general expression that can be expanded for the general case.

Anyone see any problems with my analysis in post #9?

 

[Badged01]

Easily solved, its a quadratic based equation solution with a power load simulation at the front end feeding into it but you need a computer programme to help the re-iterations ..... or lots of time, paper and pencils!