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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 geting 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?
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 geting 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?
In reply to 999thAndy:
Lamps wired in parallel (not stated explicitly by the OP but seems pretty clear).
The current in each section of cable is constant, but different sections carry different currents.
Lamps wired in parallel (not stated explicitly by the OP but seems pretty clear).
The current in each section of cable is constant, but different sections carry different currents.
In reply to Hooo:
If they are in series then the current will be the same in all of the lamps.
Current can only be created by the power supply and if a function of the total resistance of the circuit and the voltage powering it.
If they are in parallel all of the lamp will have 100v applied to them.
Edit: This is true for DC, Not sure about AC
If they are in series then the current will be the same in all of the lamps.
Current can only be created by the power supply and if a function of the total resistance of the circuit and the voltage powering it.
If they are in parallel all of the lamp will have 100v applied to them.
Edit: This is true for DC, Not sure about AC
Post edited at 16:59
In reply to Hooo:
You can calculate it numerically on paper or with a spreadsheet by working out current & resistance losses at each stage so a mathematical series rather than a differential equation.
That could be laborious so you could estimate a worst case for voltage drop and resistive losses where all the bulbs are at the end of the wire and assume the full current flows along the full length of wire. That might tell you that the voltage drop is insignificant or it might tell you that you need the proper calculation.
You can calculate it numerically on paper or with a spreadsheet by working out current & resistance losses at each stage so a mathematical series rather than a differential equation.
That could be laborious so you could estimate a worst case for voltage drop and resistive losses where all the bulbs are at the end of the wire and assume the full current flows along the full length of wire. That might tell you that the voltage drop is insignificant or it might tell you that you need the proper calculation.
Post edited at 17:50
In reply to dapoy:
Nope. You're forgetting the voltage drop in the cables between each lamp.
> If they are in parallel all of the lamp will have 100v applied to them.
Nope. You're forgetting the voltage drop in the cables between each lamp.
In reply to elsewhere:
That's what I did! The voltage drop is significant, so I've done a worst case estimate. What I'm interested in is how I'd do the maths properly. Just out of curiosity.
I've done a sequential calculation in Excel, but every lamp affects all the others, so it ends up circular.
To the others, yes they are in parallel. Otherwise it would be trivial.
That's what I did! The voltage drop is significant, so I've done a worst case estimate. What I'm interested in is how I'd do the maths properly. Just out of curiosity.
I've done a sequential calculation in Excel, but every lamp affects all the others, so it ends up circular.
To the others, yes they are in parallel. Otherwise it would be trivial.
In reply to Hooo:
I'd work backwards from the last lamp in the chain, and declare an arbitrary voltage 'V' across it. Then work backwards through the chain, calculating currents, voltage drops and lamp voltages at each node. You'll eventually find a formula for the voltage at the input. Re-arrange* to find V in terms of Vin, then put your V into the equations.
* this might be tricky: cheat by using Excel solver to vary V until Vin=100...
I'd work backwards from the last lamp in the chain, and declare an arbitrary voltage 'V' across it. Then work backwards through the chain, calculating currents, voltage drops and lamp voltages at each node. You'll eventually find a formula for the voltage at the input. Re-arrange* to find V in terms of Vin, then put your V into the equations.
* this might be tricky: cheat by using Excel solver to vary V until Vin=100...
In reply to Hooo:
The problem is that there is more than one solution. Consider the following:
Constant voltage source of 10v
Constant line resistance of 9ohm
Constant load of 1W
What's the line current?
The obvious answer is 1A (with 1v across the load), but 1/9 A (with 9v across the load) will also satisfy this.
In this case, the two solutions can come from solving the quadratic:
10v = (9ohm*i)+ (1W/i)
so
9i^2 -10i +1 = 0
In practice, such larger problems would be solved using iterative methods (software), usually assuming load voltage would be significantly higher than line voltage drop.
The problem is that there is more than one solution. Consider the following:
Constant voltage source of 10v
Constant line resistance of 9ohm
Constant load of 1W
What's the line current?
The obvious answer is 1A (with 1v across the load), but 1/9 A (with 9v across the load) will also satisfy this.
In this case, the two solutions can come from solving the quadratic:
10v = (9ohm*i)+ (1W/i)
so
9i^2 -10i +1 = 0
In practice, such larger problems would be solved using iterative methods (software), usually assuming load voltage would be significantly higher than line voltage drop.
In reply to captain paranoia:
Yes, that was my next step if I needed a more accurate answer. In fact the last lamp V is not arbitrary - it has a minimum voltage below which it won't work, so I'd use this. This will work fine and give me an accurate enough answer.
Like I said, I have a solution, I'm just curious if there's a proper mathematical way of calculating the voltages in the series, without decalring an arbitrary value and working backwards from there. I have a feeling that there should be...
Yes, that was my next step if I needed a more accurate answer. In fact the last lamp V is not arbitrary - it has a minimum voltage below which it won't work, so I'd use this. This will work fine and give me an accurate enough answer.
Like I said, I have a solution, I'm just curious if there's a proper mathematical way of calculating the voltages in the series, without decalring an arbitrary value and working backwards from there. I have a feeling that there should be...
In reply to knthrak1982:
Now you're getting somewhere. Of course, there are multiple solutions. But I assume there is only one steady state that satisfies the solution for all lamps simultaneously?
Now you're getting somewhere. Of course, there are multiple solutions. But I assume there is only one steady state that satisfies the solution for all lamps simultaneously?
In reply to Hooo:
What you're describing there is actually a proper mathematical way. It's basically this:
http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method
> I'm just curious if there's a proper mathematical way of calculating the voltages in the series, without decalring an arbitrary value and working backwards from there. I have a feeling that there should be...
What you're describing there is actually a proper mathematical way. It's basically this:
http://en.wikipedia.org/wiki/Gauss%E2%80%93Seidel_method
In reply to Hooo:
I've covered this kind of problem in a degree course when dealing with HV busbar loads. IIRC, when using the Gauss method I've linked to above, the solution will converge on the one with the lowest line drop (ie the busbar voltages are close to the generator voltages as one would expect) though I'm not sure why the maths will always work this way.
> Now you're getting somewhere. Of course, there are multiple solutions. But I assume there is only one steady state that satisfies the solution for all lamps simultaneously?
I've covered this kind of problem in a degree course when dealing with HV busbar loads. IIRC, when using the Gauss method I've linked to above, the solution will converge on the one with the lowest line drop (ie the busbar voltages are close to the generator voltages as one would expect) though I'm not sure why the maths will always work this way.
In reply to knthrak1982:
That's interesting. I assumed there must be way of solving it algebraically. So, it seems that my Excel-based iteration is as close as I can get to a mathematical solution? It just seems like such a bodge It's only one step removed from actually wiring the things up and measuring it!
That's interesting. I assumed there must be way of solving it algebraically. So, it seems that my Excel-based iteration is as close as I can get to a mathematical solution? It just seems like such a bodge It's only one step removed from actually wiring the things up and measuring it!
In reply to Hooo:
Yeah. In practice there will only be one solution that satisfies the max/min conditions of your lamps. There'll be a specific range of V and I in which your lamps behave as a constant power load, outside which all bets are off.
Yeah. In practice there will only be one solution that satisfies the max/min conditions of your lamps. There'll be a specific range of V and I in which your lamps behave as a constant power load, outside which all bets are off.
In reply to Hooo:
You can write a series of simultaneous equations representing the system and solve those. Classic V=IR on each cable, slightly more unusual V=CONSTANT_POWER/I for the lamps, summing currents to zero at the nodes and setting up a fixed voltage source.
Because of the constant power nature of the lamps a direct linear solver is out, but you could use something itterative like Newton Raphson. You would also need a semi-sensible initial guess to iterate from, perhaps assuming negligible cabling resistance.
So that's some maths that gets you the solution numerically. Unless you're happy programming your Excel method is probably easier. I can probably adapt some code to do this easily enough - PM me if you want it.
As for an algebraic solution, I haven't the foggiest although it puts me in mind of the futility of the rocket equation and exponentials.
> That's interesting. I assumed there must be way of solving it algebraically. So, it seems that my Excel-based iteration is as close as I can get to a mathematical solution? It just seems like such a bodge
You can write a series of simultaneous equations representing the system and solve those. Classic V=IR on each cable, slightly more unusual V=CONSTANT_POWER/I for the lamps, summing currents to zero at the nodes and setting up a fixed voltage source.
Because of the constant power nature of the lamps a direct linear solver is out, but you could use something itterative like Newton Raphson. You would also need a semi-sensible initial guess to iterate from, perhaps assuming negligible cabling resistance.
So that's some maths that gets you the solution numerically. Unless you're happy programming your Excel method is probably easier. I can probably adapt some code to do this easily enough - PM me if you want it.
As for an algebraic solution, I haven't the foggiest although it puts me in mind of the futility of the rocket equation and exponentials.
Post edited at 23:13
In reply to Hooo:
If you start from the last bulb N, with V across it, it must carry current I. Using P=IV, 100 = I * V, so I = 100/V.
The voltage drop between the last and penultimate bulb is I * R, (R==1), so
VN = V + 100/V.
Rinse and repeat to get V0, the voltage across the zeroth bulb (aka the EMF).
Chuck this in a spreadsheet, and use trial and error to find the actual PD at the last bulb, or set up some sort of Newton Raphson iteration if doing it manually (if you can for this, not sure).
If you start from the last bulb N, with V across it, it must carry current I. Using P=IV, 100 = I * V, so I = 100/V.
The voltage drop between the last and penultimate bulb is I * R, (R==1), so
VN = V + 100/V.
Rinse and repeat to get V0, the voltage across the zeroth bulb (aka the EMF).
Chuck this in a spreadsheet, and use trial and error to find the actual PD at the last bulb, or set up some sort of Newton Raphson iteration if doing it manually (if you can for this, not sure).
In reply to Hooo:
The constant power bit is where I'm struggling.
As the volts drop, the lamps will try to pull more current in the cable, as you pull more current in the cable, the volts will drop.
How can this ever work?
The constant power bit is where I'm struggling.
As the volts drop, the lamps will try to pull more current in the cable, as you pull more current in the cable, the volts will drop.
How can this ever work?
In reply to Hooo:
Don't forget Kirchoff's law.
10x100W lamps gives 1000W from the source plus what ever the length of cable resistance uses.
P=1000+W where W is power loss on cables.
I would work it back in terms of power losses at each junction rather than voltage drops. Using I P and R.
Don't forget Kirchoff's law.
10x100W lamps gives 1000W from the source plus what ever the length of cable resistance uses.
P=1000+W where W is power loss on cables.
I would work it back in terms of power losses at each junction rather than voltage drops. Using I P and R.
In reply to DancingOnRock:
Constant power means the lamps are regulated, just like a typical laptop power supply which still works when the voltage halves when you move from UK to US. Provided the resistance in the mains lead is small enough the power supply works despite the greater current, greater resistive loss and reduced supply voltage in the US.
Eventually if you add enough lamps to a long enough cable it will stop working when the lamps have under or over voltage at different ends of the cable.
> The constant power bit is where I'm struggling.
> As the volts drop, the lamps will try to pull more current in the cable, as you pull more current in the cable, the volts will drop.
> How can this ever work?
Constant power means the lamps are regulated, just like a typical laptop power supply which still works when the voltage halves when you move from UK to US. Provided the resistance in the mains lead is small enough the power supply works despite the greater current, greater resistive loss and reduced supply voltage in the US.
Eventually if you add enough lamps to a long enough cable it will stop working when the lamps have under or over voltage at different ends of the cable.
In reply to elsewhere:
Ok. So in that case the voltage drop is irrelevant. As long as the voltage stays over the 100v.
So all you need to do is provide 100v at end of line.
Each lamp will draw the same current.
The problem becomes trivial.
Ok. So in that case the voltage drop is irrelevant. As long as the voltage stays over the 100v.
So all you need to do is provide 100v at end of line.
Each lamp will draw the same current.
The problem becomes trivial.
Post edited at 11:02
In reply to DancingOnRock:
No. If you supply a certain voltage at one end then the voltage at the other end will be different due to the resistive losses.
No. If you supply a certain voltage at one end then the voltage at the other end will be different due to the resistive losses.
In reply to elsewhere:
Yes . Which should be trivial because you're working back from the furthest point with known currents and voltages, rather than trying to work away from the source.
Yes . Which should be trivial because you're working back from the furthest point with known currents and voltages, rather than trying to work away from the source.
In reply to DancingOnRock:
No, because you *don't know* the current and voltage applied to the furthest lamp, only the power that it's drawing and that the voltage across it is less than the 100V supply.
No, because you *don't know* the current and voltage applied to the furthest lamp, only the power that it's drawing and that the voltage across it is less than the 100V supply.
In reply to elliptic:
No. Because you have to supply it with 100v for it to work. So you are working back from that to find out what your minimum source voltage needs to be.
No. Because you have to supply it with 100v for it to work. So you are working back from that to find out what your minimum source voltage needs to be.
In reply to DancingOnRock:
No. Read the original problem statement again:
The lamps can work with less than the full 100V and in fact none of them actually receive it. But they're regulated to constant *power* so draw more current to compensate.
No. Read the original problem statement again:
> 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 geting a lower voltage. Every lamp current depends on the voltage that all the others are getting.
The lamps can work with less than the full 100V and in fact none of them actually receive it. But they're regulated to constant *power* so draw more current to compensate.
Post edited at 12:06
In reply to elliptic:
What I mean is that if you say the furthest lamp is seeing V volts and work back towards the source, adding currents to the equation, it's a simpler equation than starting at the source and trying to create the equation.
What I mean is that if you say the furthest lamp is seeing V volts and work back towards the source, adding currents to the equation, it's a simpler equation than starting at the source and trying to create the equation.
In reply to DancingOnRock:
The trouble is that the voltage that the furthest lamp sees depends on the voltage drop across the wires supplying it, but that depends on the currents drawn by all the other lamps.
And the currents drawn by each of those lamps depend on the voltage across them, which depends on the voltage drop over the wires, which depends on......
You just can't separate out the problem in the way you're trying to do.
The trouble is that the voltage that the furthest lamp sees depends on the voltage drop across the wires supplying it, but that depends on the currents drawn by all the other lamps.
And the currents drawn by each of those lamps depend on the voltage across them, which depends on the voltage drop over the wires, which depends on......
You just can't separate out the problem in the way you're trying to do.
Post edited at 12:12
In reply to elliptic:
I'm not suggesting separating it out. If you start at the furthest lamp with i1=P/V then at the second lamp you get i2=P/(V+i1R) and so on. It becomes quickly obvious that this is a series and can be simplified and solved using P and R. If you start at the first lamp it's a lot more complicated to build the equation.
I'm not suggesting separating it out. If you start at the furthest lamp with i1=P/V then at the second lamp you get i2=P/(V+i1R) and so on. It becomes quickly obvious that this is a series and can be simplified and solved using P and R. If you start at the first lamp it's a lot more complicated to build the equation.
In reply to DancingOnRock:
Go on then, show us how simple it is. A simple four lamp case would be fine.
Go on then, show us how simple it is. A simple four lamp case would be fine.
Post edited at 12:33
In reply to DancingOnRock:
Good to see you have i1 and i2 differ because each lamp draws a different current.
If it really is trivial, please write down the equation for a few lamps distributed along the wire.
Good to see you have i1 and i2 differ because each lamp draws a different current.
If it really is trivial, please write down the equation for a few lamps distributed along the wire.
In reply to DancingOnRock:
Don't forget that V is not a constant and for each lamp depends on the current drawn by *every* other lamp not just the ones upstream or downstream.
In other words, you can't easily simplify it by starting from *either* end.
Don't forget that V is not a constant and for each lamp depends on the current drawn by *every* other lamp not just the ones upstream or downstream.
In other words, you can't easily simplify it by starting from *either* end.
In reply to Hooo:
If God had intended man to solve problems like this analytically he would not have invented SPICE.
If God had intended man to solve problems like this analytically he would not have invented SPICE.
Post edited at 12:54
In reply to elliptic:
V is the voltage the far lamp 'sees'. All other voltages and currents will depend on this. Or more properly it will depend on all the other voltages and currents.
So your current in the furthest lamp is i1, which is P/V. Where P = 100 Watts.
i2 the current in the second lamp etc...
Using R as the resistance of your cable.
i2= P/(V+ i1R)
i3 = P/(V + i2R)
i4 = P/(V + i3R)
The current flowing from your source is i1+i2+i3+i4.
The rest is fairly simple combining the equations and adding known constants.
V is the voltage the far lamp 'sees'. All other voltages and currents will depend on this. Or more properly it will depend on all the other voltages and currents.
So your current in the furthest lamp is i1, which is P/V. Where P = 100 Watts.
i2 the current in the second lamp etc...
Using R as the resistance of your cable.
i2= P/(V+ i1R)
i3 = P/(V + i2R)
i4 = P/(V + i3R)
The current flowing from your source is i1+i2+i3+i4.
The rest is fairly simple combining the equations and adding known constants.
In reply to DancingOnRock:
i2= P/(V+ i1R)
i3 = P/(V + (i2+i1)R + i1R)
i4 = P/(V + (i3+i2+i1)R + (i2+i1)R + i1R)
i1R voltage drop in last section section of cable
(i2+i1)R voltage drop in penultimate section of cable
(i3+i2+i1)R voltage drop in second from last section of cable
i2= P/(V+ i1R)
i3 = P/(V + (i2+i1)R + i1R)
i4 = P/(V + (i3+i2+i1)R + (i2+i1)R + i1R)
i1R voltage drop in last section section of cable
(i2+i1)R voltage drop in penultimate section of cable
(i3+i2+i1)R voltage drop in second from last section of cable
In reply to DancingOnRock:
Yes it's simple enough if you assume that the end voltage is a known quantity. But it isn't. The supply voltage is 100v not the final load voltage.
Yes it's simple enough if you assume that the end voltage is a known quantity. But it isn't. The supply voltage is 100v not the final load voltage.
In reply to DancingOnRock:
Yes, which is the whole problem (and noting that the voltage drop across each section depends on the *total* current through it, as elsewhere pointed out).
> V is the voltage the far lamp 'sees' .... it will depend on all the other voltages and currents.
Yes, which is the whole problem (and noting that the voltage drop across each section depends on the *total* current through it, as elsewhere pointed out).
In reply to knthrak1982:
Yes.
You now have a whole set of simpler simultaneous equations.
i1xV=100 is an easier/better start to begin substituting variables.
Yes.
You now have a whole set of simpler simultaneous equations.
i1xV=100 is an easier/better start to begin substituting variables.
Post edited at 14:48
In reply to elsewhere:
Yes. It's still simpler to work this way as it gives a single V value in terms of R and the power which are known. Otherwise you end up with some horrendous equations.
> i2= P/(V+ i1R)
> i3 = P/(V + (i2+i1)R + i1R)
> i4 = P/(V + (i3+i2+i1)R + (i2+i1)R + i1R)
> i1R voltage drop in last section section of cable
> (i2+i1)R voltage drop in penultimate section of cable
> (i3+i2+i1)R voltage drop in second from last section of cable
Yes. It's still simpler to work this way as it gives a single V value in terms of R and the power which are known. Otherwise you end up with some horrendous equations.
In reply to DancingOnRock:
No it doesn't. As you said earlier it depends on all the other voltages and currents, which is the whole problem.
I'm going to reverse the numbering convention (so it's easier to generalise) and start with V0 for the supply voltage = 100V, V1 for the unknown voltage across the first lamp, V2 for the second etc.
For the four lamp case, substituting P/Vn back in for the currents I get:-
V1 = V0 - PR( 1/V1 + 1/V2 + 1/V3 + 1/V4 )
V2 = V0 - PR( 1/V1 + 2/V2 + 2/V3 + 2/V4 )
V3 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 3/V4 )
V4 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 4/V4 )
Why don't you show us how to solve it from here?
> It's still simpler to work this way as it gives a single V value in terms of R and the power which are known.
No it doesn't. As you said earlier it depends on all the other voltages and currents, which is the whole problem.
I'm going to reverse the numbering convention (so it's easier to generalise) and start with V0 for the supply voltage = 100V, V1 for the unknown voltage across the first lamp, V2 for the second etc.
For the four lamp case, substituting P/Vn back in for the currents I get:-
V1 = V0 - PR( 1/V1 + 1/V2 + 1/V3 + 1/V4 )
V2 = V0 - PR( 1/V1 + 2/V2 + 2/V3 + 2/V4 )
V3 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 3/V4 )
V4 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 4/V4 )
Why don't you show us how to solve it from here?
Post edited at 15:38
In reply to DancingOnRock:
In my opinion, your equations are wrong.
For example, your equation for i4 only has current i3 for lamp 3 and does not include the voltage drop in the cable between lamps 4 & 3 due to the current i2+i1 for supplying lamps 2 & 1.
In my opinion, your equations are wrong.
For example, your equation for i4 only has current i3 for lamp 3 and does not include the voltage drop in the cable between lamps 4 & 3 due to the current i2+i1 for supplying lamps 2 & 1.
Post edited at 15:41
In reply to elsewhere:
Yes, I was agreeing with you.
Once you have everything in terms of i1, and V theoretically you can use the known Power P, and resistance R to calculate V.
The big problem you eventually arrive at is that we are using active power sources for the lamps and we're trying to use ohms law which is only really any good for nice resistive loads.
Yes, I was agreeing with you.
Once you have everything in terms of i1, and V theoretically you can use the known Power P, and resistance R to calculate V.
The big problem you eventually arrive at is that we are using active power sources for the lamps and we're trying to use ohms law which is only really any good for nice resistive loads.
In reply to DancingOnRock:
Which is much harder to achieve than you seem to think. Go on, show us how.
Luckily we're only applying Ohm's law to losses in the cables which *are* nice resistive loads.
> Once you have everything in terms of i1, and V
Which is much harder to achieve than you seem to think. Go on, show us how.
> The big problem you eventually arrive at is that we are using active power sources for the lamps and we're trying to use ohms law which is only really any good for nice resistive loads.
Luckily we're only applying Ohm's law to losses in the cables which *are* nice resistive loads.
In reply to Hooo:
In fact the problem is even impossible to solve even with one lamp and a length of cable.
V and I can be anything you chose as long as IV=100 and V<100.
In fact the problem is even impossible to solve even with one lamp and a length of cable.
V and I can be anything you chose as long as IV=100 and V<100.
In reply to DancingOnRock:
The thing is, we're now all agreeing that iterative methods are required. Your method is what captain paranoia said yesterday at 1805.
Hooo then asked if there was a simpler algebraic expression. ie, something you add all the known values to and which returns the right answer. ie no iterations required. I think we're all in agreement that the answer is no.
The thing is, we're now all agreeing that iterative methods are required. Your method is what captain paranoia said yesterday at 1805.
Hooo then asked if there was a simpler algebraic expression. ie, something you add all the known values to and which returns the right answer. ie no iterations required. I think we're all in agreement that the answer is no.
In reply to knthrak1982:
Yes. But captain paranoia hasn't shown the algebra. Which I'm working on paper and will try to type later.
Yes. But captain paranoia hasn't shown the algebra. Which I'm working on paper and will try to type later.
In reply to DancingOnRock:
For one lamp, If the supply voltage and cable resistance are both known quantities, then there are two possible solutions, as it's a 2nd order quadratic equation.
See my post at 1810.
> In fact the problem is even impossible to solve even with one lamp and a length of cable.
> V and I can be anything you chose as long as IV=100 and V<100.
For one lamp, If the supply voltage and cable resistance are both known quantities, then there are two possible solutions, as it's a 2nd order quadratic equation.
See my post at 1810.
Post edited at 16:53
In reply to DancingOnRock:
I haven't shown the algebra because it's probably tedious and complicated, and ICBA. It starts:
V1
V2 = V1 + R.P/V1
V3 = V2 + RP(1/V1+1/V2)
I guess you could go back and do repeated substitutions to express it all in terms of V1, which would get very messy, and I'm not good at tedious manipulation...
But I have created an Excel spreadsheet to address the problem, as I suggested. This is much easier to create, since, apart from the last lamp, it's a matter of copy & paste of formulae from previous stages.
The spreadsheet starts with the voltage across the last lamp, finds the current in the wire to the last lamp, and the voltage drop in the wires to the last lamp. From this voltage drop, I find the voltage across lamp 2, and the total current in the wire to lamp 2 etc. and work back to the supply. There are ten lamps and ten wire drops.
I then cheat and use Solver (again, as suggested) to modify V1 to attempt to make Vsupply=100
Solver fails to find a solution, and produces the same result as my manual iteration; the best it can achieve is V1=58.69V, and Vs=137.29 (at Is=12.69); Vs increases if V1 is increased or decreased from this value.
In order to allow Vs to reach 100V, I have to reduce the lamp power to about 53W.
I also calculate the power loss in the wires, and the lamp power, and check that they sum to the input power.
I've assumed the wire resistance accounts for the two conductors (i.e. 0.5ohm in supply and ground wires).
In the real world, a 'constant power' load using a switching regulator will take a variable power from the supply, as the efficiency is likely to change with input voltage...
I haven't shown the algebra because it's probably tedious and complicated, and ICBA. It starts:
V1
V2 = V1 + R.P/V1
V3 = V2 + RP(1/V1+1/V2)
I guess you could go back and do repeated substitutions to express it all in terms of V1, which would get very messy, and I'm not good at tedious manipulation...
But I have created an Excel spreadsheet to address the problem, as I suggested. This is much easier to create, since, apart from the last lamp, it's a matter of copy & paste of formulae from previous stages.
The spreadsheet starts with the voltage across the last lamp, finds the current in the wire to the last lamp, and the voltage drop in the wires to the last lamp. From this voltage drop, I find the voltage across lamp 2, and the total current in the wire to lamp 2 etc. and work back to the supply. There are ten lamps and ten wire drops.
I then cheat and use Solver (again, as suggested) to modify V1 to attempt to make Vsupply=100
Solver fails to find a solution, and produces the same result as my manual iteration; the best it can achieve is V1=58.69V, and Vs=137.29 (at Is=12.69); Vs increases if V1 is increased or decreased from this value.
In order to allow Vs to reach 100V, I have to reduce the lamp power to about 53W.
I also calculate the power loss in the wires, and the lamp power, and check that they sum to the input power.
I've assumed the wire resistance accounts for the two conductors (i.e. 0.5ohm in supply and ground wires).
In the real world, a 'constant power' load using a switching regulator will take a variable power from the supply, as the efficiency is likely to change with input voltage...
In reply to DancingOnRock:
No, we're using P=IV to find the current required for each lamp; not a problem (neglecting the efficiency problem I mentioned above). Then we use Ohm's law to find the voltage drop in the wire; that's no problem.
The big problem is that the algebra ends up as a big, ugly clusterf*ck (I suspect a 10th order equation in V1), although I'm sure there's a cunning mathematical solution... But I'm not good at cunning mathematics.
> The big problem you eventually arrive at is that we are using active power sources for the lamps and we're trying to use ohms law which is only really any good for nice resistive loads.
No, we're using P=IV to find the current required for each lamp; not a problem (neglecting the efficiency problem I mentioned above). Then we use Ohm's law to find the voltage drop in the wire; that's no problem.
The big problem is that the algebra ends up as a big, ugly clusterf*ck (I suspect a 10th order equation in V1), although I'm sure there's a cunning mathematical solution... But I'm not good at cunning mathematics.
In reply to captain paranoia:
Yes. I'm part way through the clusterf*ck.
Looks like if I can get a simple term for i1 for n lamps. I can the use a quadratic to solve for i.
Power = i^2 R in the wire as knthrank points out.
All interesting stuff. Just need to find my Stroud book in the garage...
Yes. I'm part way through the clusterf*ck.
Looks like if I can get a simple term for i1 for n lamps. I can the use a quadratic to solve for i.
Power = i^2 R in the wire as knthrank points out.
All interesting stuff. Just need to find my Stroud book in the garage...
In reply to elliptic:
I agree with these equations (in fact I'm certain they are correct!)
They are also clearly a nightmare* and probably impossible to solve algebraically. It is also clear they have multiple solutions. Even if the solutions were found numerically, I don't think it would be at all clear which would happen in practice (maybe only one of them is stable?).
*The one lamp case gives a quadratic with two solutions which is fine, but the 2 lamp case gives a sixth degree polynomial (though I did square both sides of the eqaution at one point to get rid of a square root and so introduced extra invalid solutions).
> V1 = V0 - PR( 1/V1 + 1/V2 + 1/V3 + 1/V4 )
> V2 = V0 - PR( 1/V1 + 2/V2 + 2/V3 + 2/V4 )
> V3 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 3/V4 )
> V4 = V0 - PR( 1/V1 + 2/V2 + 3/V3 + 4/V4 )
> Why don't you show us how to solve it from here?
I agree with these equations (in fact I'm certain they are correct!)
They are also clearly a nightmare* and probably impossible to solve algebraically. It is also clear they have multiple solutions. Even if the solutions were found numerically, I don't think it would be at all clear which would happen in practice (maybe only one of them is stable?).
*The one lamp case gives a quadratic with two solutions which is fine, but the 2 lamp case gives a sixth degree polynomial (though I did square both sides of the eqaution at one point to get rid of a square root and so introduced extra invalid solutions).
Post edited at 18:20
In reply to RichardAWatson:
No. The reason each lamp, despite being connected in parallel, will not have the same voltage across them, is because of the voltage drop in the cables between them.
No. The reason each lamp, despite being connected in parallel, will not have the same voltage across them, is because of the voltage drop in the cables between them.
In reply to knthrak1982:
I didn't realise this thread was still going! I think knthrak1982 nailed it on Tuesday when he declared that there isn't an algebraic solution, and it has to be solved iteratively. I think all the subsequent posts are confirming this. Those who still think there's a solution ( like I did on Monday), haven't fully understood it.
I might put it in Spice just to see what happens... Also, this should actually get built in a couple of weeks, although not to the original design, as the last lamp would get about 2V (according to Excel).
I didn't realise this thread was still going! I think knthrak1982 nailed it on Tuesday when he declared that there isn't an algebraic solution, and it has to be solved iteratively. I think all the subsequent posts are confirming this. Those who still think there's a solution ( like I did on Monday), haven't fully understood it.
I might put it in Spice just to see what happens... Also, this should actually get built in a couple of weeks, although not to the original design, as the last lamp would get about 2V (according to Excel).
In reply to Hooo:
You end up with a recursive non linear equation.
If you take the current flowing through the resistive cable to be f(x) you end up with:
0= Rf(x).(f(x)-f'(x)) + V (f(x)-f'(x)) + P
Where f'(x) is the current flowing down into the rest of the circuit.
Which as Robert says turns into a horrendous polynomial very quickly.
In the case of the single lamp f'(x)=0 so
0 = Rf(x)^2 + Vf(x) + P
V=100, P=100, R=1
0 = f(x)^2 + 100.f(x) + 100
f(x) = 98.99 or 1.01
Note: I've not done this sort of thing for about 20years so feel free to rip that apart.
You end up with a recursive non linear equation.
If you take the current flowing through the resistive cable to be f(x) you end up with:
0= Rf(x).(f(x)-f'(x)) + V (f(x)-f'(x)) + P
Where f'(x) is the current flowing down into the rest of the circuit.
Which as Robert says turns into a horrendous polynomial very quickly.
In the case of the single lamp f'(x)=0 so
0 = Rf(x)^2 + Vf(x) + P
V=100, P=100, R=1
0 = f(x)^2 + 100.f(x) + 100
f(x) = 98.99 or 1.01
Note: I've not done this sort of thing for about 20years so feel free to rip that apart.
In reply to DancingOnRock:
Yes, thank god you've finally realised what we've been trying to tell you all along
Note that knthrak1982 first solved it for one lamp on Tuesday, which you've now restated in rather bizarrely non-standard notation.
For the four-lamp case I've already shown you the full set of equations you need to solve, which generalise easily to any number of lamps.
The trouble is the order of polynomial involved grows rather rapidly (along with the multiplicity of possible solutions) and polynomials above degree four aren't guaranteed to be algebraically solvable at all.
Reminds me of https://xkcd.com/356/ ...
> You end up with a recursive non linear equation.... Which as Robert says turns into a horrendous polynomial very quickly.
Yes, thank god you've finally realised what we've been trying to tell you all along
Note that knthrak1982 first solved it for one lamp on Tuesday, which you've now restated in rather bizarrely non-standard notation.
For the four-lamp case I've already shown you the full set of equations you need to solve, which generalise easily to any number of lamps.
The trouble is the order of polynomial involved grows rather rapidly (along with the multiplicity of possible solutions) and polynomials above degree four aren't guaranteed to be algebraically solvable at all.
Reminds me of https://xkcd.com/356/ ...
In reply to elliptic:
I thought he was looking for an elegant algebraic solution. I think my single formula is a bit more simple than your four equations which still need to be merged into one and given 10lamps would become very long and unwieldy. Which you can't do because you're trying to do everything in terms of volt drop and not current. You need the current to determine the volt drop rather than the other way round as it's the current that is common factor around the circuit. knthrank started by demonstrating a certain set of values but never produced the equation.
It's not a nonstandard notation.
I thought he was looking for an elegant algebraic solution. I think my single formula is a bit more simple than your four equations which still need to be merged into one and given 10lamps would become very long and unwieldy. Which you can't do because you're trying to do everything in terms of volt drop and not current. You need the current to determine the volt drop rather than the other way round as it's the current that is common factor around the circuit. knthrank started by demonstrating a certain set of values but never produced the equation.
It's not a nonstandard notation.
In reply to DancingOnRock:
Yes, and my equations just have P/Vn substituted back in for the currents through each lamp. The thing is, we know what voltage the supply is delivering (V0 in my notation) but we don't know *any* of the currents, so imho it makes more sense to eliminate the currents and work entirely in voltages. It's all equivalent though.
f(x) usually refers to a function f() of an independent variable x, and f'(x) usually refers to the first derivative of f(x).
You're using them as variable names for the currents in different cables. They're not functions of x, whatever "x" is supposed to be. Why not just use I1 and I2?
[edit] Good luck if you're going to keep chipping away at this btw, but I'm done here.
> You need the current to determine the volt drop rather than the other way round as it's the current that is common factor around the circuit.
Yes, and my equations just have P/Vn substituted back in for the currents through each lamp. The thing is, we know what voltage the supply is delivering (V0 in my notation) but we don't know *any* of the currents, so imho it makes more sense to eliminate the currents and work entirely in voltages. It's all equivalent though.
> It's not a nonstandard notation.
f(x) usually refers to a function f() of an independent variable x, and f'(x) usually refers to the first derivative of f(x).
You're using them as variable names for the currents in different cables. They're not functions of x, whatever "x" is supposed to be. Why not just use I1 and I2?
[edit] Good luck if you're going to keep chipping away at this btw, but I'm done here.
Post edited at 13:35
In reply to elliptic:
Ok. You're correct it's been awhile. I should probably use f(n) and f(f(n)) instead.
Ok. You're correct it's been awhile. I should probably use f(n) and f(f(n)) instead.
In reply to Hooo:
To add to the learned replies. Draw a diagram, that will explain exactly what you are doing and it becomes a bit more intuitive. There is only one set of solutions to this scenario. No complex maths involved, but you need the basic parameters, voltage at each lamp and voltage at the power source. You could work backwards or forwards from all of this. NO diff equations etc necessary. Break it down into sections and simple Ohm's Law works. EG the leg from power source to first lamp carries all the current, the leg to the next lamp carries all the current less that of the first lamp, and so on. The last leg carries only the current to the last lamp. There will be a voltage drop from lamp to lamp due to the resistance of the wires, but to work that out you would need the wire resistance and the current. Since typically the wires are of almost negligible resistance they 'could' be ignored though. But if you used a voltmeter you could check what you are getting at each point and armed with the current in each lamp all the other values can be calculated. Put simply, the maximum current drawn would be the sum of the currents of all the lamps. If the wires were sufficient to carry this current then there is no problem. The effect of the longer wires would be to reduce the actual current at the lamp due to the increased resistance. Using Excel is more complicated than a pen and paper! How far you go or how complicated you make the calculation depends on what you are trying to achieve. Equal brightness of each lamp for example. This would entail different wires for each leg and or compensating resitances at each lamp and so on. The lamps are of a nominal constant resistance, therefore the lower the voltage available the lower the current through them. No need for trial and error or making assumptions, which will inevitably give you the wrong answer. And for lamps and a system of this nature it can be considered immaterial whether it is AC or DC.
To add to the learned replies. Draw a diagram, that will explain exactly what you are doing and it becomes a bit more intuitive. There is only one set of solutions to this scenario. No complex maths involved, but you need the basic parameters, voltage at each lamp and voltage at the power source. You could work backwards or forwards from all of this. NO diff equations etc necessary. Break it down into sections and simple Ohm's Law works. EG the leg from power source to first lamp carries all the current, the leg to the next lamp carries all the current less that of the first lamp, and so on. The last leg carries only the current to the last lamp. There will be a voltage drop from lamp to lamp due to the resistance of the wires, but to work that out you would need the wire resistance and the current. Since typically the wires are of almost negligible resistance they 'could' be ignored though. But if you used a voltmeter you could check what you are getting at each point and armed with the current in each lamp all the other values can be calculated. Put simply, the maximum current drawn would be the sum of the currents of all the lamps. If the wires were sufficient to carry this current then there is no problem. The effect of the longer wires would be to reduce the actual current at the lamp due to the increased resistance. Using Excel is more complicated than a pen and paper! How far you go or how complicated you make the calculation depends on what you are trying to achieve. Equal brightness of each lamp for example. This would entail different wires for each leg and or compensating resitances at each lamp and so on. The lamps are of a nominal constant resistance, therefore the lower the voltage available the lower the current through them. No need for trial and error or making assumptions, which will inevitably give you the wrong answer. And for lamps and a system of this nature it can be considered immaterial whether it is AC or DC.
In reply to nscnick:
The cable between each lamp is long, it has resistance, 1ohm.
The voltage and current through each lamp are different and to compound this the current through each lamp is inverse to the voltage across it as they are constant power lamps.
If they were normal lamps then yes it's a trivial problem.
The cable between each lamp is long, it has resistance, 1ohm.
The voltage and current through each lamp are different and to compound this the current through each lamp is inverse to the voltage across it as they are constant power lamps.
If they were normal lamps then yes it's a trivial problem.
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