Bright Sparks
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Go out for a couple of drinks and everyone's been very busy.
John - I'm not the spark here, everyone else is - your queetion seems to me to be a spark question not a maths question - never did that in my degree, but if you'd like to give me a proper MATHS question, Pure, Statistics, Mechanics and Game Theory are my specialist subjects - I'd try to oblige ... ?
However, in respect of your spark question, I think I see the issue - As the resistive load increase, (which contributes to higher power), current falls (which contributes to lower power). I'd therefore investigate creating a function for power that I could differentiate. Once differentiated, I'd set the first derivative to zero to the find maximum power, then check whether it's a maximum or minmum by producing a second derivative. If the second derivative was less than zero, I'd have confirmed the maximum exists and provide you with the answer.
But prey tell master what is the answer and the way to proceed?
John - I'm not the spark here, everyone else is - your queetion seems to me to be a spark question not a maths question - never did that in my degree, but if you'd like to give me a proper MATHS question, Pure, Statistics, Mechanics and Game Theory are my specialist subjects - I'd try to oblige ... ?
However, in respect of your spark question, I think I see the issue - As the resistive load increase, (which contributes to higher power), current falls (which contributes to lower power). I'd therefore investigate creating a function for power that I could differentiate. Once differentiated, I'd set the first derivative to zero to the find maximum power, then check whether it's a maximum or minmum by producing a second derivative. If the second derivative was less than zero, I'd have confirmed the maximum exists and provide you with the answer.
But prey tell master what is the answer and the way to proceed?
There your rather wrong. I'm not an electrician, and nor is BAS - and, to the best of my knowledge, nor are several of the otehrs who have particpated ti this thread.
It's a bit odd that you say that, since you go on to give a very good mathematical answer to it!
Spot on. If you sketched the function, or merely thought about the nature of the problem (as you've done above), you'd realise that there is only one turning point, which is a maximum, so you wouldn't really need to look at the second derivative. The only bit you haven't done is the 'spark' bit - actually 'creating the function', but that's really very easy (I can demonstrate if you wish). The resultant function is a bit tedious to differentiate, but one just has to plod through it.
However, for those who have come across this electric issue (or done the maths once) before, none of that maths is necessary. ....
.... maximum power is always delivered to a load when its impedance is the same as that as the 'internal impedance' of the power source. In the example I gave, the maximum current of 1150A occurs when the supply is 'shorted', hence the only impedance in that circuit is the supply's internal impedance. At 230V, that means that the supply impedance (voltage/current) is 230/1150, namely 0.2Ω. Hence, for maximum power delivery, we also have a 0.2Ω load. Total impedance in circuit is then 0.4Ω, hence a current (voltage/impedance) of 230/0.4 =575A. Power in the load (current squared x impedance) is therefore 575 x 575 x 0.2 = 66125 watts. Simples
Kind Regards, John.
OK, albeit well 'off-topic', just for fun. You may well know this one already - so, if you do (without having to think), perhaps you would leave it for someone else to muse over ...
5 stacks, each of 10 visually identical coins. You are told that all the coins in 4 of the stacks weigh exactly 10g, but all of those in the fifth stack weigh 11g. You have an accurate set of weighing scales. How do you determine which stack has the heavy coins by undertaking just ONE weighing?
Kind Regards, John.
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That would probably make the problem very difficult to solve - unless, of course, you believed that the heavier ones would fall more quickly !!Am I allowed to drop them on the floor first
Kind Regards, John.
Dizz,
Earlier I asked,
1/
Two farmers meet at crossroads and get talking.
Each have a herd of sheep.
One says
"Sell me two of yours and I will have twice as many as you"
Tother says
"Sell me two of yours and we`ll each have the same"
How many sheep had each farmer?
2/
(Best spoken not written but here goes!)
If a brick weighs 7 pounds and half a brick then what does a brick and a half weigh?
Seems no one answered!
Earlier I asked,
1/
Two farmers meet at crossroads and get talking.
Each have a herd of sheep.
One says
"Sell me two of yours and I will have twice as many as you"
Tother says
"Sell me two of yours and we`ll each have the same"
How many sheep had each farmer?
2/
(Best spoken not written but here goes!)
If a brick weighs 7 pounds and half a brick then what does a brick and a half weigh?
Seems no one answered!
I could do the coin prob in two weighings but not one (yet) let me think!
Not sure if this counts as "one weighing" or not but I'd put all the stacks on the scales at the same time & then remove a coin from each stack until the weight dropped by 11g
Ok - haven't seen it, but I get it.
You might change your post to ask posters to post the "minimum" number of coins that needs to be weighed in order to solve the problem - that way you know with a high probability they know how to do it (and subsequent posters will see the minimum and confirm that they know), without the first poster spoiling it for the rest.
But thinking about your question, there appears to be a nice supplementary question to those that can do your question, that requires just a little additional thinking.
Supplementary Question
Let X be the number of stacks, and let N be the number of coins in the stacks (with N-1 stacks having 10g coins, with just 1 stack containing 11g coins as before).
If you are allowed TWO weighings, what is the maximum X and minimum N?
ADDED [binding condition - the coins you used in the first weighing are discared, and the number of cons in each stack are reduced accordingly for your seconds weighing]
Two weighings is easy peasy back of fag packet stuff.
It is the one weighing that makes it .
It is the one weighing that makes it .
Let one farmer have X sheep.
Let the other have Y sheep.
The first part of the question implies :
(X+2) = 2(Y-2) .... [A]
[you take two sheep away from one farmer and give two sheep to the other, then the gaining farmer will have twice as many as the losing farmer]
The second part of the question implies :
X - Y = 2 ....
[the difference between X and Y is 2]
From B, X = Y+2 - substitute for X from into [A], then [A] becomes :
Y + 2 + 2 = 2Y - 4
Re-arrange to get Y = 8, and sub for Y back into
so X = 10.
Solution - one farmer has 8, and one farmer has 10.
[ADDED - ERROR NOTED HERE - SO WHAT'S THE ERROR]
2/
(Best spoken not written but here goes!)
If a brick weighs 7 pounds and half a brick then what does a brick and a half weigh?
Seems no one answered!
Your question implies :
B = 7 + B/2
[A brick = 7 plus half a B]
Re-arrange equation and it becomes :
B - B/2 = 7
B/2 = 7
B = 14
So, the next part you ask what does B + B/2 weigh :
Well B + B/2 = 14 + 14/2 = 21 pounds
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One and a half bricks = 21 pounds.
Yes, two weighing with John's numbers - but now I am asking what is the maximum number of stacks and minimum number of coins in each stack where TWO weighings is still possible?
30 stacks of 40 coins, 60 stacks of 70 coins, ... ?
What I'm looking for is the MAX number of stacks and MIN number of coins in each stack where you still ONLY need to make two weighings.
[Just missed another binding condition I could have asked - which I'll now add - the coins you used in the first weighing are discared, and the number of coins in each stack are reduced accordingly for your seconds weighing]
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The sheep solution does not work.
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