Bright Sparks

[davelx]

.... 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.

Though it's worth noting that at max power delivery, efficiency is only 50% as the same amount of power is also dissipated in the source impedance.

 

[dizz]

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?

That's almost a non-question - since (given only one condition, which doesn't impose a maximum X or miniumum N, per se), my method will always give you the answer with ONE weighing.

Kind Regards, John.

I don't think it is a non-question, as I think people that get your question right may get the supplementary question wrong - and it gives them something to do!

1. You agree that there is a maximum number of piles with two weighings?
2. You agree that there is a minimum number of coins needed in each pile?

Suppose you have to discard the coins used in the first weighing, what would be the maximum number of piles and the minimum number of coins is each pile for you to be able to identify which original stack the 11g coins were in?

 

[dizz]

John - I'd ask people to confirm that they know they can do it with just 10 coins and that they couldn't do it with 12 stacks of 20 coins with one weighing! That way they don't spoil it for others that are still thinking?

Well, you've said it, so we'll see - but I think that might possibly 'spoil it' for some people.

Kind Regards, John.

Maybe maybe not - I work on probabilities - if you left the question the way it is, someone would post the method of attack, and then 100% of other people would know how to do it immediately.

My stating the above, might enable a percentage of people to get it quicker, but still a significant cohort will be none the wiser, and those that do get it will know its right by referencing the above.

 

[JohnW2]

Though it's worth noting that at max power delivery, efficiency is only 50% as the same amount of power is also dissipated in the source impedance.

Very true - and that obviously follows directly from the calculation.

However, if one wanted to maximise efficiency, one would have a problem - the efficiency approaches 100% as the load impedance approaches infinity, and the power delivered therefore approaches zero!

Kind Regards, John.

 

[JohnW2]

[I don't think it is a non-question, as I think people that get your question right may get the supplementary question wrong - and it gives them something to do!
1. You agree that there is a maximum number of piles with two weighings?

No. Albeit there is one condition, there is no maximum number for getting the answer in ONE weighing, per se - so obviously no maximum for two weighings (one's second weighing could just be a random, unnecessary, one). Can't say anything about condition without giving gtoo much away.

[2. You agree that there is a minimum number of coins needed in each pile?

No. Corresponding reason to above, with same caveat as the condition.

Kind Regards, John.

 

[ban-all-sheds]

BTW - if anybody wants to demonstrate, safely, how supply impedance affects how much power you can get out of it try powering a lamp from a telephone line.

 

[JohnW2]

My stating the above, might enable a percentage of people to get it quicker, but still a significant cohort will be none the wiser, and those that do get it will know its right by referencing the above.

Agreed. Probabilistically, you're correct. We'll see! It obvioulsy will take only one post to kill the discussion!

Kind Regards, John.

 

[dizz]

[I don't think it is a non-question, as I think people that get your question right may get the supplementary question wrong - and it gives them something to do!
1. You agree that there is a maximum number of piles with two weighings?

No. Albeit there is one condition, there is no maximum number for getting the answer in ONE weighing, per se - so obviously no maximum for two weighings (one's second weighing could just be a random, unnecessary, one). Can't say anything about condition without giving gtoo much away.

[2. You agree that there is a minimum number of coins needed in each pile?

No. Corresponding reason to above, with same caveat as the condition.

Kind Regards, John.

Ha - oh yes - there is other (obvious) info available to me that enables me to identify precisely the stack with any number of piles ...

 

[ebee]

Go on I give in - I am missing something or I am making a bad assumption

 

[JohnW2]

Go on I give in - I am missing something or I am making a bad assumption

The incredibly obvious (once one twigs it - a real 'kickself'!) solution does not require any assumptions. The more general problem, of X piles each containing N coins, involves one 'condition' (regarding the values of X and N), but again no assumptions. Does that count as a 'clue'?

Kind Regards, John.

 

[EFLImpudence]

Go on I give in - I am missing something or I am making a bad assumption

I have it.

Don't forget it's not balancing scales but a weighing machine.

 

[ebee]

I suppose if I now forget about it then it will come to me in a blinding flash.

That did happen once (The Monty Hall Conundrum) I was shaving one morning and looked in mirror and said "Eureka!" two weeks after I`d given up on it.

 

[JohnW2]

I suppose if I now forget about it then it will come to me in a blinding flash.

That's how it usually happens! At risk of of 'giving it away', let me tell the one (and only) condition which applies to the general case of X piles, each containing N coins. In that situation, the required condition (for it be be do-able with one weighing) is N ≥ X.

That did happen once (The Monty Hall Conundrum) I was shaving one morning and looked in mirror and said "Eureka!" two weeks after I`d given up on it.

I've seen heated 'discussions' (arguments) between very senior statisticains go on for weeks over that one. In the final analysis, it's often only a demo/simulation which will eventually convince them of what is an incredibly counter-intuitive answer! However, that one is almost the antithesis of a 'kickself' - since even when one has been convinced of what is the correct answer, it still seems totally counter-intuitive to most people.

Kind Regards, John.

 

[ebee]

When you look back at the monty hall it is very simple.
Your first guess gives you a 1 in 3 chance of winning therefore a 2 in 3 chance of losing.
Changing you mind after seeing the goat then gives the opposite of this therefore 2 in 3 chances of winning and 1 in 3 chances of losing.

1 in 3 becomes 2 in 3 and 2 in 3 becomes 1 in 3.

(Just hope your first guess is wrong then change your mind)

It is so simple when you kick yourself.

I thought the piles (here wrongly assuming again I am) had to stay in tens although it may be permitted to weigh more than one at any time ie 10,20,30, 40 or 50.

Hmm need a rethink now

 

[bernardgreen]

Yes it is possible, and the answer confirms that for the method to work there must be more coins in a stack than there are stacks.