Comments on: I2R https://wimminz.wordpress.com/2014/07/19/i2r/ Wimminz Sun, 08 Apr 2018 01:13:44 +0000 hourly 1 http://wordpress.com/ By: bob k. mando https://wimminz.wordpress.com/2014/07/19/i2r/#comment-8070 Wed, 23 Jul 2014 13:37:36 +0000 http://wimminz.wordpress.com/?p=4666#comment-8070 actually, afor’s example works.

only, you have to assume a 1 ohm impedance work load in all examples. which i wouldn’t consider to be best engineering practices.

a larger part of the problem understanding this essay is that afor keeps jumping between different power formulas with no explanation as to what he’s doing.

1 – the essay is titled I^2 * R
2 – first and second examples are V * I
3 – third and fourth exa are back to I^2 * R

the logic can be followed … but if you don’t already know that V = I * R you’re going to be completely out to sea.

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By: Joe https://wimminz.wordpress.com/2014/07/19/i2r/#comment-8054 Sun, 20 Jul 2014 04:29:07 +0000 http://wimminz.wordpress.com/?p=4666#comment-8054 India had massive grid down back in 2012 but it was prompty restored. Even with what you describe here, much of the third world is even more “fly by the seat of your pants”. ZH is fun to read but it can also be alarmist.

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By: Digger Nick https://wimminz.wordpress.com/2014/07/19/i2r/#comment-8053 Sun, 20 Jul 2014 04:21:15 +0000 http://wimminz.wordpress.com/?p=4666#comment-8053 >Wire of 5 Ohms resistance
>100 x 100 = 10,000, 10,000 x 5 = 50,000, so you’ll get *nothing* out the other end, in fact, to get 10,000 Watts out the other end, you’ll have to put 10,000 + 50,000 = 60, 000 watts in one end, and you’ll get 84% transmission losses.

No afor. Resistance only serves to limit current. Hence the max you can pass trough a 5r wire (if you short it), at 100v, given the wire’s resistance is 5R is 20A. Because U=I*R.
Now say a 20% drop at the end user in voltage is acceptable. 5*5=25, so you’d need a total R of 25, 5 being the wire 20 being the appliance, allowing you to pass 4 amps at 100v, with only a 20% loss. Aint much, but thats just how shit works.

You could put a perfect current supply on the supply end, and still only get 20 amps max out at the other. That is just assuming resistive losses, of course. A more accurate calculation would involve inductive and corona lossses, like so http://large.stanford.edu/courses/2010/ph240/harting1/
Notice how you do not technically put watts in, you have either a current, or a power source. Typically a generator is a power source.

Furthermore centralizing control over appliances would be orwellian, and unneccesary, you could simply make substations that have a stepped transformer allowing them to step the voltage down a bit more if an unauthorized load gets too heavy, and a chain of power stations that broadcasts it’s energy requirements to eachother over a secure connection. Each power sink could then request an amounth of power from a power source higher up the chain. You could even do the same with house appliances requesting an amounth of power from the substation before turning on, anonymously.

There would however be no viable way of securing this system either. And the only way to prevent shit from falling apart given a malicious actor would be the transformers individual ability to step down the available voltage at their ends.

But giving the government full control over who gets to turn on what? All of my nope. Not that that cant be hacked of course, but it’s fucking orwellian for the average user.

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