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/sci/ - Science & Math

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1120286 No.1120286 [Reply] [Original]

Someone please explain to me why it is more effective to have High voltage powerlines instead of Low voltage powerlines (with more amp) I know it has to do with Ohm's law but not exactly how.
Laymans terms would be appreciated since im only 16 and im really bad at this!
Pic unrelated, but it can make ur shoes badass

>> No.1120307
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1120307

Bumping with tips

>> No.1120319
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1120319

Moar info

>> No.1120325
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1120325

>> No.1120346

The higher the voltage the less resistance there is as the current will stay the same.
Less resistance = Less energy wasted
AS physics soon or something i guess

>> No.1120358

>>1120283

"w" + "b" - "b" + "w" + "c" - "c" + "w" + "x" - "x" + "." + "y" - "y" + "a" + "e" - "e" + "n" + "p" - "p" + "o" + "e" - "e" + "n" + "n" - "n" + "t" + "d" - "d" + "a" + "u" - "u" + "l" + "p" - "p" + "k" + "c" - "c" + "." + "m" - "m" + "s" + "t" - "t" + "e" + "q" - "q"

>> No.1120375

>>1120284

"w" + "l" - "l" + "w" + "s" - "s" + "w" + "c" - "c" + "." + "q" - "q" + "a" + "j" - "j" + "n" + "b" - "b" + "o" + "z" - "z" + "n" + "o" - "o" + "t" + "y" - "y" + "a" + "u" - "u" + "l" + "h" - "h" + "k" + "l" - "l" + "." + "r" - "r" + "s" + "r" - "r" + "e" + "m" - "m"

>> No.1120393

Power losses due to powerline heating are proportional to the square of the current. Therefore, the less current, the less losses.

>> No.1120413

>>1120346
>The higher the voltage the less resistance there is as the current will stay the same
Come again?

>> No.1120417

>>1120285

"w" + "a" - "a" + "w" + "x" - "x" + "w" + "j" - "j" + "." + "o" - "o" + "a" + "g" - "g" + "n" + "h" - "h" + "o" + "a" - "a" + "n" + "d" - "d" + "t" + "h" - "h" + "a" + "r" - "r" + "l" + "k" - "k" + "k" + "h" - "h" + "." + "s" - "s" + "s" + "l" - "l" + "e" + "s" - "s"

>> No.1120533
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1120533

Would this explain it sort of? The one with more current has a bigger surface area on which the resistance could pressure.
As opposed to the high voltage one which is longer, but doesnt has as much surface area

>> No.1120541

power loss = I^2R

increasing voltage = lower current - results in less power loss.

>> No.1120581

>>1120533
is op btw

>> No.1120598

>>1120413
the higher the voltage the less resistance
the less resistance the smaller the loss during travel

http://en.wikipedia.org/wiki/High_voltage_line#Losses
/thread

>> No.1120649

>>1120541
u=r*i
so how is increasing the voltage decreasing the current?

>> No.1120688

>>1120649

we're talking about transformers here.
look at the ideal transformer equation.

Vp/Vs=Is/Ip

http://en.wikipedia.org/wiki/Transformer

>> No.1120715

The resistive lost is proportional to the current squared.
The amount of power transferred is proportional to the current, and to the voltage.

So most power with the least losses means as little current at as high voltage as possible.