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soyuppy wrote:

daydreamer0023 wrote:

soyuppy wrote: Like, "how many cans of coke does it take to power up an iPhone?"

Was that actually a question?

Yes, from N. Cal state

That's just awful. I've seen some vague question, but I think that one takes the... coke.

sorry couldn't resist

Can someone help elaborate or explain this Power loss and voltage problem show in the attach picture? This taken from one of the invitational test.
I understand that P(loss) = I*I*R or P(loss) = V*V/R, but it still does not explain the reasoning behind the answer? Base on the latter equation P(loss)=V*V/R, If R remain constant, as V increase, P(loss) would increase, wouldn't it?
Or do we just accept that this is a rule specified somewhere in electrical physic book? Whenever voltage is stepped up by X factor, P(loss) decrease by the X*X factor?

soyuppy wrote:Can someone help elaborate or explain this Power loss and voltage problem show in the attach picture? This taken from one of the invitational test.
I understand that P(loss) = I*I*R or P(loss) = V*V/R, but it still does not explain the reasoning behind the answer? Base on the latter equation P(loss)=V*V/R, If R remain constant, as V increase, P(loss) would increase, wouldn't it?
Or do we just accept that this is a rule specified somewhere in electrical physic book? Whenever voltage is stepped up by X factor, P(loss) decrease by the X*X factor?

P(loss) = P²R/V² <-- That's the key formula, explained here: http://www.bsharp.org/physics/transmission

Worth making the connection to some real-life stuff - http://www.dartmouth.edu/~news/releases/2009/01/07.html

soyuppy wrote:Can someone help elaborate or explain this Power loss and voltage problem show in the attach picture? This taken from one of the invitational test.
I understand that P(loss) = I*I*R or P(loss) = V*V/R, but it still does not explain the reasoning behind the answer? Base on the latter equation P(loss)=V*V/R, If R remain constant, as V increase, P(loss) would increase, wouldn't it?
Or do we just accept that this is a rule specified somewhere in electrical physic book? Whenever voltage is stepped up by X factor, P(loss) decrease by the X*X factor?

seems to make more sense to me, since this is why power lines are high voltage; to reduce resistive heating (and of course more resistance means more resistive heating).

EDIT: Ninja'd, and it looks like I was (mostly) right!

soyuppy wrote:Can someone help elaborate or explain this Power loss and voltage problem show in the attach picture? This taken from one of the invitational test.
I understand that P(loss) = I*I*R or P(loss) = V*V/R, but it still does not explain the reasoning behind the answer? Base on the latter equation P(loss)=V*V/R, If R remain constant, as V increase, P(loss) would increase, wouldn't it?
Or do we just accept that this is a rule specified somewhere in electrical physic book? Whenever voltage is stepped up by X factor, P(loss) decrease by the X*X factor?

The confusion is over the meaning of V (voltage).

In the question they are increasing the voltage that the power line is carrying (i.e. increasing the line voltage from 11 kV to 33 kV). If the line is still distributing the same power to the customers then the current will decrease by a factor of 3 (since P = V*I), so P = I*I*R will show that the power lost in the power line will decrease by a factor of 9 assuming the resistance stays the same.

The problem with using P=R/(V*V) is that in this equation V is referring to the voltage drop between the two ends of the power line, not the voltage that it is carrying. To see what the voltage drop along the power line is, use V = I*R. This immediately shows that if the current in the power line decreases by a factor of 3 then the voltage dropped along the power line will also drop by a factor of 3, and now you can use P = R/(V*V) and get the same answer as above.

In these type of questions it is usually easier to use P = I*I*R rather than trying to calculate the voltage drop.

soyuppy wrote:Can someone help elaborate or explain this Power loss and voltage problem show in the attach picture? This taken from one of the invitational test.
I understand that P(loss) = I*I*R or P(loss) = V*V/R, but it still does not explain the reasoning behind the answer? Base on the latter equation P(loss)=V*V/R, If R remain constant, as V increase, P(loss) would increase, wouldn't it?
Or do we just accept that this is a rule specified somewhere in electrical physic book? Whenever voltage is stepped up by X factor, P(loss) decrease by the X*X factor?

Guess the only way to see this in action is to prove it by example. So here's what one way to see why this is so.
Given a 100kW Power Plant, transmitting on .4 Ohm resistance wire. 2 voltages to be consider for transmission

• a) 1 kV, I = P/V==> I=100000/1000 = 100A, P(loss) = I*I*R ==> P(loss) = 100*100*.4 = 4000 W
  b) 10 kV, I=100000/10000 = 10A, P(loss) = 10*10*.4 = 40 W

This shows that V increase by 10 times, P(loss) decrease by 100(10*10) times.

Can someone tell me when the first electric wind turbine was invented, and by who? I'm getting mixed results.

Accidentally duplicated

freed2003 wrote:Can someone tell me when the first electric wind turbine was invented, and by who? I'm getting mixed results.

What results are you getting? I am assuming they are between Charles Brush and James Blyth. James Blyth invented the first electricity producing wind turbine in July 1887. Charles Brush, on the other hand, invented a wind turbine in 1888 in Ohio, AFTER James Blyth. Brush's was larger and had an automatic braking system.

Updated score sheet posted on soinc. Does anyone know what "Uncorrect Const. Violation" refer to? Which rule is this pertain to?