MCAT power, current, voltage question...

[DocBella]

• Power companies try to save the amount of copper needed for power lines by using thinner wires, which makes R quite high.
• To minimize P dissipated by the wires, they minimize I by maximizing V. This is why power lines transfer electricity at high voltage.

P=IV=I^2R

Can someone tell me how this makes sense...? I tried to use the Power equation to see the correlation but I don't see it.

Thanks!

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[JMMTB]

There are two different powers. 1. The power we want to deliver from the factories to the homes and 2. the power dissipated or lost along the way. P1 is a fixed value base upon P=IV. P2=I^2R=V^2/R so if we want P2 to be as small as possible we must lower I or raise V. Luckily for us both must happen because P1 is fixed.

So the most efficient way to transmit electricity is at a high voltage with small current. Then transformers step it down it a lower voltage for us to use.

 

[Hadi7183]

This doesn't answer OP's question. Basically you just reworded the question.

 

[Cawolf]

@Hadi7183 So why don't you answer it?

I understood his explanation fine.

 

[Hadi7183]

I didn't know the answer, but good for you. I hope OP is also happy with the answer.

The question, IMO, says given P=RI^2=V^2/R (i.e. P is directly proportional to V and I), how come they have opposite effects?

 

[JMMTB]

It goes back to the fact that we are talking about two different powers. 1 is the power we are transmitting and the other is the power that is lost along the way. The power that makes it to our houses would be P1-P2. We want P2 to be as small as possible.

Maybe it will make sense quantitatively.

Say we want to transmit 1000W of electrical power to a neighborhood. P=IV always so if we transmit 1000 W at 1000 V the current would be 1 amp, 500 V 2 amps and so on. As for the power lost along the way that is like you said P=I^2*R. and we'll say the resistance for the wire is 10 ohms.

Comparing the two extremes (1000V 1 amp v. 1000 amp and 1 V) we get

P-lost=1^2*10=10W lost Voltage drop across the wire is V=IR so 1*10=10 Voltage drop P-lost=(10^2)/10=10 Watts

And the other extreme

P-lost=1000^2*10= 10000000W, but wait that is greater than the Power we started with, so an important point is made. Because we start with only 1 volt we cannot drop more than that along the wire. Solving for the max current under these conditions would be Vdrop/R=I= .1 amps (1V/10 ohms). The lower voltage actually forces the system to flow at a lower current.

Say we transmit 500 V at 2 Amp

P-lost=2^2*10=40 Watts lost. Double checking with P=V^2/R Vdrop=2*10=20 Thus (20^2)/10=40 Watts lost.

Does that clarify at all?

 

[Hadi7183]

Yes, it does. Thanks for your time.

 

[DocBella]

It goes back to the fact that we are talking about two different powers. 1 is the power we are transmitting and the other is the power that is lost along the way. The power that makes it to our houses would be P1-P2. We want P2 to be as small as possible.

Maybe it will make sense quantitatively.

Say we want to transmit 1000W of electrical power to a neighborhood. P=IV always so if we transmit 1000 W at 1000 V the current would be 1 amp, 500 V 2 amps and so on. As for the power lost along the way that is like you said P=I^2*R. and we'll say the resistance for the wire is 10 ohms.

Comparing the two extremes (1000V 1 amp v. 1000 amp and 1 V) we get

P-lost=1^2*10=10W lost Voltage drop across the wire is V=IR so 1*10=10 Voltage drop P-lost=(10^2)/10=10 Watts

And the other extreme

P-lost=1000^2*10= 10000000W, but wait that is greater than the Power we started with, so an important point is made. Because we start with only 1 volt we cannot drop more than that along the wire. Solving for the max current under these conditions would be Vdrop/R=I= .1 amps (1V/10 ohms). The lower voltage actually forces the system to flow at a lower current.

Say we transmit 500 V at 2 Amp

P-lost=2^2*10=40 Watts lost. Double checking with P=V^2/R Vdrop=2*10=20 Thus (20^2)/10=40 Watts lost.

Does that clarify at all?

This cleared it up. Thank you!