# Power in a circuit?

1. Aug 10, 2006

### pivoxa15

For a given amount of power transmitted, you want the lowest dissiapated power through a resistance. This can be achieved through transmitting at the highest voltage and lowest current. But to be at a high voltage implies a high current. The are directly correlated. So if one was to create a high voltage but make low current flow then one has to do work to 'slow' the current flow? This translates to wasted power? But that is not taken into account in textbooks. Why? Is my concept wrong?

i.e P=iV

P lost = i^2R

If for the second time, one transmitted current at i/2 and at a higher potential of 2V then the power lost is 1/4 less but the power delivered is still the same. But for me work needs to be done to reduce the current follow because voltage has doubled. The charges 'want' to move more in the second case but you are not letting it. Why isn't this extra work taken into account or is my thinking wrong?

2. Aug 10, 2006

### Staff: Mentor

The wires aren't the load and the load does not get its voltage increased. High voltage is only for transmission lines.

3. Aug 10, 2006

### pivoxa15

I assume load means anything with resistance. Anything with a current implies some resistance. No resistance also means no current. Therefore, current and resistance are the same thing. Assuming ordinary, everyday equipment.

If transmission lines does not transmit current, what does it transmit?

4. Aug 11, 2006

### Staff: Mentor

Huh? Current and resistance are most certainly not the same thing, and line losses in high voltage transmission lines are only something like 7%, so they do not make up much of the load on the power grid.

5. Aug 11, 2006

### pivoxa15

So transmission lines does trasmit current but the power lost due to its resistance is only 7%?

In reality, in wire like copper (are they the materials in transmission lines?), whenever a current passes, resistance occurs, as not all of the electrons are moving in the same direction, resulting in power dissapation. But we can model it as a perfect current and a resistor in a small section that acts to dissiapate all the power that would have been lost in this circuit in reality.

6. Aug 11, 2006

### rbj

pivoxa, the load impedance (a more general concept than resistance) is not the same as the resistance along the transmission line. usually, an increased load (in terms of power) means less resistance (because virtually constant voltage sources are more common than current sources). also, being that it is a property of the transmission line, there is resistance even if there is no current.

7. Aug 11, 2006

### Staff: Mentor

No, load is the reason the power grid exists: to run your lights, computer, and air conditioning. Its where the other 93% of the electricity goes.

8. Aug 12, 2006

### pivoxa15

I assume Transmission lines transmit current and for efficiency, they are transmitted at very high voltages.

Load is where the current goes which translates to power used to do useful work whereas resistance is power lost without any useful work done.

Is this correct?

How about my original question that started this thread. It is still very much a mystery to me. Or is my question problematic? If so could you point it out clearly because as you can see already, I am a beginner at electricity.

Last edited: Aug 12, 2006
9. Aug 12, 2006

### pivoxa15

Do you mean random agitations of electrons in the power line?

10. Aug 12, 2006

### cesiumfrog

This may be the simpler explanation you're after. The line effectively has a fixed resistance R (perhaps imagine it will resist each unit of current equally). The whole point of the line is to get power to a load (since your TV needs a certain amount of power L). Say we want the power to arrive at a particular voltage V, so we know how much current must be pumped through (L=VI).

However, as you noted originally, the extra power used by the lines is P=IIR. Rearranging that math, the wasted power is P=R(L/V)^2. If you ponder that 1/V^2, I hope it's clear why higher voltage transmission lines are more efficient.

11. Aug 12, 2006

### rbj

no, i mean like a 220 ohm resistor that you can buy from Radio Shack, that the resistance is primarily a property of the object, not a property of the current passed through it. when a current is forced through it, the result is that a voltage drop exists. but, under normal circumstances, the resistance is relatively constant.

a more abnormal circumstance is when resistors (or power lines) start to measurably heat up because of a large amount of current passed through it. in that case, the resistance increases (due to the temperature increase) which can be attributed to the applied current. this is what is happening when there is a catatrophic failure of a power line that is carrying more current than it was ever meant to carry. it gets warm, the resistance increases, even more power is disappated in the line making it warmer and more resistance, etc. the other consequence of the line getting warmer is that the length of the steel/copper line expands (as all solid metal does when it gets warmer) and the lines droop lower and lower until one of them contacts a tree or something and shorts out and trips a circuit breaker. that's when blackouts happen.

12. Aug 12, 2006

### myself

ah, but the resistance is always constant, whether there is voltage passing through it or not. If you have the resistor hooked up to a 6 volt battery, the resistance is still 220 ohms. True, there is a 6 volt voltage drop across the resistor, but the resistance is constant. It's harder to picture a "current being forced through it" because it is more difficult to make a constant-current source than a constant-voltage source. I hope this clarified something.

The power line doesn't start heating up by that much, because of two things: first, it is low resistance, and second current is low. Why is current low? You look at POWER consumed by the people using electricy. Since energy is conserved, the power is constant, P = IV. But the line carries high voltage, so I is very small within the line. Then we use P = IR^2 within the line itself (since we want to see power consumed within the line), and since R is small and I is small, power dissipated in the line is small as well.

The idea about power lines drooping is true, but you have to realize that thermal expansion doesn't occur on that large of a scale (the expansion coefficient for steel is 12 * 10^-6 (1/K)) which isn't that large. And you also have to remember that power lines dissipate heat through air. But what you say is true; things like that do happen when power lines carry too much power.

13. Aug 12, 2006

### rbj

it depends on the physical size of the resistor (it's wattage rating) in comparison to the power you are forcing it to dissapate (which is a function of the current). and it depends on the composition of the resistive element (carbon does different things than metal, when heated up). the resistance, under adverse circumstances, is not always constant. saying that it is "always" is one of those sweeping statements that is pretty damn hard to prove (one valid counter-example will disprove it and anyone who has done volt-amp characteristics of a physical part and pushed it to the limit knows of such a counter example.) there is also some well known physics of the properties of metal and other materials that shows that the resistivity is a function of temperature. see http://en.wikipedia.org/wiki/Resistivity

for power mains? that's a silly statement.

yeah, but cities are big. i dunno if it is still the case, but when that big blackout of NYC happened in the 70s or early 80s, i remember reading that the whole of the city was supported by 5 main trunk lines. it was hot and muggy, lot'sa air conditioning, thunderstorms were around, lightning took out 2 of those lines, and the the current load was transferred to the 3 remaining, and they heated up. when one drooped low enough to contact a tree or some structure, that was it. blackout.

i'm not saying that they heated up to anything like melting or softening of the metal, but i'll bet they got warm to the touch. over ambient temperature.

but, even in normal circumstances, those lines are carrying nearly 1000 amperes (less than 1/2 MV and hundreds of MW, you do the math) and even though the conductors are pretty large in cross-section, in AC lines (nearly all) the great majority of the current is carried very near to the surface of the conductor (called "skin effect").

"small" is a relative term. i read that transformers are about 99% efficient (1% loss), yet the total distribution loss in the power grid and delivery to the consumer is about 7%. where does that other 4% go (assuming 3 levels of distribution, each coupled with a transformer)?

large enough to cause drooping of several meters that cause contact and shorting in these adverse situations. you cannot claim that both the temperature change is "small" (in the adverse circumstances) and that the coefficient of expansion is "small". there would be nothing left to explain the drooping which hasn't been small in these adverse circumstances.

of course they do, just like the resistive element on my electric stove top.

14. Aug 12, 2006

### pivoxa15

I understand that tramitting at higher voltage and lower current is more effcient but to me, high voltage implies higher current (since resistance is constant). Is this not correct?

So how do you lower current when the voltage is higher? It takes work to do that? Which means power waster?

15. Aug 12, 2006

### rbj

put a transformer in between the higher voltage and the resistance (impedance) that was meant to be connected to a lower voltage.

transformers can change the AC voltage and current in such a way that their product (which is the power) remains nearly constant. and they're efficient (like 99%).

16. Aug 13, 2006

### pivoxa15

I am concerned about trasmission lines and the current and voltage in them when no load are present, so during long distance transmissions. You want high voltage and low current in order to reduce waste. My original quetions still stand.

17. Aug 13, 2006

### Staff: Mentor

Yes.
Resistance in transmission lines causes power to be lost. Resistance isn't itself power.
You asked many questions and all were answered. Could you be more specific about what you are having trouble with? Perhaps you mean this:
As already stated, no, that is not correct in this case. Simply put, you are applying an equation to a situation where it does not apply.

Power lines are a static case. You aren't upping the voltage on a system that previously had a lower voltage (which is what that equation you are using is for). The system was designed correctly in advance.
When the load is lower (ie, at night), the generators produce less energy (because less energy is put into them) and less amperage flows through the power lines.
It really looks to me like you are clinging to an incorrect understanding and not accepting - perhaps not even thinking about - the answers given.

Last edited: Aug 13, 2006
18. Aug 13, 2006

### Staff: Mentor

Consider a generator feeding power to a load, via a pair of transformers that increase the voltage for transmission, then decrease the voltage again near the load. It might help to realize that there are actually three electrical circuits in this system:

The first circuit connects the generator with the input (low voltage) terminals of the first transformer.

The second circuit connects the output (high voltage) terminals of the first transformer to the input (high voltage) terminals of the second transformer.

The third circuit connects the output (low voltage) terminals of the second transformer to the load.

As far as the overall flow of energy is concerned, they form a single system, but as far as the flow of electric current is concerned, they form three separate circuits that are coupled together by the electromagnetic fields in the transformers. The two coils in a transformer are electrically insulated from each other, so their currents remain distinct.

So you don't have a single current which somehow decreases (or increases) as it flows through a transformer, and you don't have to worry about "where did the extra electrons go to (or come from)?" Each circuit carries a steady current, or rather, an alternating current with a steady amplitude. The second circuit simply carries less current than the other two, at a correspondingly higher voltage.

19. Aug 14, 2006

### pivoxa15

I had always assumed that voltage and current were dependent (which seems to make sense when considering a basic circuit with battery and resistence. i.e. when electrons sense a higher voltage, they 'want' to move faster towards the lower potential so current is increased) in any textbook but as some of you say, it is indepedent, which would mean my question was in error. They can be controlled separately? Then it would seem very counterintuitive, although a detailed study of the three circuits explanation given by jtbell may prove not to be.

Last edited: Aug 14, 2006
20. Aug 14, 2006

### rbj

pivoxa15, there is a lot more than Ohm's Law

$$v(t) = i(t) R$$

and the fundamental electrical power expression

$$p(t) = v(t) i(t)$$

and the common variants

$$p(t) = \left( i(t) \right)^2 R$$

$$p(t) = \frac{\left( v(t) \right)^2}{R}$$

that you need to learn.

you need to learn about Alternating Current (AC), a little about inductance (coils), and then what you really need to learn about is mutual inductance, that is two coils wrapped around the same core. this is what transformers are. transformers do not create power (in fact, they are about 99% efficient, not 100%, so slightly more electrical power goes in than comes out). since power is the product of voltage and current, if we can trade one for the other, in such a way that this product is the same, then we are not creating (or destroying) energy from nothing. transformes do that for us. they have a way of trading voltage for current, or vise versa.

those little "wall-warts" that you plug into the wall and connect to your cell phone to charge it up are transformers. 120 volts goes in, but 4 volts come out. and it isn't simply "wasting" the 116 volts in between, it's trading it for increased current.

the power distribution system has circuits at different voltage levels so that in long hauls, they can decrease the current (reducing $i^2 R$ losses) by increasing the voltage by the same factor. but since 360,000 volts is maybe not what we want in our houses, there are transformers to "step down" that voltage level to a lower one (but, at the same time, increasing current). essentially, only at the load, is voltage and current proportional to each other.

you need to start thinking about the answers given more deeply and just let it set a while before coming back to say what's wrong with them. otherwise we might just give up.

21. Aug 14, 2006

### Integral

Staff Emeritus
You seem to be familiar with the basics of Ohms la and you have mentioned power losses in the transmission line as given by $i^2 R$ this is all fine. The piece you seem to be missing is correct consideration of the power consumed by the load. This, the load, can be seen as a constant since it is not affected by the parameters of the generation system. Now the power to the load is expressed as P = I E. Can you see that if the power consumption by the load is fixed that a larger E will result in a smaller I? This is where the reduction in current with increased voltage comes in. As far as the generator is concerned, the transmission lines are a part of the load. The generator knows nothing of the coupling to the transmission lines or the type of work being done by the load it only knows the load. Since the voltage is fixed, the generator produces the amount of current necessary to drive the load.

22. Aug 14, 2006

### Staff: Mentor

In particular, in an AC circuit, you have to take into account the effects of the inductance of any coils in the circuit. By combining the effects of resistance and inductance, you can come up with a quantity called the reactance X of the circuit, which we use instead of resistance in the AC version of Ohm's Law.

I haven't taught this stuff in many years, and I'm not at the office where my textbooks are, so I'd better not try to go any further. Most first-year general physics textbooks (especially the calculus-based ones) have a chapter on AC circuits that discusses this.

23. Aug 14, 2006

### Staff: Mentor

Right - but that doesn't explain what a transformer does. On one side of a transformer you have a high voltage and low amperage and on the other, a low voltage and high amperage. The reason is simple: the amount of power (Integral's equation) is constant.

24. Aug 14, 2006

### pivoxa15

So it is the power that is demanded by the load. It sounds plausible but for me, it would seem more intuitive if it was the amount of current that is the main source needed by the load. So the voltage changes to accomodate the current needed by the load.

But this is not so as all of you suggest.

Last edited: Aug 14, 2006
25. Aug 14, 2006

### pivoxa15

I understand that transformers allow the transfer of electricity be practical so that we get the right amount while also keep transmitting electricity efficient. I am more concerned about how voltage and current can be independent. If we treat the transmission line as a conducting material then Ohm's Law applies. We have a linear relationship V=RI where R is constant. So it is not possible to increase V and reduce I. But the latter is possible and we do it to reduce thermal dissiaption so the material in the transmission line is not a conductor and made out of something so that R increases when V increase so that I can decrease. Which is what we want. This is one way to settle the issue. Is it correct?

Last edited: Aug 14, 2006