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- Thread starter Rockazella
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- #2

russ_watters

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Bottom line is there are two ways to get a certain number of watts from a power line (which is the goal - delivering power). Either you can deliver a low voltage and high amperage or a high amperage and low voltage.

Did your teacher explain WHY high voltage is advantageous? Line losses (that buzzing sound is power escaping into the air) are a function of amperage alone. So to reduce line losses, you reduce the amperage (and increase the voltage).

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We're talking about source voltage here. E=IR (I=R/E) is the RESULT of that voltage when applied to a load of a given R. Cause vs effect. The two equations are describing two separate things.

Whats the difference between source voltage and voltage obtained by a transformer? I mean if you look at a generator (common source) it just uses coils of wire and magnets, a transformer also uses coils and a magnetit field. If current is the flow of electron, why would electons care where the voltage is coming from? I thought voltage was best described as a "push", and a greater push would always result in a larger flow.

Where am I mixed up?

- #4

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The answer is, use 2 separate circuits. The device linking them is called a transformer.Originally posted by Rockazella

How can you boost the voltage without also boosting the current?

You can construct a transformer so that it boosts voltage. But it will have a high reactance on the secondary side, so the current is lower. Theory says P2 <= P1.Isn't current just a result of a voltage (higher voltage always means higher current)?

- #5

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You can construct a transformer so that it boosts voltage. But it will have a high reactance on the secondary side, so the current is lower.

Arcnets,

would you mind defining reactance?

- #6

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http://whatis.techtarget.com/definition/0,,sid9_gci283998,00.html

- #7

BoulderHead

Let me give this the old ‘High School try’;

P=IE

Okay, this was probably meant to show you that the same amount of power can be delivered to a load even if the values of I and E are changed. For example, if the current was 10 Amps and E was 100 Volts then multiply together for a power of 1000 Watts. But, if you had 100 Amps and 10 Volts servicing your load what would your power be? Well, it would still be 1000 Watts. So you can play with these two values while still delivering the same amount of power.

Now, in the equation E=IR a problem arises. The problem is that the resistance of the transmission line will produce a voltage drop across the line itself whenever a current passes through it. You want the voltage across your load where it can do you some good, not across the transmission line. This has a huge impact across transmission lines that are many miles long, but because of the relation between power, current, and voltage, it is possible to reduce the line loss by increasing the voltage and decreasing the current. So, you step up the voltage at the sending end (imagine a half-million volts), then step it back down to less dangerous levels prior to sending it inside someone’s home.

The same power can be delivered without wasting energy heating up the transmission lines which would have been caused by using lower voltages and higher current levels.

Hope that helps a little.

- #8

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I had a thought, would it be possible to use the old 'water in a bucket' analogy (potential energy of water in bucket relates to voltage) to help with this concept? I've been trying to think of ways it could be set up, but havnt been able to come up with anything that would work. Can any of you think of a way, or is this just a foolish attempt?

- #9

BoulderHead

That’s probably not the best way to think about it, so let me try again and I hope I don’t screw you up too badly;

What is it that makes a wire get hot? (consider the heating elements of many electric stoves). Those babies get glowingly hot don’t they?

Well, is not the degree to which the element heats related to the amount of current passing through it? If you want it hotter you cause more current to pass through the element, and so forth. Now, there is just about no physical thing you will likely deal with in your life that doesn’t have some inherent value of resistance to it. If you were to manipulate an experiment such that an equal amount of current were to pass through an element with a low value of resistance and then through an element with a higher value of resistance, which one do you think would feel the warmest to the touch?

It would be the element with the higher value of resistance.

I spoke about ‘manipulating the experiment’ and the reason for that was because the goal was to maintain an equal amount of current passing through the two elements. What would have actually taken place in order to maintain that identical amount of current through the two unequal resistances is that you would have had to use a higher value of voltage across the higher resistance element than you would have for the lower valued element. Now imagine that the element represented a power line which carried electricity from the local utility company to your home. Would the power company want to waste their valuable energy heating up the transmission wires when they know that nobody in their right mind would be climbing up a pole and warming their coffee there, and even if someone did there would be no meter with which to assign a dollar value to the amount of power used? So, the goal then is to get the power into your house while minimizing the amount of power wasted across the transmission lines.

It might seem counter intuitive to you at this time to see the point of increasing the voltage when as demonstrated above this led to increased heating of the element but that is only because something has been ignored. The answer to this is found by taking a look at power. Up to now we have mostly been dealing with voltage, current, and resistance… now it’s time to think about power for a moment. In the power equation you sited; P=IE, I is the current and E is the voltage. According to Ohm’s Law the voltage is also given as the product of resistance across some element multiplied by the current passing through it, that is; IR=E (don’t worry for the moment about whether an E or a V gets used in these equations, just consider them interchangeable for the time being). Watch how IR can be substituted for E in the first equation;

P=IE, but because E=IR you also have;

P=I(IR)

P=I^2R (sorry about the symbols, it should read; P equals I squared R)

So it turns out that all these critters relate to one another and the resistance of the transmission line ends up playing a role in the form of P=I^2R. But understanding that P also equals IE means that there is a way to get around the resistance of the lines while delivering the same amount of power. What took place in the experiment above was that more power was required to deliver the same current through the element with the higher resistance. The thing that ends up wasting the money is the current (I) because it produces useless heat in the wires, but because of P=IE we have a way to reduce I while maintaining the value of P… and this is done by increasing E. The voltage isn’t creating any heat and so why not raise the value of it dramatically while simultaneously reducing the amount of current and still retaining the same value of P only without all the waste.

What is it that makes a wire get hot? (consider the heating elements of many electric stoves). Those babies get glowingly hot don’t they?

Well, is not the degree to which the element heats related to the amount of current passing through it? If you want it hotter you cause more current to pass through the element, and so forth. Now, there is just about no physical thing you will likely deal with in your life that doesn’t have some inherent value of resistance to it. If you were to manipulate an experiment such that an equal amount of current were to pass through an element with a low value of resistance and then through an element with a higher value of resistance, which one do you think would feel the warmest to the touch?

It would be the element with the higher value of resistance.

I spoke about ‘manipulating the experiment’ and the reason for that was because the goal was to maintain an equal amount of current passing through the two elements. What would have actually taken place in order to maintain that identical amount of current through the two unequal resistances is that you would have had to use a higher value of voltage across the higher resistance element than you would have for the lower valued element. Now imagine that the element represented a power line which carried electricity from the local utility company to your home. Would the power company want to waste their valuable energy heating up the transmission wires when they know that nobody in their right mind would be climbing up a pole and warming their coffee there, and even if someone did there would be no meter with which to assign a dollar value to the amount of power used? So, the goal then is to get the power into your house while minimizing the amount of power wasted across the transmission lines.

It might seem counter intuitive to you at this time to see the point of increasing the voltage when as demonstrated above this led to increased heating of the element but that is only because something has been ignored. The answer to this is found by taking a look at power. Up to now we have mostly been dealing with voltage, current, and resistance… now it’s time to think about power for a moment. In the power equation you sited; P=IE, I is the current and E is the voltage. According to Ohm’s Law the voltage is also given as the product of resistance across some element multiplied by the current passing through it, that is; IR=E (don’t worry for the moment about whether an E or a V gets used in these equations, just consider them interchangeable for the time being). Watch how IR can be substituted for E in the first equation;

P=IE, but because E=IR you also have;

P=I(IR)

P=I^2R (sorry about the symbols, it should read; P equals I squared R)

So it turns out that all these critters relate to one another and the resistance of the transmission line ends up playing a role in the form of P=I^2R. But understanding that P also equals IE means that there is a way to get around the resistance of the lines while delivering the same amount of power. What took place in the experiment above was that more power was required to deliver the same current through the element with the higher resistance. The thing that ends up wasting the money is the current (I) because it produces useless heat in the wires, but because of P=IE we have a way to reduce I while maintaining the value of P… and this is done by increasing E. The voltage isn’t creating any heat and so why not raise the value of it dramatically while simultaneously reducing the amount of current and still retaining the same value of P only without all the waste.

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- #10

Dave_3of5

V = I*R

so I = V/R

i.e. if voltage is high and resistance stays the same the current must be lower for that equation (the name of escapes me) to work.

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