The V in the equation is not the line voltage, it is the line voltage
*loss*. That's equal to V(at generator) - V(at load). And that quantity goes down when you step up to high-voltage, low-current transmission.John
-- Because for the same line resistance, if V increases, I will go down. Say the power company was sending 120V through one mile of of #0 AWG into a 1000 watt load. Then the circuit would look like this: +-------+ | 120AC|-----[0.519R]---
^^^ Oops... two
-- John Fields Professional Circuit Designer
Power lost in the transmission lines is the current squared times the line resistance (called I squared R). Drop the current (by boosting the voltage) and you can eliminate much of the power lost in the transmission lines. Up the voltage so you can deliver the same power, at less current. 500 kV, or there abouts, is used for long-distance transmission.
Some numbers: 120 volts at 100 amperes = 12 kW and 500,000 volts at 24 mA =
12 kW. So, suppose it is your choice: send energy to your customers at 100 amperes or 24 mA. It is an easy choice as you can also use smaller gage wires (beside saving the heat loss in your transmission lines). By the way, this is the basic point that Edison missed and he went so far as to publicly electrocute animals with ac to "prove" that ac was too dangerous. Tesla was way ahead of Edison on this particular point.Some loss numbers: Assuming 100 ohms of transmission line resistance with
100 amperes = 1 MW! With 24 ma = 0.06 W!I have a simple question. The way the power company delivers power from the power plant to the homes is using stepped-up transformers and high voltage lines. The reason is that they want to minimize the power loss which is P=I * I * R. But P can also be equal to P=V*I. If you increase the voltage, why don't you have the same power loss which equals to P=I*I*R=V*I?
Perhaps the OP is confused about the P = V*I part.
Power is developed as V(ckt) * I(ckt). On a power line, the V(ckt) is quite small (actually derived from V = I*R). So on a power line, think of the voltage **across the cable** as being a result of the current, rather than the other way around. Certainly for any power line, I^2 R will equal V * I *for the power line*
Looked at from that perspective, then by raising the transmission voltage, we lower the current (P remains roughly constant) so the voltage drop across the line decreases, thus reducing V * I (by a square function) at the same proportion as I^2 R.
Cheers
PeteS
The "V" in your equation is the voltage lost in the lines, not the line voltage. As you increase transmission voltage, both V and I will go down. In the limiting case, both V and I go to zero and there are no losses.
-Bill
you do, but the significant voltage in this case (for calculating the loss) is the difference between the voltage at each end of the transmission line not the difference between the line and ground.
-- Bye. Jasen
Yes, it's the voltage drop from one end of the wire to the other, which will be constant, based on the resistivity of the wire and its length. By using a higher primary voltage, that portion dropped by the wires themselves is a smaller proportion of the whole loss set, so they save money.
Cheers! Rich
You will? ;-P ;-D ;-P
-- Cheers! Rich ------ "Sure eating yoghurt will improve your sex life. People know that if you\'ll eat that stuff, you\'ll eat anything."
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