Why is it preferable transmit A.C. at high voltage?

Question

I read in the books that when A.C. is transmitted at high voltage and low current, the loss in form of heat is minimized. The books say that the H = I^2*R*T and hence, the power loss is minimized, if the current I is minimized. What I fail to understand is --- Why not use the formula H = (V^2/R)*T and say that the loss is more when V is more? (exactly the opposite conclusion!)

More precisely, When a step-up transformer increases V and reduces I, what happens to the formula V=I*R ?

What I mean is this : When we increase the number of windings in the "secondary" in a transformer, the voltage increases but current decreases, though the resistance remains the same. Why is the formula V=I*R violated?

Answer 1

Heating is really power dissipated in the resistor. Power=V*I. In a simple battery and resistor circuit, where V=I*R, if V doubles then I doubles too and the power (and heating) goes up. But if you want the same power at a different voltage, it is different. A 120 watt, 120 volt appliance draws 1 amp, so it has a resistance of 120 Ohms. A 120 watt, 240 volt appliance had a resistance of 240 ohms and draws half an amp. The power is kept the same at increased voltage by increasing the resistance of the appliance which decreases the current. So even though I^2*R is the same as V^2/R, you cannot change V without also changing R. But that is a simple EMF and load example, no transformer and no transmission resistance in series with the appliance.

In a transformer, V doubles while I is cut in half, without changing the resistance at all. Very different from what V=I*R says. In fact, as you can see from any schematic of a transformer, you are really dealing with two different circuits. One circuit goes from the EMF to the transformer’s primary coil and back to the EMF. The other circuit goes from the secondary coil to the load resistance and back to the secondary. The secondary is acting like a separate EMF in the second circuit. So when you substituted V/R for I to change I^2*R into V^2/R and said that higher V would also mean higher heating, your higher V came from a different circuit that had a different resistance.

Now, to handle transformers requires generalizing Ohm's law to AC circuits. In that case resistance becomes impedance and is a complex number. And complex is right!. AC theory is much more complicated than DC theory.

Answer 2

The reason is not a simple resistance problem. The power line has a reactance with the AC and can be made to conduct it very efficiently, like a tuned circuit. One thing to consider is that the thickness of the wire limits the amount of current it can carry, but it does not limit the voltage. So by raising the voltage, more power can be delivered at a given current without paying for thicker, heavier copper wire. The advantages are great enough to offset the cost of the transformers and transmission towers.

Since V=IR, then V^2 / R exactly equals I^2 R. But raising voltage has a smaller part because of that division by R. This implies that raising the voltage makes the heat change in a linear manner, but raising the current makes it increase as the square.

Answer 3

What everyone who has answered thus far doesn't know is the answer.

The reason is much much simpler than they are making it out to be. It has to do with power loss to resistance.

Power: P=R*I^2 or P=V*R

You can see that power lost in the resister (the wiring) goes up much quicker (as the square) if electricity is sent at high currents. This is why the voltage is stepped up and current is stepped down before transmission.

If you don't believe me, look it up.

Answer 4

You are missing a very important point. It isn't the square of the voltage divided by the resistance but the square root of E/R. This leads to the exact same conclusion as I^2*R. If resistance is constant, current is inversely proportional to voltage. Power is at the inverse square of the current.

Answer 5

You are forgetting about the frequency. We need to maintian 60 hertz or we start breaking stuff. At 55 hertz impetence increases, current increases and stuff melts.
The higher the voltage, the better the chance of maintaining 60 hertz.