After electricity is generated in a power station, it has to be delivered to the consumer via power cables. To avoid energy loss, the voltage is kept very high, and the current is kept very low. Now, if Ohm’s law states that
V = I * R
and the desirable resistance for a power cable is very low, then how does it all add up?
I consulted my physics textbook for answers (Physics by Patrick Fullick), and it stated that
Provided the temperature and other physical factors remain constant, the current through a wire is proportional to the p.d. across its ends.
Now, I know that an alternating current is used to avoid energy loss as heat, so could another physical factor be involved? It stumped my old science teacher…
I’m not exactly sure what you’re asking about, but energy losses can be examined by looking at the power consumption of the wires. The equation is P = I[sup]2[/sup]R, and since R is inherently small we only care about current, I. With large voltages, according to Ohm’s law the current will be small, and hence the power loss is minimized by large voltages.
Alternating current is used because you can step the voltage up or down with a simple transformer, which is just a couple of coils of wire. The voltage goes up or down depending on the ratio of the turns of wire in each coil. Actual transformers get slightly more complicated because they often use an iron core and may be filled with oil, but they are still very simple devices.
In most electrical systems, the system attempts to be a voltage source, which means that it tries to keep the voltage constant throughout the system while varying the current. The potential difference across the wire is rarely the important factor, because the impedence of the wire will still be much lower than that of the load, so the overall current is determined more by the load than the wire. Still, it is true that the current through the wire is proportional to the voltage, through the old V=IR equation. The important thing here is that V is the voltage drop across the wire, not the voltage of the system. In say a 120 volt system, when it enters your house, you wouldn’t expect the voltage across the wires to drop more than a couple of volts between the service entrance and the outlet in your living room.
The wires are in series with the load. The current through all will be the same, and the voltage will be divided up between the wires and the load. Google Kirchoff’s Laws and Voltage Divider for more details.
If you have a certain power (wattage), then you can deliver it as a high voltage (high and low current or a low voltage and high current. You put your finger on it when you said “to avoid energy loss” – energy is the conserved quantity here, and power is just the rate of energy flow (watts = Joules/sec).
If you’re wondering why a voltage of 250 kV doesn’t automatically create a current of 250 kA on a 1 ohm stretch of line, the cartoon answer is "the generator isn’t producing enough energy to pump that many amps at that many volts. 250kV times 250 kA would take 62.5 Gigawatts (Standard transmission lines operate at up to 765 kV, which would carry 585 GW, in principle – but as you an imagine, there’s no way a transmission line could handle even a tiny fraction of that current or power)
The technically correct answer would be a lot more complicated, but you can think of it in these terms: if voltage and current had to be locked together, they’d simply be unable to reach that voltage in the first place. Worse, there would be exactly one allowable voltage/current for any power level over a given wire. You can see how this would make modern electronics impossible, and completely change the character of common phenomena like “carpet sparks” (when you touch a doorknob, the contact resistance can be an ohm or less – Western civilization would collapse overnight as the hypothetical kA of a static discharge killed all of us. Polyester would be a weapon of mass destruction.
I’m sure you’ve been exposed to the "water’ analogy for electricity. If voltage (aka potential difference or elecrtotmotive force) is the pressure of the water, then current is the volume of water moving under that pressure and conductance (the inverse of resistance) is analogous to the crossection of the pipe. Think of a low current transmission line as a pipe with water moving slowly through it. (This analogy has definite limits, but it is often very useful)
The objective in power transmission is to transmit the maximum amount of
POWER =IIR to the enduser or load.
with a minimum amout of loss in the transmission cables, i.e. power=iir.
Now what is to understand?
PS The transmission line is also three phase in order to minimize capital cost of the line and be consistent with system standards.
No. Three-phase transmission is employed because it is more efficient than single-phase, but since you have to run three wires instead of two, the initial cost of the line is actually higher.
iamalemur: I read your OP three times, and I’m still not sure what you’re asking.
As others have pointed out, it’s really quite simple.
The power company delivers you electrical power §. Because P = IV, it can deliver you high voltage at a low current, low voltage at a high current, or medium voltage at a medium current. Theoretically it doesn’t matter. But practically and economically speaking, it is best to deliver high voltage at a low current. Why? Because implementing isolation and insulation components (which is required for high voltage) is a lot cheaper than using gobs of copper & aluminum (which is required for lots of current).
So the power company delivers high voltage at relatively low current. But there are a couple problems with this:
The generator doesn’t produce high voltage. How can the power company efficiently boost the generator’s voltage before it goes out on the wires?
All of your appliances require low voltage. How can you efficiently reduce the voltage on the wires wires before it enters your house?
Good questions. It can’t be done with DC. (Though switching techniques are getting better. But I won’t get into that.) But it can be done with AC with the use of transformers. (Transformers only work with AC.) So that’s the primary reason our electrical distribution systems are based on AC – because with AC we can use transformers to efficiently increase or reduce the voltage. Another reason is motors. But I’ll stop right there.
Now back to the wires.
As mentioned above, the power company produces high voltage at a relatively low current. The wires are sized so there’s an acceptable amount of voltage drop on the wires. If the wires are too small (in cross section), there will be too much voltage drop along the wires. If the wires are too big, they will be too heavy, too big, and too expensive. So the engineers specify a “happy medium size.” The voltage drop along the wires is governed by Ohms law (V = IR), where I is the current through the wire and R is the resistance of the wire. The power dissipated by the wire is P = I[sup]2[/sup]R.
We’ve got our equation, V=IR, and let’s say we put our numbers in. Voltage will be very high. Current will be very low. Resistance will also be very low.
You see, there are two voltages that come into play, the voltage ovet the load U[sub]load[/sub] and the voltage over the transmission line U[sub]trans.[/sub]. In order to minimize the transmission losses one tries to minimize U[sub]trans.[/sub], by increasing U[sub]load[/sub].
The purpose of the transmission line is to maximize the power-out, i.e. P[sub]load[/sub]=IU[sub]load[/sub], and minimize the losses, P[sub]trans.[/sub]=IU[sub]load[/sub].
(The current is the same in line and load - where else would the electrons go?)
By increasing U[sub]load[/sub] for a given P[sub]load[/sub] we can decrease I, which in turn will decrease U[sub]load[/sub]. It’s a win-win situation, really. (Untill you run into other high-voltage problems… But that’s outside the scope of this post.)
(OK, on re-reading this I realize that it’s not very clear - but I did at least draw a colourfull picture! No doubt one of our regulat science popularisers will describe it much more succinctly shortly.)
The voltage is high. I know this is going to be confusing (and appears to contradict what I and others have said), but it’s incorrect to say “the current is low.”
For discussion purposes the power company can be modeled as a constant voltage source. This means that the current draw is dynamic; if the load requires a lot of power (in other words, if the load has low resistance), then the power company will deliver lots of current. If the load requires a small amount of power(in other words, if the load has high resistance), then the power company will deliver small amount of current. All the while, the voltage stays relatively constant.
This is a very simplified example, and is only meant to convey the concept of what’s going on:
Let’s say the output voltage at a very small power company is 240 V. Let’s say the customer load varies from 100 Watts to 1,000,000 Watts during a typical year. This means the current will vary from 0.4 A to 4167 A. This sucks, because it will require some big, heavy, and expensive transmission wires to handle 4167 A. Then an engineer at the power company has a brilliant idea: he boosts the generator’s voltage to 11,000 V using a transformer. The load still varies from 100 Watts to 1,000,000 Watts. But now the current will vary from 0.009 A to 91 A. Much better.
Yes, you’re missing something. Electronerd gave you the answer in the very first reply to this thread.
But to expand on it a bit, there’s more than just one equation. Besides:
V = I x R,
there’s also:
P = I x V.
That is, power equals current times voltage.
You can substitute one equation into the other to give:
P = I[sup]2[/sup] x R, and
P = V[sup]2[/sup] / R.
The trick is in knowing which equation to use where.
Let’s try an example. We’ve got a 1000 W load, and a 1 [symbol]W[/symbol] transmision line. We have the choice of supplying the load at 100 V, or 1000 V.
First, at 100 V:
P = I x V
1000 = I x 100
I = 10 A.
The current in the load, and the current in the transmission line is 10 A. What’s the power lost in the transmission line?
P = I[sup]2[/sup] x R
P = 10[sup]2[/sup] x 1 = 100 W.
So 100 W gets lost in transmission. The generator will have to produce 1000 W for the load, plus 100 W for the transmission losses, for a total of 1100 W.
Next, at 1000 V:
P = I x V
1000 = I x 1000
I = 1 A.
The current in the load, and the current in the transmission line is 1 A. What’s the power lost in the transmission line?
P = I[sup]2[/sup] x R
P = 1[sup]2[/sup] x 1 = 1 W.
Now only 1 W gets lost in transmission. The generator will have to produce 1000 W for the load, plus 1 W for the transmission losses, for a total of 1001 W.
By increasing the transmission voltage by a factor of ten, we’ve reduced the transmission losses by a factor of a hundred, from 100 W lost to only 1 W lost.
Yes you are. Your “small number * small number” computes the voltage drop on the wires of the transmission line itself. The total voltage from the generator equals the voltage drop on the wire from one terminal of the generator to one terminal of the load plus the voltage across the load plus the voltage drop on the wire from the other terminal of the load back to the other terminal on the generator. You want the voltage drops on the two wires to be as small as possible leaving the maximum possible voltage for the load.
Your equation should be “small number*small number = small number.”