# How power lines use high voltages with a low current?

I've read that power lines use high voltages and low currents to reduce power loss due to resistance. Looking at the formula for power -

P = VI

So to increase P, you increase V rather than I for efficiency, which makes sense. The formula to calculate voltage though is -

V = IR

This seems like a contradiction. To increase the power in power lines they increase the voltage, but to increase the voltage you either have to increase the current or the resistance.

We have established power lines use a lower current, so does that mean to increase the voltage they increase the resistance, that seems counter productive and the reason we don't increase the current to increase power...

How do power lines use high voltages and a low current when a high current (or high resistance) is needed for a high voltage?

• physics.stackexchange.com/q/79573 the second answer down might help
– user81619
Commented Jul 24, 2015 at 22:38
• Supply voltage and voltage drop along the circuit are not the same quantity. See also physics.stackexchange.com/q/144315 Commented Jul 24, 2015 at 22:59
• @AcidJazz Thanks, I'm sure it explains it well but it's a bit over my head to be honest. This sort of thing isn't really my area I was just curios and couldn't find a clear explanation online.
– user86680
Commented Jul 24, 2015 at 22:59
• Copy the answer , reference the link, and post it as a new question, asking for explanation of the terms you don't follow, seriously, that's what I do. Lots of people are interested in the subject so chances are good you will get an answer that suits you. Regards
– user81619
Commented Jul 24, 2015 at 23:19

# The short answer

This is not 100% true since it assumes DC transmission, but it gives the simplest form of the idea: even if the transmission lines are themselves at high voltages, that doesn't directly mean anything, since voltages are not defined relative to anything special (they're defined relative to some other line which is in parallel with your transmission line). So for a schematic diagram, consider this:

Some current $I$ flows through the top wire, it causes $V_1 = V_0 - I R$. Now there are three voltages we're talking about, and they're all very different: $V_0$ on the left, where the power is coming from, and $V_1$ on the right, where the power is being used, and $I R$, which is the loss through the lines. (We could also use two resistors of resistance $R/2$, one on each side: it doesn't change a thing.)

Now the power lost via the resistor is $P_L = I (I R) = I^2 R$, while the power used at the distant terminal is $P_U = I V_1,$ and they trivially sum up to that total power $P_T = I V_0$. If we're minimizing $P_L$ for a given $P_T,$ then we solve for $I = P_T / V_0$ and find $P_L = R P_T^2 / V_0^2$, so in the important case, we should raise the voltage to lower the losses.

# The true answer

Okay, that's cheating and if you think too much about DC transmission you're going to struggle with it: "after all, the current that's flowing is only flowing because of some resistance placed across $V_1$ and if you don't configure things just right with $R$ then you have the wrong voltage and things explode, so do we even really have that tradeoff? We'd need to create a voltage-reduction circuit and in DC that usually means some resistors in series adding to $R$," and so forth. It gets across the most important part of the idea which is where the resistor is, but it lacks true force because it's not AC current. For AC current, you need a transmission line. For all of this, you need multi-variable calculus and partial derivatives. Sorry if that goes over your head.

The simplest generic transmission line looks like this: divide the length $L$ of the line into segments of size $\delta x$, then model them each as an L-R-C circuit:

A transmission system usually contains two conductors near each other, with some capacitance-per-unit-length $c$ and inductance-per-unit-length $\ell$ as well as some resistance-per-unit-length $\rho.$

A static analysis of this circuit gives two equations:\begin{align}V(x + \delta x) = &V(x) - \ell ~ \delta x~ \frac{\partial~I}{\partial~t} - \rho~\delta x~I(x + \delta x, t) \\ I(x + \delta x) =& I(x) - c ~\delta x~\frac{\partial}{\partial t} (V(x) - \ell~\delta x~I(x))\end{align} If we choose $\delta x$ small enough then terms like $(\delta x)^2$ get arbitrarily small while $[V(x + \delta x) - V(x)] / \delta x \mapsto \frac{\partial V}{\partial x}$. The governing equations for this are therefore:\begin{align}{\partial V \over \partial x} = & - \ell ~ \frac{\partial~I}{\partial~t} - \rho~I(x, t) \\ {\partial I\over \partial x} =& - c ~\frac{\partial V}{\partial t} \end{align}Combining these two leads to a wave equation:$${\partial^2 V\over \partial x^2}= \ell~c~{\partial^2 V \over \partial t^2} + \rho ~c ~{\partial V \over \partial t}.$$

Now we have to drive this system with the input at $x = 0$, $V_0 \cos(\omega t)$, then in general at the output you will see some output $V_1 \cos(\omega t + \phi)$ for some phase difference $\phi$ and amplitude difference $V_1$.

The loss of voltage from $V_0$ to $V_1$ comes from $\rho$ and is a transmission loss. This is different from the value $V_1$ which can certainly be used to extract power. Hook up a resistor on the other end and measure the power output through that resistor: while holding this constant, you discover that the proper way to lose less energy is to use higher $V_0.$ I'm pretty sure that this applies even if we add a transformer to "step down" the output to a constant voltage.

• Hi @Chris thanks for the answer. I should have put in the question that this sort of thing isn't my area and have little knowledge in the subject, so your answer is over my head to be honest. I appreciate the time you took to answer and I'm sure it will be useful for people in the future with more knowledge on the subject.
– user86680
Commented Jul 25, 2015 at 18:19
• @RJSmith92 the V in the question is the voltage difference along the cable = "wasted" voltage. By having a high voltage/low current cable then with a given R you have a low V difference = low wasted power. Commented Jul 25, 2015 at 21:38
• @RJSmith92 I've added some diagrams which may help. Commented Jul 27, 2015 at 7:10

There are two different $V$'s here. Suppose the power station outputs at 10,000 V. By the time the wire makes it to your house, this may have dropped to, say, 9,000 V.

The $V$ in the first equation refers to the voltage difference you can use, which is 9,000 V (between the wire you receive and ground). The $V$ in the second equation refers to how much voltage was lost on the way to your house, which is 1,000 V. They're totally different things.

In general, be careful about plugging equations into each other just because they have the same letters. You can do that in math, since $x$ will mean only one thing in a math problem, but a $V$ (or an $F$, or an $a$, etc.) in a physics equation could mean tons of things.

Voltage is a measure of the electric potential difference across two points in a circuit. It may be considered the work done to transport an electric charge. Power lines are made of thick easily conductive material in order to minimize resistance and power loss to heat. But resistance within power lines is fixed, and power is delivered through the line according to this formula:

P = ∆V * Q/t = ∆V * I

P is power; V is voltage; Q is electric charge; t is time; I is current (charge per unit time)

Ohm's law describes how power is lost: ∆V = I * R, where R is resistance. If you combine Ohm's law with the power equation, you find P = I^2 * R, and P = ∆V^2 / R.

Because R is fixed, you can deliver a given amount of power using either greater current or more voltage. But because high current results in more power lost to resistance in the power lines, transformers are used in high voltage power lines to step down the voltage. Between the transformers, high voltage in the lines delivers electric power with less loss than if high current flowed through the lines.

• Hi @Ernie, thanks for the answer. You say "Because R is fixed, you can deliver either greater current or more voltage for a given amount of power", so for example using arbitrary values, we have a fixed residence of 2.5 and want to deliver 1000W. What choice do we have for the voltage and current, doesn't it have to be 50V and 20I? P = VI - 1000 = 50 x 20 and then V = IR - 50 = 20 x 2.5. If power companies want to use a higher voltages on power lines to reduce waste, do they have to increase the amount of power they send (i.e not fixed at 1000W)?
– user86680
Commented Jul 25, 2015 at 15:24
• also, if the resistance is fixed, isn't it impossible to increase the voltage without also increasing the current? Wouldn't this mean that the only way to increase the voltage on the power lines and have a low current be to increase the resistance, according to V = IR? Are power cables meant to be high resistance?
– user86680
Commented Jul 25, 2015 at 18:31
• @RJSmith92: You deliver power against resistance with either great current or great voltage. In your example, the ratio of V to I is 2.5, but is this in the line, or at the transformer? Transformers allow you to step down voltage, while retaining high voltage in the line. In your example, you have 50V and 20I in the line. The transformer's primary coil has high impedance (resistance), so you step down the voltage there. To deliver more watts through the line without increasing voltage, use a thicker wire. See this & scroll to end: bsharp.org/physics/transmission Commented Jul 25, 2015 at 20:20
• @RJSmith92: Power cables are meant to be low resistance. At the transformers, change in resistance changes the ratio of voltage to current. Commented Jul 25, 2015 at 20:29
• @RJSmith92: The line resistance always should be as small as possible. Then there's load resistance, which varies as customers use more or less electric power. This may be the resistance you are asking about. When the load is great during peak periods of the day, more power needs to be delivered through the lines. Commented Jul 26, 2015 at 1:50

Let's assume that the power company is supplying a neighborhood with 1000 A of current at 120 V. Since P = IV, the neighborhood is receiving 120 kW of power, which is the "load" seen by the power company. To maximize efficiency, the power company wants to minimize the losses involved with transmitting power to the neighborhood, which occur due to resistance heating of the transmission lines. For the transmission lines alone, this loss corresponds to the formula P = I^2(R), meaning that the losses are proportional to the amount of current squared. Thus, the power company wants to minimize the transmitted current in order to minimize the transmission losses.

When current goes through a "step-up" transformer, the voltage is increased and the amperage is decreased, due to conservation of energy considerations. Taking advantage of this, the power company generates 1000 A of current at 120 V (it's actually different than this, but assume this for the sake of the argument), runs this current through a step-up transformer to convert the current to 120,000 V at 1 A, and sends the power to the neighborhood. At the neighborhood, a step-down transformer converts the power back to 1000 A at 120 V (assuming no loss) and each individual house uses a portion of that power. Due to this power distribution method, very low electrical transmission losses are incurred because a very low current was transmitted to the neighborhood.

Wire resistance causes losses in power supply transmission. If you keep resistance constant, losses are linearly proportional to the square of the current. So if you double the voltage for the same power you have half the current and power dissipation is effectively half for the same power. Another reason is weight. In order to transmit more current and keep losses in control one would need a bigger wire with a larger cross-sectional area.

• Linearly proportional to the square of current? That’s gone over my head; how is it linear if theres a quadratic current term? Commented Dec 12, 2019 at 17:20
• @UbaidHassan read as "directly proportional"
– josh
Commented Apr 20, 2021 at 7:31

The thing you are missing is that there are two parts of the circuit – the transportation wires and the load. Since they are separated by a transformer, Ohm's Law does not apply between them. Imagine this pseudo-circuit:

Here, $$R_a$$ is the resistance of the wire, $$R_b$$ is the load we want to power. We will look at voltage and current at the points 1, 2, 3 and 4.

First a hand-wavy explanation, then we can do some math: We need to transport a fixed amount of energy to the transformer, i.e. the amount of energy the load is consuming. Thus $$P_T=I_2V_2$$ is constant and the more voltage we bring to the transformer, the less current we need. The power dissipation across the wire is $$P_a=I_2(V_2-V_1)=I_2^2R_a$$, where we used Ohm's Law with $$(V_2-V_1)=I_2R_a$$. Thus higher voltage means less current means less power dissipation.

## Complete calculation

We set $$V_4=0$$ and assume our load voltage $$V_3$$ to be fixed (e.g. $$V_3 = 230V$$). As the power company, we can choose $$V_1$$ to our liking.

We are interested in the power dissipation of the wire, which is $$P_a=I_2(V_2-V_1)$$. Due to Ohm's Law, $$I_2=\frac{(V_1-V_2)}{R_a}$$, thus $$P_a=\frac{(V_1-V_2)^2}{R_a}$$. $$V_1$$ and $$R_a$$ are assumed to be known, so we are looking for $$V_2$$:

The transformer must output the same amount of energy as was put in, thus: $$I_2V_2=I_3V_3$$ Due to Ohm's Law, $$I_3=\frac{V_3}{R_b}$$ and again $$I_2=\frac{(V_1-V_2)}{R_a}$$, thus: $$\frac{(V_1-V_2)}{R_a}V_2=\frac{V_3^2}{R_b}$$ Rearranging this equation gives $$V_2^2 - V_1V_2+\frac{R_a}{R_b}V_3^2=0$$ We can solve this quadratic equation to get $$V_2=\frac{V_1}{2}+\sqrt{\frac{V_1^2}{4}-\frac{R_a}{R_b}V_3^2}$$ With this formula for $$V_2$$, we can plot $$P_a$$ for various $$V_1$$ (let's fix $$R_a=R_b=10$$):

This plot clearly shows how increasing $$V_1$$ reduces the transport loss in the wire.