Power = Voltage x Current

Question

I understand the equation that Power = VxI. Due to the Joule heating effect, engineers prefer to keep V high and I low to minimize the heating losses.

When I think of current, it is the number of electrons passing a given point per unit time. The effect of decreasing or increasing the current to change the power is understandable.

How does changing voltage affect the power physically? What physical phenomenon happens inside a circuit when the voltage is increased and at the same time resistance is increased to keep the current low. Thus producing the same power but with less losses. How do those fewer electrons per unit time carry more energy with them?

What physical attribute is increased by the increased voltage so that the power (energy per unit time) is increased? In what form does this energy exist?

Answer 1

Each charge that traverses the length of some conductor begins at one end with a certain amount of potential energy, and exits the other end with lower potential energy. The potential difference (voltage) between the two ends is the amount of potential energy, in electron-volts (eV), that each charge loses on its journey between the two points. Electrons, being negatively charged, have their highest potential energy at the point of lower potential.

For instance if one end, X, of some device is at 2V, and end Y is at 10V, then all electrons have 10eV - 2eV = 8eV more potential energy at X than those at Y. An electrons will travel through the device under the influence of the electric field caused by the applied 8V potential difference, beginning its journey at X, where it has 10eV, and emerging at Y with a potential energy of 2eV.

In other words, each electron will have lost 8eV of energy to the device it travelled through, and by conservation of energy that lost energy must reside somewhere else now, or have a different form.

If the device has purely resistance, then all that energy has become heat, due to interactions with stationary charges in atomic nuclei. These "collisions" impart any accumulated kinetic energy of moving electrons (accelerated by the field of the potential difference) to the stationary atoms, which we recognise as heat. This also prevents the electrons accelerating unimpeded, resulting in an average speed of all the electrons to be somewhat fixed and predictable, current.

Clearly then, the energy imparted by each charge to the device after it makes a complete journey through the device, is proportional the voltage across the device.

Also, by extension, the rate at which energy is imparted, called "power" will be proportional to the rate at which charges complete this journey, and that will depend on the average velocity of a charge during its journey.

As you alluded to, current is the amount of charge passing by some point each second, which (assuming a constant uniform charge density in the conductive path) will be proportional to the average speed of the charges. We have two ways of influencing that. We may either we change potential difference, or we change the path's resitance, both will result in a change in current.

It is supremely interesting to me, from a purely physical perspective, that by doubling the voltage across something, you are simultaneously doubling the potential energy lost by each charge on its journey through that thing, and the speed at which it travels (current). Both of these factors individually contribute to the rate of delivery of energy, called "power". This gives rise to the square term in the famous equation \$P = \frac{V^2}{R}\$.

However, it must be noted that the power I refer to above, is the power delivered to the current carrying element, and not the power delivered via this element to some remote recipient. This is where most confusion stems from.

Our task as energy providers is to minimise loss in the delivery system, and maximise power arriving at the destination. Remembering that increased current results in greater heating (power loss) in the cables carrying those charges, we aim to keep that current to a minimum. The most common way to do this is by increasing the potential difference across the entire circuit loop, which is very different from increasing the potential difference across sections of charge-carrying cables themselves. Consider these two scenarios, which deliver (almost) exactly the same power to a remote load Rd:

^{simulate this circuit – Schematic created using CircuitLab}

R1 and R2, which I have chosen arbitrarily to be 4Ω in total, represent the resistance in the cables delivering power. In the first (top) circuit, I have chosen to provide power from a 12V source. The voltage "lost" across each delivery cable is 1V, and that corresponding power (heating) in those cables is 0.5W.

In the second circuit, I have increased the source potential, but also increased Rd such that it still dissipates 5W of power. Importantly, the current around the loop is now much lower. The voltage across the cables has correspondingly fallen to 0.43V each, even though the source potential has risen. This is not so difficult to understand when you remember that Rd is much greater now.

Since power companies and their customers are concerned purely with power delivery, and not the "resistance" of their connected appliances, this approach makes sense; to design appliances that operate from greater voltages, and increase the voltage at the source, so that current in the delivery path is minimised, as will be power lost therein.

Answer 2

I'll give the mathematical answer first and then try to explain it with your electron analogy:

Higher voltage and lower currents are typically used to minimize conduction loss during power transmission. If a certain amount of power has to reach from point A to point B, it will require a certain length of conductor/wire. Assume that the resistance of this wire is R.

Now you have two choices:

1. Send a higher amount of current at a lower voltage

2. Send a lower amount of current at a higher voltage

Power dissipation in the wire will be iiR

Thus it makes sense to have lower current during transmission. This is done by using step-up transformers to increase the voltage. Since transformer cannot add power, the current will automatically come down.

For ex - assume that wire has a resistance of 1 ohm and you want to deliver 100 watts of continuous power at the other end.

Case 1: 10V and 10A: Power dissipation in wire = 10x10x1 = 100 watts

Case 2: 100V and 1A: Power dissipation in wire = 1x1x1 = 1 watt

Coming back to your electron analogy, think of it as electrons will dissipate some energy as they move through the conductor. More the electrons moving, more the dissipation, so you'd rather have less electrons moving to reduce this power wastage.

How does changing voltage effects the power physically?

It doesn't change the power physically. It does change the circuit requirements to handle the power. For ex - your circuit might not be able to handle a higher voltage that is being used during transmission. So now you need to add another component (step down transformer) to reduce the voltage to acceptable levels.

What physical phenomenon happens inside a circuit when the voltage is increased and at the same time resistance is increased to keep the current low.

I think you misunderstood something here. Voltage is increased but resistance is not increased to reduce the current. It happens on its own because power is not being created. Step-up transformers are typically used for this purpose.

How do those less electrons per unit time carry more energy with them.

Think of it as each electron carrying more energy when voltage is increased. For example - I can throw 10 bullets with my hand and it won't do any damage. I throw one bullet with a gun and it becomes deadly.

What physical attribute is increased by the increased voltage so that the power(energy per unit time) is increased?

Power is not increased. The relationship P = VxI still holds true. You increase one. The other decreases automatically.

Answer 3

I understand the equation that Power = VxI. Due to joules heating effect, engineers prefer to keep V high and I low to minimize the heating losses.

... when they are trying to deliver power to a remote load, they keep I low to minimise heating in the supply cable

When I think of current, it is the number of electrons passing a given point per unit time. Effect of decreasing or increasing the current to change the power is understandable.

reasonable intuition

How does changing voltage effects the power physically ? What physical phenomenon happens inside a circuit when the voltage is increased and at the same time resistance is increased to keep the current low. Thus producing the same power but with less losses. How do those less electrons per unit time carry more energy with them.

The most important thing you've said here is 'resistance is increased', which enables the voltage across the load to be increased when you want it to be.

Consider this 100 ohm resistor R1, with 1 amp flowing through it, dissipating 100 watts. If you connect two in series, R3 and R4, you now have 200 V drop, and 200 watts dissipation. But that's equivalent to a resistor of 200 ohms, R3+4. It's just that you can't see the internal connection.

The heat produced is increased not because there are more electrons flowing, but because each electron carries more energy, drops through more potential difference (voltage) as it flows from one end of the circuit to the other. In this case, each electron does 100 V of work dropping through each of resistor R3 and R4.

In the hydraulic analogy, which has its limitations, but is very good in this respect, each resistor is like a pipe that requires a pressure difference across the ends (voltage, or potential difference) for water (electrical charge) to flow through it (electrical current). It delivers more energy when the pressure is higher. Pressure (or voltage) is just a measure of the amount of energy needed to move the charge.

^{simulate this circuit – Schematic created using CircuitLab}

What physical attribute is increased by the increased voltage so that the power(energy per unit time) is increased ? In what form does this energy exists ?

The increased voltage requires an increase in resistance to be supported at the same current. The increase in resistance effectively means each charge carrier does some work in the first part of the resistor (R3), then does some more work in the next part (R4).

Answer 4

How does changing voltage effects the power physically ?

• Start with the power equation you wrote (\$P = V\times I\$)
• Then introduce ohm's law (\$V = I\times R\$)
• Then combine them: -

$$P = I^2\times R$$

If power remains constant and \$I\$ decreases, then recalculate resistance to accommodate that change: -

$$R = \dfrac{P}{I^2}$$

So, you understand power = V x I as you wrote: -

I understand the equation that Power = VxI

And, if you understand ohm's law, then it should be clear.

Answer 5

What physical attribute is increased by the increased voltage so that the power(energy per unit time) is increased ?

the energy of each electron.

In what form does this energy exists ?

One can understand it as a kinetic energy.

In reality, it is rarely a "pure" kinetic energy, but if you want a pure kinetic energy example, see the vacuum diode. The electron is accelerated in the electric field and the higher the voltage, the more energy the electron gets before it hits the anode.

In our quantum world the energy of an electron in a solid state matter cannot be really separated to "kinetic" and "potential" energy.

Answer 6

When I think of current, it is the number of electrons passing a given point per unit time. The effect of decreasing or increasing the current to change the power is understandable.

Fundamentally, a voltage difference is a number that tells you how much energy would be released if a positive charge were moved the higher potential point to the lower potential point (or how much would be required to move a charge the opposite direction).

So increasing the voltage but decreasing the current means that you move fewer charges through the circuit, but each one releases more energy. If you keep the product \$V\times I\$ constant it means you've balanced these two effects to keep the net power consumption equal.

How does changing voltage affect the power physically?

It depends on how you're applying that voltage.

Maybe you're running a generator faster, creating stronger magnetic fields, that induce a greater emf in the windings of the generator.

Maybe you're combining chemical cells in series, using multiple instances of a chemical reaction to separate ions of a dielectric, creating multiple potential differences in series.

Maybe you're reducing the resistance of a pass element in a linear regulator, allowing current to flow with less potential drop between some fixed higher supply and the circuit of interest.

Etc.