# Electrical energy and power

#### Prerequisites

- Quantifying electric current
- The electric potential
- Resistive electric flow -- Ohm's law
- Kinetic energy and the work-energy theorem

We know that when you hook something up to a battery it will run something for a while and then "run out of steam." Whatever energy was stored in the battery as chemical energy has been used up. What is using this energy and where does it go?

One of the ways of seeing where it goes is to realize that when we have a current pushing through a resistance, there is a viscous drag force on the charges tending to slow it down. Establishing a voltage drop across a resistance creates an electric force that pushes the charges through, so the electric force is what's doing the work.

We can quantify how fast work is being done in the resistor by the field — and therefore being used up by the battery — by calculating the work done by the electric force.

The work-energy theorem tells us that the amount of work done by a force on an object is the force times the distance the object moves while feeling that force. For a charge in an electric field this will be the electric force, $qE$, times the distance, $L$, that the object moves while feeling the force — the length of the resistor. So

$$\mathrm{work} = FL = (qE)L = q(EL) = qΔV.$$

We've just regrouped the $E$ with the $L$ to make the voltage difference, $\Delta V$. This shouldn't surprise us, as we know that the charge times the voltage difference is the change in the energy of the charge.

So the rate at which work is done (the ** power**) is the total amount of charge passing through the resistor in a time $Δt$ times the voltage difference it passes through. Let's call that amount of charge $Δq$. (It's what we called the amount of charge passing through a surface in our quantification of the concept of current.) As a result, we have

$$\mathrm{power} = \mathrm{rate\;at\;which\;work\;is \;done} = \frac{Δq}{Δt} ΔV = IΔV$$

the current through the resistor times the voltage drop across the resistor.

If we use Ohm's law, we can express this relation in different ways that might be convenient, depending on what we know or what we are comparing. (But it's useful to remember the original form, which is more directly connected to the work-energy theorem.) There forms come from using $ΔV=IR$ to replace either the $ΔV$ or the $I$ in the power relation:

$$\mathrm{power} = IΔV = I (IR) = I^2R$$

or

$$\mathrm{power} = IΔV = \bigg(\frac{ΔV}{R}\bigg) ΔV = \frac{(ΔV)^2}{R}.$$

This lets us calculate the rate at which the energy is being used up in a circuit.

## Units

Since power is current times voltage, the units are [charge/time] x [energy/charge] = energy/time, the rate at which energy is used/delivered. The standard unit for this is the *Watt:*

1 Watt = 1 Joule/second

This means that the standard units on you electrical bill — kiloWatt-hours — is a Watt times a time, so it is simply proportional to the total number of Joules of energy delivered. The time cancels.

Joe Redish 2/28/12

Last Modified: May 3, 2019