2-3 Voltage

Energy Gains and Drops

Let's look at a simple electric circuit which has a light bulb, a switch, a 2-cell battery and some wires.

We can show where the electrons are flowing: they are repelled away from the negative terminal, are pulled through the circuit, and are attracted to the positive terminal.

All the charges go around and around, never getting out of the wires. Their job is to do two things:

1. Pick up energy at the source(s).

2. Drop energy off at the load(s).

That's it! They don't get eaten-up by the load, they don't get made by the source out of thin air. The charges are in the wires already, and all they do is pick up, and drop off, energy. The charges go around and around on an endless loop.

The source -- in this case a 2-cell battery -- gives out electrical energy. The load takes in electrical energy and changes it to something else. In this case that electrical energy gets changed into light (and heat) energy in the light bulb.

The battery here can be called an energy gain, and the load can be called an energy drop. Energy is given the symbol E, and is measured in joules (J). So, if you wanted to say that something had an energy of fifteen joules, you would write it like this: E = 15 J

An energy gain is something that gives energy to the electric current flowing through it: any source like a cell or battery. An energy drop is something that takes electrical energy out of the flowing current and turns it into something else: any load like a light bulb, speaker, heater, etc.

Defining Voltage

Let's take a closer look at the light bulb, and let's label two points: A is just before the bulb, and B is just after the bulb.

Take a second and calculate the current through the bulb, using the equation I = Q/t. (You should get that the current I = (20 C)/(5.0 s) = 4.0 A.)

Since light bulbs are energy drops, they take energy out of the current. They turn this electrical energy into light and, unfortunately, some heat. Here, heat is a form of waste energy, because it is not needed. We will be looking at energy efficiency later. But here, let's think a little more deeply about points A and B.

The current will give up some energy to the bulb. How we usually describe this by talking about how much energy each coulomb of charge gives up to the bulb as it passes through.

(There's a reason we do it this way. If a light bulb is on for a few seconds, not much charge is passing through it, so not much energy is given to it. But if it's on for a whole day, a lot of charge passes through it, and this current gives up a lot of energy. Dividing-out the amount of charge lets us examine the bulb in a fairer way. Think of it this way: if a sports stadium sells tickets at $50 each, if one fan buys a ticket, the stadium won't earn much money. But if fifty thousand people attend, they take in a lot of money. Dividing by the number of people lets you see how much money the tickets cost per person, which is useful information.)

The amount of energy gained or lost by each coulomb of charge is called the voltage, often also called either voltage difference or, if you want to be really fancy, electric potential difference. (We will usually be calling it "voltage" here.)

Voltage is given the symbol V. But here's where it gets confusing: it's measured in units of volts, which are also given the symbol V. So, if the voltage change across the light bulb is 6.0 volts, we write it like this: V = 6.0 V

Since voltage (V) is the energy change (E) per coulomb of charge (Q), we can come up with an equation to describe this:

Example: What is the voltage drop across an electric motor, which takes 56 J of energy out of the current as 8.0 C of charge flows through it?

E = 56 J

Q = 8.0 C

V = ?

V = E/Q = (56 J)/(8.0 C) = 7.0 V

Voltage Gains and Drops

If you look closely at cells and batteries, they will tell you their voltage rating.

If we zoom-in a bit, we can see the voltage rating of each of the batteries and cells.

Two of these have voltages of 1.5 V, another has a rating of 3 V, another is 9 V, and another is 12 V. Each of these is a different size, so it would be difficult to put a cell or battery into the wrong device. That's a good thing, because the wrong voltage could damage a device.

Each of these batteries or cells is a voltage gain, since when every coulomb of charge goes through it, it picks up energy. Since a volt is a joule per coulomb, that means the "12 V" battery gives each coulomb of charge 12 J of energy. The "1.5 V" cell only gives each coulomb of charge 1.5 J of energy.

Loads are voltage drops because they take energy out of the current, and turn it into something else.

Here's the next video in the TVO Electricity series, which deals with voltages. Remember, "voltage" is sometimes called "electric potential difference."

Practice

The Basics

1. In your own words, describe the difference between voltage and current.

2. An electric fan takes in 1200 J as 10 C of charge pass through it. What is the voltage drop through it?

Extensions

1. Compare the units of current and voltage. How are they the same? And how are they different?

2. Find the charge that passes through a flashlight, if it's a 6.0 V drop across the bulb and the bulb takes 2.5 J of energy.