Objectives:

• To understand that work is done when charge flows in a circuit.
• The amount of energy transferred by electrical work can be calculated using the equations:
$P = \frac{W}{t}$
$V = \frac{W}{Q}$
• To understand that the power of a device is related to the potential difference across it and the current through it by the equation:
$P = IV$
• To identify the links between everyday electrical appliances and how they are designed to bring about energy transfers.
• To describe how different domestic appliances transfer energy from batteries or a.c. mains to the kinetic energy of electric motors or the energy of heating devices.
• To understand that the amount of energy an appliance transfers depends on how long the appliance is switched on for and the power of the appliance.

What is Voltage?

You will be familiar with the term voltage from GCSE, but what is it and what does it actually mean?

When electrons pass through a component in a circuit, they usually give up some of their energy (and hence allow the component to work). The voltage is simply the term used to describe the amount of energy PER unit of charge. Before a component each unit of charge will have some energy , they will then give up (some) of this energy to the component and therefore the voltage will be lower after that component.

Potential difference (p.d.)

The amount of voltage that has been given to the component can be measured using a voltmeter, the voltmeter must be placed in parallel to a component. A voltmeter works by measuring the voltage at two points (before and after a component), one of these points is a reference point, it then measures the difference in voltage. Voltage is the energy per unit of charge, so the charged particles have some potential energy. A voltmeter measures the difference in potential energy from one side of a component to another – hence the term potential difference.

Understanding the definition of potential difference is vital for almost all future work on electricity (many students take this definition for granted and don’t learn it as the definition – this is ill advised!)

Potential difference is the energy per unit charge;     $V = \frac{W}{Q}$
where;
$V$  is the voltage, measured in volts,  $V$
$W$  is the work done, measured in joules,  $J$
$Q$  is the charge, measured in coulombs,  $C$

Voltage in an electrical system can be thought of as the same thing as pressure in a water system; the cell being the pump;

If we know the amount of energy given to each unit of charge, we can rearrange the equation to determine the total energy given to a circuit; $W = VQ$.

You will (hopefully) recall from the energy topic, what power is:

The accurate definition of power is how much energy something does in a certain amount of time. This is said more appropriately as:

Power is the rate of doing work

This means that it is the amount of energy that is applied, or used up, in a given amount of time. In Physics we choose this amount of time to be one second. The definition ‘power is the rate of doing work’ is therefore the same as:

Power is the amount of work done per second

Work done is the amount of energy transferred as discussed in a previous topic. The following equation can then be used:

$P = \frac{W}{t}$

where;
$P$ is the power, measure in Watts, $W$
$W$ is the work done, measure in Joules, $J$
$t$ is the time the work done is supplied over, measure in seconds, $s$

In this electricity topic you have learned that electricity comprises of voltage and current; since voltage is the word done per unit charge, if the voltage increased it makes sense that the power would increase too. Similarly, if the current increased, the more charge is flowing each second, therefore it makes sense that the power would increase too. So we can therefore write a new equation for power using current and voltage:

$P = IV$

where;
$P$ is the power, measure in Watts, $W$
$I$ is the current, measure in Amps, $A$
$V$ is the potential difference, measure in Volts, $V$

Physicists don’t just make up equations, they derive them and produce them from other principles. If they didn’t is it really fair to say the power from the equation $P = \frac{W}{t}$ is the same as the power from the equation $P = IV$ ? Or are they really different things altogether that we just hope are the same?

Deriving $P =IV$

We can take two equations you have learned previously for current and for work done; $I = \frac{Q}{t}$ and $W = VQ$ and substitute these into $P = \frac{W}{t}$ :

Rearranging $I = \frac{Q}{t}$ for $t$ gives:

$t = \frac{Q}{I}$

Substituting this along with $W = VQ$ into $P = \frac{W}{t}$ gives:

$P = \frac{VQ}{\frac{Q}{I}}$

Notice how the charge cancels? This leaves:

$P = \frac{V}{\frac{1}{I}}$

Re-written as $P = V \div \frac{1}{I}$

It becomes $P = V \times I$

$P = IV$

Electrical devices should always have numbers and symbols written on them, some of these are useful to us especially if we are travelling outside of the country.

The value for the potential difference on the white plug below says   $250 V \sim$. This means that a potential difference of 250 V is required to get the correct power output from this plug. The wavy symbol $\sim$ tells us that an alternating current is required. The UK mains is set at approximately $230 \pm 10 V$ and is alternating therefore this plug almost fits into this bracket and so should work well. If you go abroad to say the United States, they have a voltage of $110 V \sim$ in some places. This means that your appliance would not gain sufficient power to run effectively.

Have you ever gone abroad and tried to charge your mobile phone and it seems to take ages to charge (far longer than normal)? It is because of the above reasons.

Where does the electrical energy go?

The electrical energy store in our mains, when connected to a device or appliance, is transferred into a different energy store depending on the type of appliance.

• When charging a battery of some kind, the electrical energy store will go into a chemical potential store.
• When running a motor, the electrical energy store will go into kinetic energy stores.

It is important to note that in almost all cases, when energy is transferred from one form to another there is usually waste energy produced (due to resistance in the wires for example, or friction with moving parts). Therefore the most usual type of (unwanted) energy produced is thermal energy. So we can re-write the above to state:

• When charging a battery of some kind, the electrical energy store will go into a useful chemical potential store but some will be wasted as thermal energy and dissipated into the surroundings.
• When running a motor, the electrical energy store will go into kinetic energy stores, but wasted energy in the form of thermal energy will be dissipated into the surroundings

It is very difficult to reuse the energy that is lost into the surrounding. This tends to contribute towards global warming.

The difference between batteries and mains electrical supplies

Which sort of electrical energy supply do you think is most advantageous and why?

The mains electricity – this is so convenient, we plug in an appliance and it sets off to work – we almost don’t think about it.

Batteries however allow us to be mobile, they allow us to carry out gadgets around with us, but ow would we charge them without the mains?

How does the amount of energy an appliance uses depend on the time the appliance is being run for?

Hopefully you are looking at this question thinking that is is an easy answer. The longer an appliance runs for the more energy it requires – this is absolutely correctly! The equations we used above demonstrate this too…

If we rearrange $P = \frac{W}{t}$ for work done (the amount of energy transferred), we get:

$W = Pt$

This tells us that for an appliance that’s requires a certain power input, $P$, if time $t$ increase, then the work done must increase too…

$W\uparrow = P \times t\uparrow$

This can also be seen if we substitute the equation linking power, current and voltage, $P = IV$ into this equation too. Became now we can write: $W = IVt$

This tells us that for an appliance that has a specific potential difference across it and a set current running through it, if it is switched on for a longer amount of time, then the energy supplied to it will increase (if time increases, work done increases):

$W\uparrow = IV \times t\uparrow$

Putting this to some use: Determining the Efficiency of a motor

In order to determine the efficiency of an object, you need to be able to measure the amount of energy you are putting in as well as the amount of energy you get out of the system. If you want to determine the efficiency of a Bunsen Burner heating up a beaker of water for example, it is very difficult because it would be a struggle to determine how much gas is used.

So, by using a motor we can determine the electrical energy input into the system. If the motor is used to lift an object through a certain height, we can measure the new gravitational potential energy of the system. To sum up… we can measure the energy input and output using this example.

The electrical energy input can be determine if we connect a voltmeter and ammeter up to the motor. Using the following equation we can then work out the power:

$P = IV$

If the motor is on for a specific amount of time, which we can record using a stopwatch, the energy input can be measured by using:

$P = \frac{W}{t}$

This rearranged for work done gives us:

$W = Pt$

The first two equations can actually be combined to give us on equation to calculate the energy input:

$W_{input} = IVt$

To determine the efficiency of the motor we now need to determine the useful energy we obtain from the motor. In this case, the purpose of the motor was to raise the mass up through a certain height – work was done against gravity! This means the object gained useful energy in the form of gravitational potential energy, the equation for which is:

$E_{useful} = mgh$

We now have methods to calculate the input and output energies, these can be substituted into the energy efficiency equation to give:

$Efficiency = \frac{mgh}{IVt} \times 100 \%$

Where might the non-useful energy go?

In most systems, there is wasted energy which normally dissipates in the form of height and sound.

we can use Sankey Diagrams to help us to visualise how efficient an object is, here is an examples of one for a light bulb that is only $10 \%$ efficiency:

Try some questions on the work done on a charged particle or the power of appliances here – link to IsaacPhysics