Objectives:

• Be able to describe power as the rate of energy transfer
• Be able to select and use the power equations;  $P = IV$  $P = I^{2}R$  and  $P = \frac{V^{2}}{R}$
• Be able to select and use the equation  $W = IVt$
• To define the kilowatt-hour (kW h) as a unit of energy
• To be able to calculate energy in kWh and the cost of this energy when solving problems

Power

Power is a term used to describe the supply of mechanical or electrical energy to component or circuit.

Take two electrical appliances, appliance A and appliance B. If they both receive the same amount, say 1000J, of electrical energy but both receive this energy over different time frames then the device that has used this energy the fastest will have required a greater power. For example, if appliance A spent this energy in 10 seconds and appliance B in 100 seconds, the amount of energy consumed is the same but the power required is different. In this particular case appliance A would require 10 times the amount of power that appliance B would require because the time frame is 10 times smaller.

Power is therefore related to both the amount of energy (or work done) and the time that that energy is supplied of used over;

$Power = Work \ done \ / \ time \ taken$

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

The unit for power is the Watt,  $latex$

We can instead substitute S.I units into the equation above to get alternative unit for power;

$\Delta W \rightarrow joules, J$
$t \rightarrow seconds, s$

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

$P \rightarrow \frac{J}{s} = Js^{-1}$

This tells us that one Watt is equivalent to 1 joule of energy per second;

$1 Watt = \frac{1 joule}{second}$

Deriving equations for electrical power

Deriving  $P = IV$

Using the definition of p.d;                    $V = \frac{\Delta W}{\Delta Q}$

Since  $\Delta Q = I \Delta t$ ;                             $V = \frac{\Delta W}{I \Delta t}$

Rearranging for  $\frac{\Delta W}{\Delta t}$;                         $\frac{\Delta W}{\Delta t} = IV$

Since  $P = \frac{\Delta W}{\Delta t}$ ;                                 $P = IV$

This tells us that electrical power can be calculated by multiplying the p.d. and the current. This should make sense as the current is a quantity that is dependent on the amount of charge that passes a point in a second – so time is involved, also p.d. is the term used to input the amount of energy a unit of charge has – so energy is also involved.

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Deriving  $P = I^{2}R$

Since $R = \frac{V}{I}$ and therefore $V = IR$$V$ can be substituted into the equation  $P = IV$ ;

$P = IV$

$P = IIR$

$P = I^{2}R$

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Deriving  $P = \frac{V^{2}}{R}$

Since $R = \frac{V}{I}$ and therefore $I = \frac{V}{R}$$I$ can be substituted into the equation  $P = IV$ ;

$P = IV$

$P = \frac{V}{R} V$

$P = \frac{V^{2}}{R}$

‘Measuring’ the power and/or energy used by an appliance in a circuit

One method used to measure the power used by an appliance, such as a bulb, is by connecting a voltmeter across the device and an ammeter in series to it.*

A stopwatch would be required to measure the length of time the bulb is illuminated for.

From the moment the bulb is connected to the cell the stopwatch needs to begin so as to time how long the bulb is illuminated for. The voltage and current can also be recorded. The following equations can then be used to calculate the power;

Deriving  $P = IV$  or Deriving  $P = \frac{\Delta W}{\Delta t}$

In order to calculate the total amount of energy used by the bulb, we can use $P = \frac{\Delta W}{\Delta t}$ rearranged for;

$\Delta W = P \Delta t$

Substituting $P = IV$ in gives;                   $\Delta W = IV \Delta t$

This same method can be used for determining how much energy a power supply or battery uses, instead of the voltmeter being put across the appliance it would need to be put across the power source (i.e. to measure the e.m.f. instead of the p.d.)

Measuring electricity usage in our homes

This is an example of an electricity meter that will be found in any house in the UK. It is used by electricity company to determine how much electricity (and therefore energy) a consumer has used and to therefore determine how much to charge them.

Electricity companies use the unit kiloWatt-hours  $kWh$  to determine how much energy consumers use – this is instead of using the joule. The reason for this is that measuring in joules would result in huge numbers being calculated for a household each month.
Think about how much energy (in terms of electricity) your household uses each month;  theres all the lights, the televisions, oven, heating, fridge freezer, laptop chargers, mobile phone chargers etc.

The  $kWh$  is another unit to express energy and it is derived as shown;

Power is usually measured in Watts, W. Because we use so much power it is more convenient in this case to measure it in kW  ( $1kW = 1000 W$ ).
Time is usually measure in seconds, s. Because we usually use our appliances for hours at a time, it is more convenient in this case to measure it in hours, hrs, h   ( $1hr = 60 \times 60 s = 3600 s$ ).

Subbing these units into the equation allows the units  $kWh$  to be shown;

$\Delta W = P \Delta t$

$\Delta W = kW \times hrs = kWh$

$latex 1 kWh$  is also referred to as a unit of energy (when dealing with electricity companies), so in the image above, you could say “I have used 6471 units” or you could say “I have used 6471 kWh’s”.

How many joules are in 1 kWh?

$1 kWh$ is equivalent to a device using 1000 Watts for 1 hours. 1 hour is equivalent to 3600 seconds and so;

$1 kWh = 1000 Wh = 1000 \times 3600 Ws = 3600,000 Ws = 3.6 \times 10^{6} J$

*Another method to measure the amount of energy used by an appliance is to use a jouletmeter. The power can then be found by simply using the joulemeter for a selected period of time and to the use the equation \$latex P = \frac{\Delta W}{\Delta t