Objectives:

• To understand that power is defined as the rate at which energy is transferred or the rate at which work is done. This can be written in the form $P = \frac{W}{t}$
• To identify that an energy transfer of 1 joule per second is equal to a power of 1 watt, this can be done using the equation $latex P = \frac{W}{t} • To be able to give examples that illustrate the definition of power, e.g. comparing two electric motors that both lift the same weight through the same height but one does it faster than the other. We all come across the term power frequently. You could say that a car is more powerful than a motorbike or that a plane is more powerful than a helicopter, you may hear people saying ‘person 1 is more powerful than person 2′. But what do these comments really mean? Definition of power 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 (believe it or not) we like to make life easy… 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, $s$ Using this equation, can you answer the following questions? 1. On the motorway and driving at the speed limit, which has the greater energy store: A Triumph Street Triple (Motorbike) or an Audi A3 (car)? 2. On the motorway and driving at the speed limit, which has the greatest power: A Triumph Street Triple (Motorbike) or an Audi A3 (car)? 3. Driving at whatever speed is necessary, which consumes the most power, a Street Triple covering 20 miles in 20 minutes or a street triple covering 10 miles in 20 minutes? Answers 1. The Audi has the greatest energy store, this is because it has the greatest mass (assuming the vehicles have not been altered in any way!), they will both be travelling at the same speed, so it is only the mass that differs 2. The Audi has the greatest power, this is because it has the greatest work done per second, it has the greatest work done because it has the greatest mass and therefore the most kinetic energy (it has the greatest energy store) 3. The motorbike that covers 20 miles in 10 minutes has the greatest power because it travels the fastest and therefore has the greatest kinetic energy store. If this is the case, it uses the greatest amount of energy per second. You may think that the two vehicles would have the same power but in fact we only consider the question when discussing the object when it is moving Power is incredibly important to us as we need it to power our electronic devices, cars, homes, even our own bodies! Take a sporting event for example, if you want to throw a javelin further than your competitor, only two things are of importance – the angle and the power. If you both throw it with the same angle then the person with the greatest power transfers the greatest amount of energy to the javelin per second and therefore will make it speed up the most and make it travel furthest! If you want to make an object more efficient, giving it more power is not always the best way, you could instead try making it more efficient. By decreasing the total input power, but still getting the same output power, the efficiency improves. In fact the equation for efficiency involving energy is: $Efficiency = \frac{useful output energy}{total input energy} \times 100%$ Using the equation for power (as stated above) but rearranging for Energy gives: $E = Pt$. we can substitute this into the equation for efficiency to get: $Efficiency = \frac{useful output power \times t}{total input power \times t} \times 100%$ Which reduces to give: $Efficiency = \frac{useful output power}{total input power} \times 100%$ The Watt The Watt is the unit that is given to Power. To understand what $1 W$ really is we can substitute the useful units for each quantity into the equation for power, where power is measured in watts, word done is measured in joules and time is measured in seconds: $P = \frac{W}{t}$ $W = J/s$ What this really means is that $1 W$ is equivalent to $1 J$ of work done in $1 s$: $1 S = 1 J/ s$ Putting power to use Imagine you have a motor which has been setup to life a set of masse from one height to another. Work is done against gravity to lift it up (gravity is pulling it down so a force in the opposite direction is required). The amount of useful energy that is put into this system, would ultimately go into gravitational potential energy. The input energy is how much electrical energy is supplied to the motor. $electrical energy \rightarrow gravitational potential energy$ In an ideal world, all of the electrical energy would go into the gravitational potential energy, making it $100 %$ efficient, however, undoubtedly some will be lost in the forms of heat and sound in the motor. – – – – – – – – – – – – – – – – – – – – 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$ – – – – – – – – – – – – – – – – – – – – ‘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