Jump to a key chapter

In this article, we will discuss how electrical energy is transferred between appliances and the factors it depends on. We will also look into how the power consumed by an electrical appliance is calculated and work on a few examples. You will also be able to make informed decisions when you're out shopping for any new electrical device. So keep reading till the end to find out - oh, and don't forget to turn your lights off!

## Power ratings definition

Power ratings are what we see on our home appliances, defining how much **energy** is being transferred from the mains to power the device. The power rating helps consumers choose between different appliances based on their power consumption. It also highlights the **maximum power** at which the appliance can safely operate, which the cable and plug also need to be able to handle.

A mobile phone charger typically has a power rating in the range of 5-25 watts. This means that the charger draws a maximum of 25 watts or 25 joules-per-second from the mains electricity supply. An electric kettle on the other hand has a power rating of 3 kilowatts. That is 3000 joules per second, which is 120 times the power consumed by the charger! This means that with the same amount of energy, you could boil a kettle for 1 minute or charge your phone for 2 hours (120 minutes)! Let us now look at how to calculate the power using the current drawn and the voltage. Now let us look at different equations to calculate the amount of power consumed by appliances using the current drawn and the voltage.

Not all the electrical energy transferred into an appliance is converted into **useful work**. A part of the energy is almost always converted into heat energy or some other form of waste in electrical devices. The **efficiency **of the appliance tells us how much of the** input energy** is converted into **useful** **work**. We'll speak more about the term efficiency and what it means in a later section.

## Power rating formula

**The Electric power** **rating **formula in a circuit can be calculated using the formula:

$P=VI$

or in words:

$Power=Electricpotential\times Current$

Where**$P$**is the **power** in watts$\left(\mathrm{W}\right)$,$V$is the **potential**** difference **in volts$\left(\mathrm{V}\right)$between the points of energy transfer and$I$is the **current**** **in amperes$\left(\mathrm{A}\right)$. Therefore, **1 watt **of electric power can be defined as the energy transferred when a current of$1\mathrm{A}$flows through a potential difference of$1\mathrm{V}$**.**

There is also another method to calculate the power rating of an appliance. It can also be calculated using the **work done **or **energy transferred** by an appliance in a given amount of** **time.

$P=\frac{W(\mathrm{or}E)}{t}$

or in words,

$Power=\frac{workdone}{Timetaken}$

Here$W$is the work done or$E$the energy transferred in **joules**, and$t$is the time in **seconds**. The power rating unit is **watts**. For appliances that consume higher values of power, we use **kilowatts**$({10}^{3}\mathrm{watts})$or **megawatts**$({10}^{6}\mathrm{watts})$.

From the above equations, we can see that the power consumed by an appliance depends on the total energy transferred and the time the appliance is on for. The above equation can be re-arranged to obtain the energy consumed by an appliance. Another way to calculate the rate of energy transfer to an appliance is by measuring how many **coulombs** of charge flow through a given potential difference. This is given by:

$E=QV$

or in words,

$\mathrm{energy}\mathrm{transferred}=\mathrm{charge}\mathrm{flow}\times \mathrm{potential}\mathrm{difference}$

where$Q$is the charge in **coulombs**$\left(\mathrm{C}\right)$and$V$is the potential difference in **volts**$\left(\mathrm{V}\right)$**. **Now lets look at the efficiency; this will not only help you in your GCSEs but also when you're out shopping for any new electrical appliance.

## Power Rating of appliances

This is an efficiency label that you'll find on devices. The different coloured bands allow you to compare which device is more efficient in operating. Wikimedia.

When you turn on an electric device it is designed to convert the electricity into whatever useful work it was intended to perform. During this conversion, some energy is always lost, typically as heat or noise. An efficient device is one that minimizes this energy loss. So if we have two devices with the same power consumption rating, examining their efficiencies tells you how much of the power consumed is being converted into useful work. Efficiency can be calculated as follows.

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{output}\mathrm{work}}{\mathrm{total}\mathrm{input}\mathrm{energy}\mathrm{transfer}}$

It can also be calculated as

$\mathrm{efficiency}=\frac{\mathrm{useful}\mathrm{power}\mathrm{output}}{\mathrm{total}\mathrm{power}\mathrm{input}}$

Calculating efficiency will give you a decimal value less than or equal to one - a useful way to represent this is by using a percentage. When you multiply the efficiency with$100\%$, we get what is called a percentage efficiency. A theoretical device with$100\%$efficiency converts all the supplied power into useful power. A percentage efficiency of$20\%$means that the device is only converting$20\%$of the supplied power into useful power or work.

A well-known example of an **inefficient device** is a filament lightbulb - these are designed to produce light, but over 95% of the input energy is converted into waste heat instead!

Now let's work on a few examples to practice what we just learned.

## Power rating examples

A$3000\mathrm{W}$kettle can boils a litre of water in$5\mathrm{minutes}$; how long would it take a$3\mathrm{W}$phone charger to transfer the same amount of energy as the kettle?

**Step 1: List out the given values**

${P}_{Kettle}=3000\mathrm{W},{t}_{Kettle}=5\mathrm{min}\phantom{\rule{0ex}{0ex}}{P}_{charger}=3\mathrm{W},{t}_{charger}=?$

**Step 2: Convert quantities**

$\mathrm{Time}=5\mathrm{min}=5\times 60\mathrm{seconds}=300\mathrm{seconds}$

**Step 3: Calculate the energy transferred to the kettle by rearranging the equation for power.**

$P=E/t\phantom{\rule{0ex}{0ex}}{E}_{kettle}={P}_{kettle}\times t\phantom{\rule{0ex}{0ex}}{E}_{kettle}=3000\mathrm{watts}\times 300\mathrm{seconds}=900000\mathrm{joules}=900\mathrm{kJ}$

**Step 4: Calculate the time taken for a phone charger to transfer the same energy as the Kettle.**

**$P=\frac{E}{t}$, **

rearrange this equation for time.

**${t}_{charger}=\frac{{E}_{charger}}{{P}_{charger}}\phantom{\rule{0ex}{0ex}}{t}_{charger}=\frac{900\times {10}^{3}\mathrm{joules}}{3\mathrm{watts}}=300,000\mathrm{seconds}$**

Finally, converting the time in seconds to minutes:

$\frac{300,000\mathrm{seconds}}{60\frac{\mathrm{seconds}}{\mathrm{minute}}}=5000\mathrm{minutes}$

The charger will take$5000\mathrm{minutes}$to transfer the same amount of energy transferred by the kettle in$5\mathrm{minutes}.$

Calculate the amount of energy transferred if an electric bulb connected to a supply of$20\mathrm{V}$has$140\mathrm{C}$of charge pass through it?

**Step 1: List out the given values**

$\mathrm{Potential}\mathrm{difference}=20\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{Charge}=140\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{Energy}\mathrm{transferred}=?$

**Step 2: Calculate energy transferred using the right equation**

**$E=QV\phantom{\rule{0ex}{0ex}}E=140\mathrm{C}\times 20\mathrm{V}\phantom{\rule{0ex}{0ex}}E=2800J$**

The energy transferred by moving a charge across a potential difference of**$20\mathrm{V}$is$2800\mathrm{Joules}$.**

## Power rating of the resistor

The **power rating** of a resistor gives the maximum amount of power it can dissipate before it will fail and break the circuit.

Every resistor comes with a specific power rating. A resistor gets **heated up** as it obstructs the **flow of current** through it. Thus if a resistor has a maximum power rating, it is to prevent it from heating beyond the limit of how much heat it can dissipate. The power rating of the resistor is usually measured in watts.

## Power Rating - Key takeaways

- The power rating that we see in our home appliances defines how much electrical energy is being transferred from the national to power the device.
- Not all the energy transferred into an appliance is converted into
**useful work**. - Electric power or the electric energy transferred in a circuit can be calculated using the electric power formula$P=VI$
- It can also be calculated using the work done or energy transferred by an appliance in a given amount of time$P=\frac{W\left(or\right)E}{t}$
- The power rating symbol is represented by the same symbol for power,
**W.** - The power rating of a resistor gives the maximum amount of power it can dissipate without failing and breaking the circuit.

###### Learn with 6 Power Rating flashcards in the free StudySmarter app

We have **14,000 flashcards** about Dynamic Landscapes.

Already have an account? Log in

##### Frequently Asked Questions about Power Rating

What is power rating?

The **power rating** of an appliance defines how much **energy** is being transferred from the mains to power the device. It also highlights the **maximum value of power** at which the appliance can safely operate.

How to find power rating of appliances?

The power rating can be found attached to most devices. It can also be calculated using the electric power formula

P=VI

What is the formula for power rating?

The power rating formula is given by ** P = IV **or

**where**

*P = E/t*,**is power in watts, I is current in amperes,**

*P***is potential difference in volts,**

*V***is the energy transferred in joules and**

*E***is the time in seconds.**

*t*What is the unit for power rating?

The unit of power rating is **watts (W)**. This is equivalent to **joules-per-second**.

How to find which appliance is using too much electricity?

The amount of electricity used is directly proportional to the power rating of the appliances. For example, a device with a power rating of 2000 W will be consuming 2000 J in one second.

To identify how much power a device is using, you could use a **voltmeter** and **ammeter** to measure the potential difference and current the device is operating at and then determine the power consumption using ** P=IV**.

##### About StudySmarter

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

Learn more