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3. Power and electric energy

Electric power

Power (P) is defined as the amount of energy (E) consumed per unit of time (t). In the international system it is measured in Watts (W).    

Energy can be related to voltage (V) as: 

Q being the load. Dividing on both sides by time, and introducing the intensity value:  I = Q /t :

  P = V · I  

Incandescent lamp , KMJ, alpha masking by Edokter,  Wikipedia , CC BY-SA 3.0 . Fluorescent lamp , Benjamín Núñez González, Wikimedia Commons , CC BY-SA 4.0 . Halogen lamp , Hellerhoff, Wikimedia Commons , CC0 1.0 . Lámpara led , Dmitry G, Wikimedia Commons , CC BY-SA 3.0 .

Understanding what power means will allow us to understand and control what we spend on electricity.

Look at the table. It shows 4 different types of light bulbs that can be used for lighting. Within each type, you can choose bulbs with a higher or lower luminous intensity (that give more or less light).

Let's say we need a light bulb for a room that requires a lot of light, for example the kitchen. If we buy an incandescent bulb, we need it to be 100 W to have the highest possible light intensity (1600 lumens), but if we choose an LED, a power of 19 W will be enough to give the same amount of light. 

The 100 W bulb consumes 100 J of energy from the mains every second, while the 19 W bulb consumes 19 J in the same time. 

Read on to see what difference it would make on your bill to use one bulb or another.

Audio:



Guided exercise 1. Fill in the gaps

Which one iluminates the most, a 100 W or a 60 W bulb? Both are connected to the domestic mains (230 V). We consider both to be the same type of bulb, only their wattage differs.

Audio: 

Datos: P1 = 100 , P2 = W, V1 = V2 =  

Unknown: I

P = I · V

I = /  

I 1 = 100 / 230 = 0.44 A

I2 = / = 0,26 A

Current through the 100W bulb is higher, so it will be its ilumination.

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Electrical energy

Electrical energy is the energy transported by an electrical circuit. It is produced in the generator and consumed in the circuit elements.  We can calculate it as:

E = P · t

E = V · I · t

Where: P is the power and its unit the Watt (W), t the time (s), V the voltage (V) and I the intensity (A)

Some household appliances, such as irons or hairdryers, convert electrical energy into thermal energy with a resistor. Considering that V = I R (Ohm's Law), we can express the electrical energy that is converted into thermal energy in a resistor of value R as:

E = I · R · I · t → E = I2 · R · t

The SI unit of energy is the Joule .

Another unit that is frequently expressed is kW·h. To calculate energy in these units, power must be expressed in kW (instead of W) and time in hours (instead of s). It is also possible to convert kW·h to J directly using the following conversion factor:

1 kW·h = 3.6 · 106 J

Audio: 

Guided exercise 2. Fill in the gaps

Fill in the blanks to solve this problem with the appropriate numerical value, magnitude, symbol or unit:

a) Calculate the power of an iron knowing that, when we plug it into 230 V, the current intensity that passes through it is 6.52 A. Write the decimals with a period, the multiplication sign with a dot (·) and the division sign with a slash (/).

b) Calculate the energy consumed by the iron in 10 minutes , in Joules and in kw·h

Audio: 

a) Data: = V,      = A

Unknown factor: P

Equation: P = I ·

We substitute the data: P = 6.52 · = 1500 W

b) t = 10 min = s = 0,17 h

P = 1500 W = kW

E = P ·

E (J)  = P (W) · t (s) = · = 9 · 105 J

E (kW · h ) = P (kW) · t (h) = · = 0,25 kW·h

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Multiple choice

Question

Look at the photos of the dryers and their labels:

Audio of the figure:

1) Which one has greater maximum power?

Hint

Power is measured in Watts (W)

Answers

a

b

c

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Question

2) If you plugged them in for the same time, which one would consume less energy?

Hint

E = P · t

Answers

a

b

c

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Question

3. With which one will you need less time to dry your hair?

Hint

P = E / t 

t = E / P 

Answers

a

b

c

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Question

4. What voltage can they be plugged into?

Answers

a) 50-60 Hz

b) 220-240 V

c) 1000-2300 W

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Guided exercise 3. Fill in the gaps

The average price of kW·h today (December 2024) is 0.2 €/kW·h. If you have 4 incandescent bulbs of 1100 lumens in your living room that you usually turn on for 4 hours a day and you replace them with LED bulbs, how much money do you save per year ?

Incandescent Fluorescent Halogen Led
450 lumens 40 W 29 In 10 in 5 in
800 lumens

60 W

 43 In 13 In 10 in
1100 lumens 75 W 53 In 16 In 15 In
1600 lumens 100 W 72 W 20 W 19 In

You can see in the table that for an incandescent bulb to provide a light intensity of 1100 lumens, its power must be 75 W, while the LED bulbs that we are going to buy to replace them must be 15 W to give us the same amount of light. 

Energy consumed by incandescent lamps in one year:

P = 75 W = 0.075 kW

Time they are on after one year: t = 4 h/day · 365 days = 1460 h

Energy they consume in one year (in kW·h): E = P · t = 0.075 · 1460 = 109.5 kW·h

Operating cost of 1 bulb = 109.5 kW h 0.2 €/kW h = 21.9 €

Operating cost of 4 bulbs = 4 · 21.9 = €87.6

Repeat the calculations for the LED bulbs:

P =  W =  kW

Time they are on after one year: t = h/day · days = h

Energy they consume in one year (in kW·h): E = P · t =  · =  kW·h

Operating cost of 1 LED bulb = · 21.9 = 4.4 €

Operating cost of 2 LED bulbs = 4 4.4 = €17.6

Savings = 87.6 -  = €70

This is how much you would save per year by changing only the light bulbs in the living room. If you do it for the whole house, it will be much more.

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Autonomous exercise

If you have two 1600 lumen incandescent bulbs in your kitchen, use the table above to calculate in your notebook how much money you would save per month by replacing them with LEDs. Suppose they are on for an average of 3 hours a day.

Energy efficiency

Etiquetado energético. MITECO. Licencia MITECO

Not all the energy provided by the grid is used by an appliance that is plugged in. For example, light bulbs convert most of the electrical energy they receive into light energy (useful energy), but some of it is converted into thermal energy and cannot be used to generate light. 

Energy efficiency is the percentage of the supplied energy that is used as useful energy:

An effective way to save energy and money is to use machines and appliances that make better use of energy. The less energy they consume to perform a given task, the more efficient they are.
The European Commission has created an energy label for household appliances that informs, among other things, about their energy efficiency. This would be given by a letter and a color, as you can see in the image. The most efficient are class A and the least efficient are class G.

Guided exercise 4. Fill in the gaps

A radiator consumes 700 W of power connected to the mains (230 V). Calculate:

a) the current intensity 

b) the resistance of the radiator

c) the energy it consumes during 1 hour of being on (in kW·h)

d) its efficiency if it produced a thermal energy of 1.25 · 106 J

Data: P = 700 W, V = V, t = 1 h, Euseful = 1.25 · 106 J

Unknowns: I, R, E, ε

a) P = I·  → I = / V = 700 / 230 = 13,7 A

b) V = I·  

    R = V/ I = / 13,7 = 16,8 Ω

c) E = P · t = · 3600 = 2,5 · 106 J

d) ε = E useful / E consumed · 100 = · 106  / 2.5 · 106   · 100 = 50 %.

Note that efficiency must always be a number lower than 100. No more energy can be converted into useful energy than is consumed. That is, the numerator is the energy that is used (in this case as heat) and the denominator is the energy consumed from the mains.

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