The electrical power is physical quantity which measures how much energy an electrical circuit needs to work during a given time, influencing thus in the electrical energy consumption of electrical devices. The greater the electrical power, the greater the energy expenditure. The electrical power can be used to calculate the energy spent on electrical installations.
Read too: Tips for saving electricity
Summary of electrical power
A power electrical measures the amount of electrical energy delivered to electrical circuits during a time interval.
The unit of measurement for electrical power is the Watt.
Electrical power can be calculated from the relationships between electrical resistance, electrical voltage and electrical current.
Electrical power can be active, reactive or apparent.
The active power is that used in the transformation of electrical energy into other useful energy, causing light, movement and heat, and measured in kilowatts (kW).
Reactive power is the useless power, which was not used by the active power, measured in kiloVolt-Ampere reactive (kVAR).
Apparent power is the resultant power in an electrical circuit, measured in kilowatt-ampere (kW A).
What is electrical power?
Electric power is a scalar physical quantity that measures the amount of energy electricity granted to electric circuits during a time interval. The greater the electrical power of the device, the greater the energy consumed by it. That's why showers and air conditioners are the biggest consumers of domestic electricity.
Unit of measurement of electrical power
According to International System of Units (SI), The unit of measurement for electrical power is the Watt., represented by the letter W, in honor of the scientist James Watt (1736-1819), who patented his copying machine, rotary engine and others, and perfected the steam engine.
What are the formulas for electrical power?
→ Electrical power related to electrical resistance and electrical current
\(P=R\cdot i^2\)
P → electrical power, measured in watts \([W]\).
R → electrical resistance, measured in Ohm \([Ω ]\).
i → electric current, measured in Ampere \([A ]\).
→ Electrical power related to electrical voltage and electrical resistance
\(P=\frac{U^2}R\)
P → electrical power, measured in watts \([W]\).
U → electrical voltage, measured in volts \([V]\).
R → electrical resistance, measured in Ohm \([Ω ]\).
→ Electrical power related to electrical voltage and electrical current
\(P=i\cdot ∆U\)
P → electrical power, measured in watts \([W]\).
i → electric current, measured in Ampere \([A ]\).
\(∆U\) → electric voltage variation, also called electric potential difference, measured in Volts \([V]\).
→ Electric power related to energy and time
\(P=\frac{E}{∆t}\)
P → electrical power, measured in kilowatts \([kW ]\).
AND → energy, measured in kilowatts per hour \([kWh ]\).
t → time variation, measured in hours \( [H ]\).
How to calculate electrical power?
The electrical power is calculated according to the information given by the statements. If it is an exercise on electrical energy consumption, we will use the formula for electrical power related to energy and time variation. However, if it is an exercise about electrical circuits, we will use the formulas for electrical power related to Electric tension, electric current and/or electrical resistance. Below, we will see examples of these two forms.
Example 1:
What is the electrical power of a shower that spends a monthly energy of 22500 Wh, being turned on every day for 15 minutes?
Resolution:
First, let's convert minutes to hours:
\(\frac{15\ min}{60\ min}=0.25\ h\)
As it is connected every day, monthly we will have:
\(0.25\ h\cdot 30\ days=7.5\ h\)
Subsequently, we will calculate the electrical power, using the formula that relates it to energy and time variation:
\(P=\frac{E}{∆t}\)
\(P=\frac{22500}{7.5}\)
\(P=3\ kW\)
The electric shower has an electrical power of 3 kW or 3000 Watts.
Example 2:
What are the electrical power and voltage in a circuit that has a 100Ω resistor that carries a current of 5A?
Resolution:
First, we will calculate the electrical power using the formula that relates it to electrical resistance and electrical current:
\(P=R\cdot i^2\)
\(P=100\cdot 5^2\)
\(P=100\cdot 25\)
\(P=2500\ W\)
\(P=2.5\ kW\)
Then, we will calculate the electrical voltage using the formula that relates it to electrical power and electrical resistance:
\(P=\frac{U^2}R\)
\(2500=\frac{U^2}{100}\)
\(U^2=2500\cdot 100\)
\(U^2=250000\)
\(U=\sqrt{250000}\)
\(U=500\ V\)
However, the electrical voltage could also have been calculated using the formula that relates it to electrical power and electrical current:
\(P=i\cdot ∆U\)
\(2500=5\cdot ∆U\)
\(∆U=\frac{2500}5\)
\(∆U=500\ V\)
See too:Ohm's first law — the relationship of electrical resistance to electrical voltage and electrical current
Types of electrical power
Electrical power can be classified as active power, reactive power or apparent power.
→ Active electrical power
Active electrical power, also called actual or useful electrical power, is the one transmitted to the charge capable of converting electrical energy into another form of energy that can be used (useful work), producing light, movement and heat. It is measured in kilowatts (kW).
→ Reactive electrical power
Reactive electrical power, also called useless electric power, is that which was not used in the process of converting electrical energy into other forms of useful energy, being stored and re-established in the generator, serving as the constant path that active energy takes to do useful work and to magnetize the windings of equipment. It is measured in KiloVolt-Ampere Reactive (kVAR).
→ Apparent electrical power
Apparent electrical power is the total power in a circuit, the sum of active power and reactive power. It is measured in kilowatt-ampere (kWA).
Solved exercises on electrical power
question 1
(PUC)
Electricity is generated using light using photosensitive cells, called photovoltaic solar cells. Photovoltaic cells in general are made of semiconductor materials, with crystalline characteristics and deposited on silica. These cells, grouped into modules or panels, make up photovoltaic solar panels. The amount of energy generated by a solar panel is limited by its power, that is, a 145 W panel, with six working hours of sunlight, generates approximately 810 Watts per day.
Source: http://www.sunlab.com.br/Energia_solar_Sunlab.htm
Check the number of hours that the panel described can keep a 9 Watt fluorescent lamp on.
A) 9 am
B) 6 pm
C) 58 hours
D) 90 hours
Resolution:
Alternative D
We will calculate the energy supplied by the electrical panel using the formula that relates it to power and time:
\(P=\frac{E}{∆t}\)
With a power of approximately 810 Watts per day, we have the energy of:
\(810=\frac{E}{24}\)
\(E=810\cdot 24\)
\(E=19\ 440\ W\cdot h\)
So, the energy consumption of the lamp during the day is:
\(9=\frac{E}{24}\)
\(E=9\cdot 24\)
\(E=216\ W\cdot h \)
Equating the amount of energy generated by the panels with the energy consumption of the lamps, we obtain:
\(19440=216\cdot t \)
\(t=90\h\)
Thus, the lamps work for 90 hours when connected to the panel.
question 2
(IFSP)When entering a building materials store, an electrician sees the following ad:
SAVE: 15 W fluorescent lamps have the same luminosity (illumination)
than 60 W incandescent lamps.
According to the ad, in order to save electricity, the electrician changes a light bulb incandescent by a fluorescent one and concludes that, in 1 hour, the electric energy savings, in kWh, will be in
A) 0.015.
B) 0.025.
C) 0.030.
D) 0.040.
E) 0.045.
Resolution:
Alternative E
To calculate the electrical energy savings, we will first calculate the energy expenditure of the fluorescent lamp and the incandescent lamp, using the formula for electrical power:
\(P=\frac{E}{∆t}\)
\(E=P\cdot ∆t\)
The energy of the fluorescent lamp is:
\(E_{fluorescent}=P\cdot ∆t\)
\(E_{fluorescent}=15\cdot1\)
\(E_{fluorescent}=15\ Wh\)
To get the value in kilowatt-hours, we need to divide by 1000, so:
\(E_{fluorescent}=\frac{15\ Wh}{1000}=0.015\ kWh\)
The energy of the incandescent lamp is:
\(E_{incandescent}=P\cdot∆t\)
\(E_{incandescent}=60\cdot1\)
\(E_{incandescent}=60\ Wh\)
To find the value in kilowatt-hours, we need to divide by 1000, so:
\(E_{incandescent}=\frac{60\ Wh}{1000}=0.060\ kWh\)
Therefore, the energy savings are:
\(Economy=E_{incandescent}-E_{fluorescent}\)
\(Economy=0.060-0.015\)
\(Economy=0.045\)
By Pamella Raphaella Melo
Physics Teacher
Source: Brazil School - https://brasilescola.uol.com.br/fisica/potencia-eletrica.htm