"Electric Power" Electric Power and Electric Power PPT Courseware 2

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"Electric Power" Electric Power and Electric Power PPT Courseware 2

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In the same period of time, which one consumes more energy, an electric fan or an air conditioner?

In the same period of time, the greater the electric power of the air conditioner consumes more electrical energy, the faster the turntable of the energy meter rotates.

Electric power

Electrical appliances consume electricity during work. Do they consume electricity at the same rate? Let's take a look at the energy meter

The physical quantity that indicates how quickly an electrical appliance consumes electrical energy is called electrical power, which is equal to the electrical energy consumed in 1 second.

unit:

A. The international unit is watt, referred to as watt, and its symbol is W.

B. The common unit is kilowatt (KW). l kilowatt = 1000 watts.

1. Electrical power of electrical appliances

1. Concept:

The amount of electrical energy consumed by current per unit time is called electrical power. Represented by the letter "P".

Electric power is a physical quantity that expresses the speed of consuming electrical energy.

2. Calculation formula: P=W/t

This is the definition formula of electric power

3. The unit and physical meaning of electrical power

Unit name: watt (referred to as watt)

Unit symbol: W

1 kilowatt = 1000 watts

The electric power of a certain motor when working is 60KW, which means that the motor consumes 60,000 joules of electric energy per second.

1 watt = 1 joule/second

2. Calculation of electrical power

1. Factors that determine the size of electrical power: U and I

2. The electrical power of an electrical appliance is equal to the product of the current passing through the electrical appliance and the voltage across the electrical appliance.

2. The origin of “kilowatt-hour”

P=W/t→W=Pt. If the units of P and t are kilowatts and hours respectively, then after multiplying, we will get another unit of electrical energy - kilowatt hours (degrees)

1 kilowatt hour is the electrical energy consumed by an electrical appliance with a power of 1 KW for 1 hour.

*Kilowatts and kilowatt-hours are two different units of physical quantities and cannot be confused.

Example 1: The electric power of a TV is 250W. If it is used for 3 hours a day, how many kilowatt-hours of electricity is used in a month? (Calculated based on 30 days)

The generally applicable formula for electric power is P=U·I

Derived formula

P= U2/R P= I2R

When using these formulas, the unit of power obtained is watts only when the units of current, voltage, and resistance are amperes, volts, and ohms.

Note: ⑴The derivation formula is also only applicable to pure resistance circuits.

⑵ Defining formulas and universal formulas are applicable to any circuit.

Example 2. The rated power of a household energy-saving fluorescent lamp is 11W. What is the current in milliamps that passes through it during use?

Solution: The voltage of household lighting circuit is 200V

It can be known from the electric power formula P=IU

I=P/U=11W/220V=0.05A

Answer: The current through the fluorescent lamp during use is 0.05A

3. Rated power

Rated power: The power of the electrical appliance working at the rated voltage.

1. Rated voltage: the voltage when the electrical appliance is working normally.

2. The lighting voltage in my country is 220V, that is, the rated voltage of electrical appliances in the lighting circuit is 220V; the power voltage is 380V

4. Actual power

1. The power achieved by the electrical appliance when working at the actual voltage.

actual voltage

2. The voltage of the electrical appliance during actual operation. Often not equal to the additional voltage.

3. The relationship between actual power and voltage:

(1) U real > U forehead, P real > P forehead, the bulb glows strongly.

(2) U real = U amount, P real = P amount, the light bulb glows normally.

(3) U real < U amount, P real < P amount, the light bulb will glow dimly.

Note and understanding: P=U2/R

Electrical power calculation

Example 1: Connect lamp L1 (PZ220-25) and lamp L2 (PZ220-60) in parallel to a voltage of 220 volts, and then connect them in series to a circuit of 220 volts. Which bulb is brighter in the two cases? Why?

Example 2: Can a small light bulb marked "6V 3W" be directly connected to a 9-volt power supply? What should I do if I want the small light bulb to glow normally after being connected?

Example 1 analysis: To determine whether two lights are bright or dark, just compare the actual power of the two lights.

Solution: When connected in parallel, the actual voltage of each lamp is 220 volts, and its actual power is equal to the rated power of the lamp. Therefore, it can be directly judged that lamp L2 is brighter than lamp L1.

When connected in series, because the voltage at both ends of each lamp is less than 220 volts, neither lamp can emit light normally. According to R=U2/P, the resistance R1 of lamp L1 is greater than the resistance R2 of lamp L2, and because the two lamps are connected in series, so The current passing through them is the same. Therefore, it can be judged according to P= UI = I2R that P1>P2, the light L1 is brighter.

Summary of the example question: In a parallel circuit, the electrical appliance with a large resistance consumes a small amount of electric power; in a series circuit, the electrical appliance with a large resistance consumes a large amount of electric power.

Analysis of Example 2: It is not difficult for students to judge based on existing knowledge, because 9 volts is already greater than the rated voltage of the light bulb, 6 volts. If it is connected directly, the actual power is much greater than the rated power and the light bulb will burn out, so it cannot be connected directly. enter. If you want to connect it, you should connect a resistor R in series with the light bulb and then connect it, so that the resistor R can share the voltage of 3 volts.

Solution: Direct access is not possible. It should be connected in series with a resistor whose resistance is R.

I = I = P / U = 3 watts / 6 volts = 0.5 amps.

∴R = (U - U amount)/I = (9 volts - 6 volts)/0.5 A = 6 ohms.

Discuss what other ways to find R for this problem.

Summary of the example: When the actual voltage applied to both ends of the electrical appliance is much greater than the rated voltage, the electrical appliance may burn out, and a resistor should be connected in series with it.

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Update Time: 2024-11-22

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