The unit of measurement for electrical power is watts (W). Power reflects the rate at which the circuit consumes energy or the power source generates energy. Power (P) is equal to voltage multiplied by current, namely

P=VI

The power can also be calculated by the formula P=I^{2}R or P=V^{2}/R.

In physics, power is the work done per unit time, that is

Power = work/time

Power is the rate of work done or the rate of energy conversion. In the metric system, the unit of energy is Joule (J), and 1 Joule per second (J/s) is equal to 1 Watt (W).

Through the water wheel (see Figure 1), you can clearly see how water is used to do work. When water hits the wheel, the wheel rotates and drives a mechanical load (such as a grinding wheel) connected to the shaft of the water wheel. If the flow of water (gallons per minute) increases, the wheels will turn faster and the work done per unit of time will increase. This relationship also applies to electric power. When the current increases, the power increases, and when the voltage increases, the power also increases.

Scientist James Watt determined the ability of the machine to do work through a steam engine experiment. In order to sell steam engines to farmers as power equipment for milling grains, he linked how many horses the machines could replace. Watt determined that, on average, a horse can do 550 feet and a pound (ft·Ib) of work per second, which is recorded as

1 horsepower (hp)=550ft·Ib/s

And 1 horsepower is equal to 746 watts, denoted as

1hp=746W

**① Energy consumption**Power companies do not sell power, they sell energy consumption, which can be expressed as

Energy = power × time

To calculate the amount of energy used or consumed, you can multiply the number of watts by the number of hours worked:

Energy=Watts×Hours=Watts·Hours

For example, if a motor with a rated power of 1hp continues to run for 1 hour, it will consume 746W·h of electrical energy.

example 1

A 1hp motor runs for 2h, the energy consumed is

Energy=746W×2h=1492W·h

Example 2

The working voltage of an electric lamp is 120V and the current is 0.83A. What is its power? If this lamp is lit for 5 hours, how much energy will it consume?

P=V×I=(120×83) W=99.6W

Energy=P×hours=(99.6×5)W·h=498W·h, or 0.498kW·h

Please note that the unit of measurement of power (P) is W, and the unit of energy consumption is W·h or kW·h.

Example 3

A 120V AC motor is powered by a wind generator, as shown in Figure 2. The output power of the wind alternator is 1200W. What is the supply current? If it runs for an average of 30 hours a week, how much electricity does it consume in a year?

I=P/V=1200/120=10A

Electric energy consumed in a year=P×hours/week×weeks=1200×30×52W·h=1872000W·h, or 1872kW·h, or 1.872MW·h.

Example 4

In Figure 3, there are two electric heaters connected in series. The current (l) flowing through each heater is 2.5A. Assume that the power of heater 1 is 100W and the power of heater 2 is 200W. Use the power formula to calculate the voltage across each heater. Then calculate the total voltage (E) by adding up all the voltage drops.

V_{1} =P_{1}/I_{1} = 100/2.5V=40V

V_{2} =P_{2}/I_{2} =200/2.5V=80V

E=total voltage=V_{1} +V_{2} =120V

A power supply voltage of 120V is required to drive the two heaters. Each heater has a different pressure drop, but the current is the same. Therefore, different heaters have different resistances (as known from Ohm’s law). Heater 1 needs a voltage of 40V to pass current, while heater 2 needs a voltage of 80V. The power supply voltage must be high enough to allow current to pass through the two heaters.

Example 5

There is a solar panel, as shown in Figure 4, which provides 120W of power for a 10Ω load. Calculate the current (I) flowing through the load and the voltage drop (V) across the load.

I=√P/R=√120/10A =3.464A

V= P/I=120/3.464V =34.6V

**②Power meter**You can use a power meter to measure power. The power meter can measure the power consumption of the resistive element in the load. Resistive power is called active power, and its unit of measurement is W. A basic power meter has two coils, namely a current coil and a voltage coil (see Figure 5).

Figure 6 shows the schematic diagram of the connection between the power meter and the load. The current coil is installed on a fixed electromagnet, and the voltage coil is connected with a rotatable hairspring to deflect the pointer of the instrument. The deflection angle of the pointer is proportional to the current passing through the load and the voltage across the load. The resistance in series with the voltage coil acts to limit the coil current.

**③kW·hour and megawatt·hour**When electric power companies charge users for electricity, kilowatt·hour (kW·h) and megawatt·hour (MW·h) are used as energy units. 1kW is equal to 1000W, and 1MW is equal to 1000000W.

1kW·h = 1000W ×1h

2kW·h = 1000W×2h

1MW·h = 1000000W×1h

2MW·h = 1000000W ×2h

Divide the wattage by 1000 to get the kilowattage. If a circuit consumes 12960W·h of electric energy, the electricity company charges 12960/1000=12.960kW·h.

Example 6

How much energy does a 1.5kW electric heater consume for 3 hours?

Energy=1.5×3kW·h=4.5kW·h

Example 7

Divide the number of watts by 10000000 to get the number of megawatts.

If the power generation power of the power company is 2000000W, it can be expressed as 2000000/1000000, that is, 2MW.

As shown in Figure 7, the measurement units of power and energy can be converted according to the application. exist

In the electrical field, the unit of measurement of power is usually W and horsepower (hp), and the unit of measurement of energy is W·h.

Electricity companies use kW·h as the unit for their meters to measure electricity.

**④Power efficiency**An important indicator of the power system is efficiency. Power efficiency refers to the ratio of output power to input power.

Efficiency = output power / input power

Many electrical equipment, such as power supply units, transformers, and DC-AC inverters, require power input. For example, plug a power supply into a 120V socket, transform the voltage to 240V and then output (see Figure 8). If the input power is 600w and the output power is 570w, what is the power efficiency?

Efficiency = output power / input power = 570/600 = 0.95, or 95%

The working efficiency of electrical equipment cannot reach 100%. The efficiency reflects the energy loss of the equipment itself. The power supply device shown in Figure 8 loses 5% of the input power due to internal losses. The main reason why the output power is less than the input power is that electrical equipment needs to consume energy to drive internal circuits, generate magnetic fields, or lose heat due to circuit resistance.