Current power formula. Actual and rated power

It is known that a perpetual motion machine is impossible. This is due to the fact that for any mechanism the following statement is true: the total work done with the help of this mechanism (including heating the mechanism and the environment, overcoming the friction force) is always greater than the useful work.

For example, more than half of the work of an internal combustion engine is wasted on heating the engine components; some heat is carried away by the exhaust gases.

It is often necessary to evaluate the effectiveness of the mechanism and the feasibility of its use. Therefore, in order to calculate what part of the work done is wasted and what part is useful, a special physical quantity is introduced that shows the efficiency of the mechanism.

This value is called the efficiency of the mechanism

The efficiency of a mechanism is equal to the ratio of useful work to total work. Obviously, the efficiency is always less than one. This value is often expressed as a percentage. It is usually denoted by the Greek letter η (read “this”). The efficiency factor is abbreviated as efficiency.

η = (A_full /A_useful) * 100%,

where η efficiency, A_full total work, A_useful useful work.

Among engines, the electric motor has the highest efficiency (up to 98%). The efficiency of internal combustion engines is 20% - 40%, and that of a steam turbine is approximately 30%.

Note that for increasing the efficiency of the mechanism often try to reduce the friction force. This can be done using various lubricants or ball bearings in which sliding friction is replaced by rolling friction.

Examples of efficiency calculations

Let's look at an example. A cyclist weighing 55 kg rode a bicycle weighing 5 kg up a hill 10 m high, doing 8 kJ of work. Find the efficiency of the bicycle. Do not take into account the rolling friction of the wheels on the road.

Solution. Let's find the total mass of the bicycle and the cyclist:

m = 55 kg + 5 kg = 60 kg

Let's find their total weight:

P = mg = 60 kg * 10 N/kg = 600 N

Let's find the work done to lift the bicycle and the cyclist:

Auseful = PS = 600 N * 10 m = 6 kJ

Let's find the efficiency of the bicycle:

A_full / A_useful * 100% = 6 kJ / 8 kJ * 100% = 75%

Answer: The efficiency of the bicycle is 75%.

Let's look at another example. A body of mass m is suspended from the end of the lever arm. A downward force F is applied to the other arm, and its end is lowered by h. Find how much the body rose if the efficiency of the lever is η%.

Solution. Let's find the work done by force F:

η% of this work is done to lift a body of mass m. Consequently, Fhη / 100 was spent on raising the body. Since the weight of the body is equal to mg, the body rose to a height of Fhη / 100 / mg.

As is known, at the moment no mechanisms have yet been created that would completely convert one type of energy into another. During operation, any man-made device spends part of the energy on resisting forces or dissipates it in vain into the environment. The same thing happens in a closed electrical circuit. When charges flow through conductors, the full and useful load of electricity is resisted. To compare their ratios, you will need to calculate the coefficient of performance (efficiency).

Why do you need to calculate efficiency?

The efficiency of an electrical circuit is the ratio of useful heat to total heat.

For clarity, let's give an example. By finding the efficiency of a motor, it is possible to determine whether its primary operating function justifies the cost of electricity consumed. That is, its calculation will give a clear picture of how well the device converts the received energy.

Note! As a rule, efficiency does not have a value, but is a percentage or a numerical equivalent from 0 to 1.

Efficiency is found using a general calculation formula for all devices as a whole. But to get its result in an electrical circuit, you first need to find the force of electricity.

Finding the current in a complete circuit

It is known from physics that any current generator has its own resistance, which is also called internal power. Apart from this meaning, the source of electricity also has its own power.

Let's give values ​​to each element of the chain:

• resistance – r;
• current strength – E;

So, to find the current strength, the designation of which will be I, and the voltage across the resistor - U, it will take time - t, with the passage of charge q = lt.

Due to the fact that the power of electricity is constant, the work of the generator is entirely converted into heat released to R and r. This amount can be calculated using the Joule-Lenz law:

Q = I2 + I2 rt = I2 (R + r) t.

Then the right sides of the formula are equated:

EIt = I2 (R + r) t.

Having carried out the reduction, the calculation is obtained:

By rearranging the formula, the result is:

This final value will be the electrical force in this device.

Having made a preliminary calculation in this way, the efficiency can now be determined.

Calculation of electrical circuit efficiency

The power received from the current source is called consumed, its definition is written - P1. If this physical quantity passes from the generator into the complete circuit, it is considered useful and is written - P2.

To determine the efficiency of a circuit, it is necessary to recall the law of conservation of energy. In accordance with it, the power of the receiver P2 will always be less than the power consumption of P1. This is explained by the fact that during operation in the receiver there is always an inevitable waste of converted energy, which is spent on heating the wires, their sheath, eddy currents, etc.

To find an assessment of the properties of energy conversion, an efficiency is required, which will be equal to the ratio of the powers P2 and P1.

So, knowing all the values ​​of the indicators that make up the electrical circuit, we find its useful and complete operation:

• And useful. = qU = IUt =I2Rt;
• And total = qE = IEt = I2(R+r)t.

In accordance with these values, we find the power of the current source:

• P2 = A useful /t = IU = I2 R;
• P1 = A total /t = IE = I2 (R + r).

Having performed all the steps, we obtain the efficiency formula:

n = A useful / A total = P2 / P1 =U / E = R / (R +r).

This formula turns out that R is above infinity, and n is above 1, but with all this, the current in the circuit remains in a low position, and its useful power is small.

Everyone wants to find increased efficiency. To do this, it is necessary to find conditions under which P2 will be maximum. The optimal values ​​will be:

• P2 = I2 R = (E / R + r)2 R;
• dP2 / dR = (E2 (R + r)2 – 2 (r + R) E2 R) / (R + r)4 = 0;
• E2 ((R + r) -2R) = 0.

In this expression, E and (R + r) are not equal to 0, therefore, the expression in brackets is equal to it, that is, (r = R). Then it turns out that the power has a maximum value, and the efficiency = 50%.

As you can see, you can find the efficiency of an electrical circuit yourself, without resorting to the services of a specialist. The main thing is to maintain consistency in the calculations and not go beyond the given formulas.

Video

That is, different types of energy. In this article we will consider and study such physical concepts as electric current power.

Current power formulas

By current power, as in mechanics, we mean the work that is performed per unit of time. A physical formula will help you calculate power, knowing the work performed by electric current over a certain period of time.

Current, voltage, power in electrostatics are related by equality, which can be derived from the formula A = UIt. It is used to determine the work performed by electric current:

P = A/t = UIt/t = UI
Thus, the formula for direct current power at any section of the circuit is expressed as the product of the current and the voltage between the ends of the section.

Power units

1 W (watt) - current power of 1 A (ampere) in a conductor, between the ends of which a voltage of 1 V (volt) is maintained.

A device for measuring the power of electric current is called a wattmeter. Also, the current power formula allows you to determine power using a voltmeter and ammeter.

An off-system unit of power is kW (kilowatt), GW (gigawatt), mW (milliwatt), etc. Related to this are some off-system units of work that are often used in everyday life, for example (kilowatt hour). Because the 1kW = 10 3 W, and 1h = 3600s, That

1kW · h = 10 3 W 3600 s = 3.6 10 6 W s = 3.6 10 6 J.

Ohm's law and power

Using Ohm's law, current power formula P=UI is written in this form:

P = UI = U 2 /R = I 2 /R
So, the power released on the conductors is directly proportional to the current flowing through the conductor and the voltage at its ends.

Actual and rated power

When measuring power in a consumer, the current power formula allows you to determine its actual value, that is, the one that is actually allocated at a given moment in time at the consumer.

The power ratings are also noted in the data sheets of various electrical appliances. It is called nominal. The passport of an electrical device usually indicates not only the rated power, but also the voltage for which it is designed. However, the voltage in the network may differ slightly from that indicated in the passport, for example, it may increase. As the voltage increases, the current in the network increases, and therefore the current power in the consumer. That is, the actual and rated power of the device may differ. The maximum actual power of the electrical device is greater than the rated power. This is done in order to prevent the device from failure due to minor changes in the voltage in the network.

If the circuit consists of several consumers, then, when calculating their actual power, it should be remembered that for any connection of consumers, the total power in the entire circuit is equal to the sum of the powers of individual consumers.

Efficiency of an electrical appliance

As you know, ideal machines and mechanisms do not exist (that is, those that would completely convert one type of energy into another or generate energy). During operation of the device, part of the expended energy is necessarily spent on overcoming unwanted resistance forces or is simply “dissipated” into the environment. Thus, only part of the energy we expend goes to perform useful work, for which the device was created.

In other words, efficiency shows how efficiently the work expended is used when it is performed, for example, by an electrical appliance.

Efficiency (denoted by the Greek letter η (“this”)) is a physical quantity that characterizes the efficiency of an electrical device and shows what part of the useful work is expended.

Efficiency is determined (as in mechanics) by the formula:

η = A P /A Z ·100%

If the power of the electric current is known, the formulas for determining the CFC will look like this:

η = P P /P Z ·100%

Before determining the efficiency of some device, it is necessary to determine what is useful work (what the device is designed to do) and what is expended work (the work being done or how much energy is expended to do useful work).

Task

An ordinary electric lamp has a power of 60 W and an operating voltage of 220 V. What work does the electric current do in the lamp, and how much do you pay for electricity during the month, at a tariff of T = 28 rubles, using the lamp for 3 hours every day?
What is the current strength in the lamp and the resistance of its coil in working condition?

Solution:

1. To solve this problem:
a) calculate the operating time of the lamp during the month;
b) calculate the work done by the current in the lamp;
c) calculate the monthly fee at the rate of 28 rubles;
d) calculate the current in the lamp;
e) calculate the resistance of the lamp spiral in operating condition.

2. We calculate the work done by the current using the formula:

A = P t

The current strength in the lamp can be calculated using the current power formula:

P = UI;
I = P/U.

The resistance of the lamp coil in operating condition from Ohm’s law is equal to:

[A] = Wh;

[I] = 1B 1A/1B = 1A;

[R] = 1V/1A = 1Ohm.

4. Calculations:

t = 30 days · 3 hours = 90 hours;
A = 60·90 = 5400 Wh = 5.4 kWh;
I = 60/220 = 0.3 A;
R = 220/0.3 = 733 Ohm;
B = 5.4 kWh 28 kW / kWh = 151 rub.

Answer: A = 5.4 kWh; I = 0.3 A; R = 733 Ohm; B = 151 rubles.

In the process of moving charges inside a closed circuit, a certain amount of work is performed by the current source. It can be useful and complete. In the first case, the current source moves charges in the external circuit, while doing work, and in the second case, the charges move throughout the entire circuit. In this process, the efficiency of the current source, defined as the ratio of the external and total resistance of the circuit, is of great importance. If the internal resistance of the source and the external resistance of the load are equal, half of the total power will be lost in the source itself, and the other half will be released at the load. In this case, the efficiency will be 0.5 or 50%.

Electrical circuit efficiency

The efficiency factor under consideration is primarily associated with physical quantities characterizing the speed of conversion or transmission of electricity. Among them, power, measured in watts, comes first. There are several formulas to determine it: P = U x I = U2/R = I2 x R.

Electrical circuits may have different voltages and charge amounts, and accordingly, the work performed is also different in each case. Very often there is a need to estimate the speed at which electricity is transmitted or converted. This speed represents the electrical power corresponding to the work done in a certain unit of time. In the form of a formula, this parameter will look like this: P=A/∆t. Therefore, work is displayed as the product of power and time: A=P∙∆t. The unit of work used is .

In order to determine how efficient a device, machine, electrical circuit or other similar system is in relation to power and operation, efficiency is used. This value is defined as the ratio of usefully expended energy to the total amount of energy entering the system. Efficiency is denoted by the symbol η, and is defined mathematically as the formula: η = A/Q x 100% = [J]/[J] x 100% = [%], in which A is the work performed by the consumer, Q is the energy given by the source . In accordance with the law of conservation of energy, the efficiency value is always equal to or below unity. This means that useful work cannot exceed the amount of energy expended in doing it.

In this way, the power losses in any system or device are determined, as well as the degree of their usefulness. For example, in conductors, power losses occur when electrical current is partially converted into thermal energy. The amount of these losses depends on the resistance of the conductor; they are not part of the useful work.

There is a difference expressed by the formula ∆Q=A-Q, which clearly shows the power loss. Here the relationship between the increase in power losses and the resistance of the conductor is very clearly visible. The most striking example is an incandescent lamp, the efficiency of which does not exceed 15%. The remaining 85% of the power is converted into thermal, that is, infrared radiation.

What is the efficiency of a current source

The considered efficiency of the entire electrical circuit allows us to better understand the physical essence of the efficiency of the current source, the formula of which also consists of various quantities.

In the process of moving electric charges along a closed electrical circuit, a certain amount of work is performed by the current source, which is distinguished as useful and complete. While performing useful work, the current source moves charges in the external circuit. When fully operational, charges, under the influence of a current source, move throughout the entire circuit.

They are displayed as formulas as follows:

• Useful work - Apolez = qU = IUt = I2Rt.
• Total work - Atotal = qε = Iεt = I2(R +r)t.

Based on this, we can derive formulas for the useful and total power of the current source:

• Useful power - Puse = Apoles /t = IU = I2R.
• Total power - Pfull = Afull/t = Iε = I2(R + r).

As a result, the formula for the efficiency of the current source takes the following form:

• η = Apoles/Atoll = Puse/Ptot = U/ε = R/(R + r).

Maximum useful power is achieved at a certain value of external circuit resistance, depending on the characteristics of the current source and load. However, attention should be paid to the incompatibility of maximum net power and maximum efficiency.

Study of power and efficiency of current source

The efficiency of a current source depends on many factors that should be considered in a certain sequence.

To determine, in accordance with Ohm's law, there is the following equation: i = E/(R + r), in which E is the electromotive force of the current source, and r is its internal resistance. These are constant values ​​that do not depend on the variable resistance R. Using them, you can determine the useful power consumed by the electrical circuit:

• W1 = i x U = i2 x R. Here R is the resistance of the electricity consumer, i is the current in the circuit, determined by the previous equation.

Therefore, the power value using the final variables will be shown as: W1 = (E2 x R)/(R + r).

Since it is an intermediate variable, in this case the function W1(R) can be analyzed for its extremum. For this purpose, it is necessary to determine the value of R at which the value of the first derivative of the useful power associated with the variable resistance (R) will be equal to zero: dW1/dR = E2 x [(R + r)2 - 2 x R x (R + r) ] = E2 x (Ri + r) x (R + r - 2 x R) = E2(r - R) = 0 (R + r)4 (R + r)4 (R + r)3

From this formula we can conclude that the value of the derivative can be zero only under one condition: the resistance of the electricity receiver (R) from the current source must reach the value of the internal resistance of the source itself (R => r). Under these conditions, the value of the efficiency factor η will be determined as the ratio of the useful and total power of the current source - W1/W2. Since at the maximum point of useful power the resistance of the energy consumer of the current source will be the same as the internal resistance of the current source itself, in this case the efficiency will be 0.5 or 50%.

Current power and efficiency problems