8.5. Thermal effect of current

8.5.1. Current source power

The total power of the current source:

P full = P useful + P losses,

where P is useful - useful power, P is useful \u003d I 2 R; P loss - power loss, P loss = I 2 r ; I - current strength in the circuit; R - load resistance (external circuit); r is the internal resistance of the current source.

Apparent power can be calculated using one of three formulas:

P full \u003d I 2 (R + r), P full \u003d ℰ 2 R + r, P full \u003d I ℰ,

where ℰ is the electromotive force (EMF) of the current source.

Net power is the power that is released in the external circuit, i.e. on the load (resistor), and can be used for some purpose.

Net power can be calculated using one of three formulas:

P useful \u003d I 2 R, P useful \u003d U 2 R, P useful \u003d IU,

where I is the current in the circuit; U - voltage at the terminals (terminals) of the current source; R - load resistance (external circuit).

Loss power is the power that is released in the current source, i.e. in the internal circuit, and is spent on the processes taking place in the source itself; for some other purpose, the power loss cannot be used.

The power loss is usually calculated by the formula

P loss = I 2 r ,

where I is the current in the circuit; r is the internal resistance of the current source.

In the event of a short circuit, the useful power goes to zero

P useful = 0,

since there is no load resistance in the event of a short circuit: R = 0.

The apparent power in the event of a short circuit of the source coincides with the power losses and is calculated by the formula

P full \u003d ℰ 2 r,

where ℰ is the electromotive force (EMF) of the current source; r is the internal resistance of the current source.

Net power has maximum value in the case when the load resistance R is equal to the internal resistance r of the current source:

R = r.

Maximum useful power:

P useful max = 0.5 P full,

where P is full - full power current source; P full \u003d ℰ 2 / 2 r.

Explicitly, the formula for calculating maximum useful power as follows:

P useful max = ℰ 2 4 r .

To simplify the calculations, it is useful to remember two points:

if with two load resistances R 1 and R 2 the same useful power is allocated in the circuit, then internal resistance current source r is related to the indicated resistances by the formula

r = R 1 R 2 ;

if the maximum useful power is released in the circuit, then the current I * in the circuit is doubled less power short circuit current i :

I * = i 2 .

Example 15. When shorted to a resistance of 5.0 ohms, a battery of cells produces a current of 2.0 A. The short circuit current of the battery is 12 A. Calculate the maximum useful power of the battery.

Solution . Let's analyze the condition of the problem.

1. When a battery is connected to a resistance R 1 = 5.0 Ohm, a current of I 1 = 2.0 A flows in the circuit, as shown in fig. a , defined by Ohm's law for a complete chain:

I 1 \u003d ℰ R 1 + r,

where ℰ is the EMF of the current source; r is the internal resistance of the current source.

2. When a battery is short-circuited, a short-circuit current flows in the circuit as shown in fig. b. The strength of the short circuit current is determined by the formula

where i is the short circuit current, i = 12 A.

3. When the battery is connected to the resistance R 2 \u003d r, a current of force I 2 flows in the circuit, as shown in fig. in , defined by Ohm's law for a complete circuit:

I 2 \u003d ℰ R 2 + r \u003d ℰ 2 r;

in this case, the maximum useful power is allocated in the circuit:

P useful max \u003d I 2 2 R 2 \u003d I 2 2 r.

Thus, to calculate the maximum useful power, it is necessary to determine the internal resistance of the current source r and the current strength I 2.

In order to find the current strength I 2, we write down the system of equations:

i \u003d ℰ r, I 2 \u003d ℰ 2 r)

and perform the division of equations:

i I 2 = 2 .

This implies:

I 2 \u003d i 2 \u003d 12 2 \u003d 6.0 A.

In order to find the internal resistance of the source r, we write down the system of equations:

I 1 \u003d ℰ R 1 + r, i \u003d ℰ r)

and perform the division of equations:

I 1 i = r R 1 + r .

This implies:

r \u003d I 1 R 1 i - I 1 \u003d 2.0 ⋅ 5.0 12 - 2.0 \u003d 1.0 Ohm.

Calculate the maximum useful power:

P useful max \u003d I 2 2 r \u003d 6.0 2 ⋅ 1.0 \u003d 36 W.

Thus, the maximum useful power of the battery is 36 watts.

Consider a closed unbranched circuit consisting of a current source and a resistor.

We apply the law of conservation of energy to the entire circuit:

Because , and for a closed circuit, points 1 and 2 coincide, the power of electric forces in a closed circuit is zero. This is equivalent to the assertion of the potentiality electric field direct current, which was already mentioned earlier.

So in In a closed circuit, all heat is released due to the work of external forces:, or , and we again come to Ohm's law, now for a closed circuit: .

full power chains call the power of external forces, it is also equal to the total thermal power:

Useful call the thermal power released in the external circuit (regardless of whether it is useful or harmful in this particular case):

(3).

The role of electrical forces in a circuit. In an external circuit, on a load R, electric forces do positive work, and when the charge moves inside the current source, they do the same negative work. In the external circuit, heat is released due to the work of the electric field. The work given in the outer chain, electric field“returns” to itself inside the current source. As a result, all the heat in the circuit is “paid for” by the work of external forces: the current source gradually loses the chemical (or some other) energy stored in it. The electric field plays the role of a "courier" delivering energy to the external circuit.

The dependence of the total, useful power and efficiency on the load resistance R .

These dependencies are obtained from formulas (1 - 2) and Ohm's law for a complete circuit:

. (4)

. (5)

You can see the graphs of these dependencies in the figure.

The total power monotonically decreases with increasing , because decreasing current in the circuit. Maximum Gross Power stands out at , i.e. at short circuit. The current source does the maximum work per unit of time, but all of it goes to heat the source itself. The maximum apparent power is

The useful power has a maximum at (which you can see by taking the derivative of the function (5) and equating it to zero). Substituting into expression (5), we find the maximum useful power:

The power developed by the current source in the entire circuit is called full power.

It is determined by the formula

Thus, the efficiency depends on the ratio between the internal resistance of the source and the resistance of the consumer.

It is customary to express the electrical efficiency as a percentage.

For practical electrical engineering, two questions are of particular interest:

1. The condition for obtaining the greatest useful power

2. The condition for obtaining the highest efficiency

The condition for obtaining the highest useful power (power in the load)

The electric current develops the greatest useful power (power at the load) if the load resistance is equal to the resistance of the current source.

This highest power equal to half of the total power (50%) developed by the current source in the entire circuit.

Half of the power is developed at the load and half is developed at the internal resistance of the current source.

If we reduce the load resistance, then the power developed at the load will decrease and the power developed at the internal resistance of the current source will increase.

If the load resistance is zero, then the current in the circuit will be maximum, this short circuit mode (short circuit) . Almost all power will be developed on the internal resistance of the current source. This mode is dangerous for the current source as well as for the entire circuit.

If we increase the load resistance, then the current in the circuit will decrease, the power at the load will also decrease. With a very large load resistance, there will be no current in the circuit at all. This resistance is called infinitely large. If the circuit is open, then its resistance is infinitely large. This mode is called idle mode.

Thus, in modes close to a short circuit and to idling, the useful power is small in the first case due to the low voltage value, and in the second due to the small current value.

The condition for obtaining the highest efficiency coefficient

Efficiency (efficiency) is equal to 100% at idling(in this case, useful power is not allocated, but at the same time, the power of the source is not consumed).

As the load current increases, the efficiency decreases in a straight line.

In the short circuit mode, the efficiency is equal to zero (there is no useful power, and the power developed by the source is completely consumed inside it).

Summarizing the above, we can draw conclusions.

The condition for obtaining the maximum useful power (R=R 0) and the condition for obtaining the maximum efficiency (R=∞) do not match. Moreover, when receiving the maximum useful power from the source (matched load mode), the efficiency is 50%, i.e. half of the power developed by the source is wasted inside it.

In powerful electrical installations, the matched load mode is unacceptable, since this results in a useless expenditure of large powers. Therefore, for power stations and substations, the operating modes of generators, transformers, rectifiers are calculated so as to ensure high efficiency (90% or more).

The situation is different in the technique of weak currents. Take, for example, a telephone. When talking in front of a microphone, an electrical signal with a power of about 2 mW is created in the circuit of the apparatus. It is obvious that in order to obtain the greatest communication range, it is necessary to transfer as much power as possible to the line, and for this it is required to perform a coordinated load switching mode. Does efficiency matter in this case? Of course not, since energy losses are calculated in fractions or units of milliwatts.

The matched load mode is used in radio equipment. In the event that a consistent mode is not provided with a direct connection between the generator and the load, measures are used to match their resistances.

Goal of the work: determine the EMF of a DC source by the compensation method, useful power and efficiency depending on the load resistance.

Equipment: current source under test, stabilized voltage source, resistance box, milliammeter, galvanometer.

THEORETICAL INTRODUCTION

Current sources are devices in which the conversion takes place various kinds energy (mechanical, chemical, thermal) in electrical energy. Separation occurs in current sources electric charges different sign. Therefore, if the source is closed to a load, for example, to a conductor, then an electric current will flow through the conductor, caused by the movement of charges under the influence of an electrostatic field. The direction of movement of positive charges is taken as the direction of current. That is, the current will flow from the positive pole of the source through the conductor to the negative. But through the source, the charges move against the forces of the electrostatic field. This can only happen under the action of forces of a non-electrostatic nature, the so-called external forces. For example, the Lorentz magnetic force in generators of power plants, diffusion forces in chemical current sources.

The characteristic of the current source is the electromotive force - EMF. It is equal to the ratio of the work of external forces to the value of the transferred charge:

Consider an electrical circuit from a current source with internal resistance r, closed to the load by resistance R. According to the law of conservation of energy, the work of external forces with stationary conductors, it turns into heat released on the load and the internal resistance of the source itself. According to the Joule-Lenz law, the heat released in the conductor is equal to the product of the square of the current strength and the resistance and the current flow time. Then . After shortening to Jt we get that the current strength in the circuit is equal to the ratio of the EMF to the total resistance electrical circuit:

. (2)

This is Ohm's law for a complete circuit. In the absence of current through the source, there is no voltage drop across the internal resistance and the EMF is equal to the voltage between the poles of the source. The unit of measurement for EMF, like voltage, is the volt (V).

EMF can be measured various methods. If, in the simplest case, a voltmeter with resistance R connect to source poles with internal resistance r, then, according to Ohm's law, the readings of the voltmeter will be . This is less than the EMF by the amount of voltage drop across the internal resistance.

In the compensation method for measuring EMF, no current flows through the source (Fig. 1). If using the PSU power supply regulator to select the voltage on the resistance box R exactly equal to the EMF of the source, then the current through the source and through the galvanometer G won't leak. Then the EMF of the source will be equal to the voltage drop across the resistance box

E = J R. (3)

The useful power of the current source with fixed conductors is thermal power allocated to the load. According to the Joule-Lenz law P \u003d J 2 R. Substituting the current strength, according to Ohm's law (2), we obtain the formula for the dependence of useful power on load resistance:

. (4)

In short circuit mode with no load, when R= 0, all the heat is released on the internal resistance and the useful power is zero (Fig. 2). As the load resistance increases, R<<r, useful power increases almost in direct proportion to the resistance R. With a further increase in load resistance, the current is limited, and the power, having reached a maximum, begins to decline. At high values of load resistance ( R>>r), power decreases inversely with resistance, tending to zero when the circuit is broken.

The maximum power corresponds to the condition that the first derivative of thermal power with respect to resistance is equal to zero. Differentiating (4), we obtain

. It follows that the useful power is maximum if R = r. Substituting into (4), we get

The operation of the current source is characterized by the efficiency. This, by definition, is the ratio of useful work to the total work of the current source: . After reduction, the efficiency formula will take the form

.(5)

In short circuit mode R= 0, the efficiency is zero, since the net power is zero. With an increase in load resistance, the efficiency increases and tends to 100% at high resistance values ( R>>r).

COMPLETING OF THE WORK

1. Set the operation mode switch to the "EMF" position. Install a resistance of 500 ohms on the store, the measurement limit of the milliammeter is 3 mA. Briefly press the button TO and notice how the galvanometer needle deviates when current flows from the source under study.

Connect the power supply to the 220 V network.

2. Press the button TO turning on the current through the galvanometer. If the galvanometer needle deviates in the same way as when only the current source is turned on, then increase the current from the power supply, controlling it with a milliammeter. If the arrow deviates in the opposite direction, then reduce the amperage of the power supply. Record the resistance value and current strength in the table. 1 .

Repeat the measurements at least five times, changing the resistance in the range of 500 - 3000 Ohm. Record the results in table. 1

3. Set the measurement mode switch to the "Power" position. Set the magazine resistance to 500 ohms. Measure the current with a milliammeter. Record the result in table. 2.

Repeat the measurements at least five times, changing the resistance in the range of 500 - 3000 Ohm. Record the results in table. 2.

Disconnect the power supply from the network.

table 2

5. Estimate the random error in measuring the EMF using the error formula for direct measurements , Where n is the number of measurements.

9. Construct graphs of useful power and efficiency versus load resistance. The size of the chart is at least half a page. Specify a uniform scale on the coordinate axes. Draw smooth curves near the points so that the deviations of the points from the lines are minimal.

10. Draw conclusions. Write result E = ± d E, P = 90%.

CONTROL QUESTIONS

1. Explain the role of a current source in an electrical circuit. Define the electromotive force of a current source (EMF).

2. Deduce, using the law of conservation of energy, and formulate Ohm's law for a complete circuit.

3. Explain the essence of the compensation method for measuring EMF. Is it possible to measure the EMF of a current source with a voltmeter?

4. Derive the formula for the useful power of the current source. Draw a graph of the dependence of useful power on the value of load resistance, explain this dependence.

5. Derive the condition for the maximum power of the current source.

6. Derive the formula for the efficiency of the current source. Draw a graph of the dependence of efficiency on the load resistance of the current source. Explain this relationship.

There are two types of elements in an electrical or electronic circuit: passive and active. The active element is able to continuously supply energy to the circuit - battery, generator. Passive elements - resistors, capacitors, inductors, only consume energy.

What is a current source

A power source is a device that continuously supplies electricity to a circuit. It can be a source of direct current and alternating current. Rechargeable batteries are direct current sources, and the electrical outlet is alternating.

One of most interesting characteristics power sources– they are able to convert non-electrical energy into electrical energy, for example:

chemical in batteries;
mechanical in generators;
solar, etc.

Electrical sources are divided into:

Independent;
Dependent (controlled), the output of which depends on the voltage or current elsewhere in the circuit, which can be either constant or changing over time. Used as equivalent IP for electronic devices.

When talking about circuit laws and analysis, electrical power supplies are often viewed as ideal, that is, theoretically capable of providing an infinite amount of energy without loss, while having the characteristics represented by a straight line. However, in real, or practical, sources, there is always an internal resistance that affects their output.

Important! Power supplies can only be connected in parallel if they have the same voltage value. The series connection will affect the output voltage rating.

The internal resistance of the power supply is represented as being connected in series with the circuit.

Current source power and internal resistance

Let it be considered simple circuit, in which the battery has an EMF E and an internal resistance r and supplies current I to an external resistor of resistance R. The external resistor can be any active load. The main purpose of the circuit is to transfer energy from the battery to the load, where it does something useful, like lighting a room.

You can derive the dependence of useful power on resistance:

The equivalent resistance of the circuit is R + r (since the load resistance is connected in series with the external load);
The current flowing in the circuit will be determined by the expression:

EMF output power:

Rout. = E x I = E²/(R + r);

Power dissipated as heat, with internal battery resistance:

Pr = I² x r = E² x r/(R + r)²;

Power transferred to the load:

P(R) = I² x R = E² x R/(R + r)²;

Rout. = Pr + P(R).

Thus, some of the battery's output energy is immediately lost due to heat dissipation on the internal resistance.

Now you can plot P(R) versus R and find out at what load the useful power will take on a maximum value. When analyzing the function for an extremum, it turns out that as R increases, P(R) will also increase monotonically until the point when R does not equal r. At this point, the useful power will be maximum, and then begins to decrease monotonically with a further increase in R.

P(R)max = E²/4r when R = r. In this case, I = E/2r.

Important! This is a very significant result in electrical engineering. Power transfer between the power supply and an external load is most efficient when the load resistance matches the internal resistance of the current source.

If the load resistance is too high, then the current flowing through the circuit is small enough to transfer energy to the load at an appreciable rate. If the load resistance is too low, then most of output energy is dissipated as heat within the power supply itself.

This condition is called agreement. One example of matching source impedance and external load is an audio amplifier and loudspeaker. The output impedance Zout of the amplifier is set from 4 to 8 ohms, and the nominal input impedance of the speaker Zin is only 8 ohms. Then, if an 8 ohm speaker is connected to the output of the amplifier, it will see the speaker as an 8 ohm load. Connecting two 8 ohm speakers in parallel to each other is equivalent to an amplifier driving a single 4 ohm speaker, and both configurations are within the amplifier's output specifications.

Current source efficiency

When doing work electric shock energy transformations take place. Full work, performed by the source, goes to energy conversions in the entire electrical circuit, and useful - only in the circuit connected to the IP.

A quantitative assessment of the efficiency of the current source is carried out according to the most significant indicator that determines the speed of work, – power:

Not all the output power of the IP is used by the energy consumer. The ratio of energy consumed and issued by the source is the formula for the efficiency factor:

η = useful power/output power = Ppol/Pout

Important! Since Ppol. in almost any case, it is less than Pout, η cannot be greater than 1.

This formula can be transformed by substituting expressions for powers:

Source output power:

Rout. = I x E = I² x (R + r) x t;

Consumed energy:

Rpol. = I x U = I² x R x t;

Coefficient:

η = Рpol./Рout. = (I² x R x t)/(I² x (R + r) x t) = R/(R + r).

That is, for a current source, the efficiency is determined by the ratio of resistances: internal and load.

Often, the efficiency indicator is used as a percentage. Then the formula will take the form:

η = R/(R + r) x 100%.

It can be seen from the expression obtained that, subject to the matching condition (R = r), the coefficient η = (R/2 x R) x 100% = 50%. When the transmitted energy is most efficient, the efficiency of the IP itself is only 50%.

Using this coefficient, the efficiency of various IP and electricity consumers is evaluated.

Examples of efficiency values: