8.5. Thermal effect of current
8.5.1. Current source power
Total power of the current source:
P total = P useful + P losses,
where P useful - useful power, P useful = I 2 R; P losses - power losses, P losses = 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 = I 2 (R + r), P full = ℰ 2 R + r, P full = I ℰ,
where ℰ is the electromotive force (EMF) of the current source.
Net power- this is the power that is released in the external circuit, i.e. on a load (resistor), and can be used for some purposes.
Net power can be calculated using one of three formulas:
P useful = I 2 R, P useful = U 2 R, P useful = IU,
where I is the current strength in the circuit; U is the voltage at the terminals (clamps) of the current source; R - load resistance (external circuit).
Power loss is the power that is released in the current source, i.e. in the internal circuit, and is spent on processes taking place in the source itself; The power loss cannot be used for any other purposes.
Power loss is usually calculated using the formula
P losses = I 2 r,
where I is the current strength in the circuit; r is the internal resistance of the current source.
During 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 total power during a short circuit of the source coincides with the loss power and is calculated by the formula
P full = ℰ 2 r,
where ℰ is the electromotive force (EMF) of the current source; r is the internal resistance of the current source.
Useful 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 Ptot is the total power of the current source;
P full = ℰ 2 / 2 r. Explicit formula for calculation maximum useful power
as follows:
P useful max = ℰ 2 4 r .
- if with two load resistances R 1 and R 2 the same useful power is released 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 half the 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 us analyze the condition of the problem.
1. When a battery is connected to a resistance R 1 = 5.0 Ohm, a current of strength I 1 = 2.0 A flows in the circuit, as shown in Fig. a, determined by Ohm’s law for the complete circuit:
I 1 = ℰ R 1 + r,
where ℰ - EMF of the current source; r is the internal resistance of the current source.
2. When the battery is short-circuited, a short-circuit current flows in the circuit, as shown in Fig. b. The short circuit current is determined by the formula
where i is the short circuit current, i = 12 A.
3. When a battery is connected to a resistance R 2 = r, a current of force I 2 flows in the circuit, as shown in Fig. in , determined by Ohm’s law for the complete circuit:
I 2 = ℰ R 2 + r = ℰ 2 r;
in this case, the maximum useful power is released in the circuit:
P useful max = I 2 2 R 2 = 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 the system of equations:
i = ℰ r, I 2 = ℰ 2 r)
and divide the equations:
i I 2 = 2 .
This implies:
I 2 = i 2 = 12 2 = 6.0 A.
In order to find the internal resistance of the source r, we write the system of equations:
I 1 = ℰ R 1 + r, i = ℰ r)
and divide the equations:
I 1 i = r R 1 + r .
This implies:
r = I 1 R 1 i − I 1 = 2.0 ⋅ 5.0 12 − 2.0 = 1.0 Ohm.
Let's calculate the maximum useful power:
P useful max = I 2 2 r = 6.0 2 ⋅ 1.0 = 36 W.
Thus, the maximum usable power of the battery is 36 W.
OHM'S LAW FOR A COMPLETE CIRCUIT:
I is the current strength in the circuit; E is the electromotive force of the current source connected to the circuit; R - external circuit resistance; r is the internal resistance of the current source.
POWER DELIVERED IN THE EXTERNAL CIRCUIT
. (2)
From formula (2) it is clear that in the event of a short circuit ( R®0) and at R® this power is zero. For all other final values R power R 1 > 0. Therefore, the function R 1 has a maximum. Meaning R 0, corresponding to the maximum power, can be obtained by differentiating P 1 with respect to R and equating the first derivative to zero:
. (3)
From formula (3), taking into account the fact that R and r are always positive, and E? 0, after simple algebraic transformations we get:
Hence, the power released in the external circuit reaches its greatest value when the resistance of the external circuit is equal to the internal resistance of the current source.
In this case, the current strength in the circuit (5)
equal to half the short circuit current. In this case, the power released in the external circuit reaches its maximum value equal to
When the source is closed to an external resistance, then current flows inside the source and at the same time a certain amount of heat is released at the internal resistance of the source. The power expended to release this heat is equal to
Consequently, the total power released in the entire circuit is determined by the formula
= I 2(R+r) = I.E. (8)
EFFICIENCY
EFFICIENCY current source is equal 
. (9)
From formula (8) it follows that
those. R 1 changes with the change in current in the circuit according to a parabolic law and takes zero values at I = 0 and at . The first value corresponds to an open circuit (R>> r), the second to a short circuit (R<< r). Зависимость к.п.д. от силы тока в цепи с учётом формул (8), (9), (10) примет вид
Thus, efficiency reaches the highest value h = 1 in the case of an open circuit (I = 0), and then decreases according to a linear law, becoming zero in the case of a short circuit.
Dependence of powers P 1, P full = EI and efficiency. current source and the current strength in the circuit are shown in Fig. 1.
Fig.1. I 0 E/r
From the graphs it is clear that to obtain both useful power and efficiency. impossible. When the power released in the external section of the circuit P 1 reaches its greatest value, efficiency. at this moment is 50%.
METHOD AND PROCEDURE OF MEASUREMENTS
Assemble the circuit shown in Fig. on the screen. 2. To do this, first click the left mouse button above the emf button. at the bottom of the screen. Move the mouse marker to the working part of the screen where the dots are located. Click the left mouse button in the working part of the screen where the emf source will be located.
Next, place a resistor in series with the source, representing its internal resistance (by first pressing the button at the bottom of the screen) and an ammeter (the button is in the same place). Then arrange the load resistors and voltmeter in the same way, measuring the voltage across the load.
Connect the connecting wires. To do this, click the wire button at the bottom of the screen, and then move the mouse marker to the working area of the circuit. Click with the left mouse button in the areas of the working area of the screen where the connecting wires should be located.
4. Set parameter values for each element. To do this, left-click on the arrow button. Then click on this element. Move the mouse marker to the slider of the regulator that appears, click on the left mouse button and, holding it down, change the parameter value and set the numerical value indicated in Table 1 for your option.
Table 1. Initial parameters of the electrical circuit
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option |
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5. Set the resistance of the external circuit to 2 Ohms, press the “Count” button and write down the readings of electrical measuring instruments in the corresponding lines of Table 2.
6. Use the regulator slider to consistently increase the resistance of the external circuit by 0.5 Ohms from 2 Ohms to 20 Ohms and, pressing the “Count” button, record the readings of electrical measuring instruments in Table 2.
7. Calculate using formulas (2), (7), (8), (9) P 1, P 2, P total and h for each pair of voltmeter and ammeter readings and write down the calculated values in Table 2.
8. Construct on one sheet of graph paper graphs of the dependence P 1 = f (R), P 2 = f (R), P total = f (R), h = f (R) and U = f (R).
9. Calculate the measurement errors and draw conclusions based on the results of the experiments.
Table 2. Results of measurements and calculations
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P full, VT |
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Questions and tasks for self-control
- Write the Joule-Lenz law in integral and differential forms.
- What is short circuit current?
- What is gross power?
- How is efficiency calculated? current source?
- Prove that the greatest useful power is released when the external and internal resistances of the circuit are equal.
- Is it true that the power released in the internal part of the circuit is constant for a given source?
- A voltmeter was connected to the flashlight battery terminals, which showed 3.5 V.
- Then the voltmeter was disconnected and a lamp was connected in its place, on the base of which it was written: P = 30 W, U = 3.5 V. The lamp did not burn.
- Explain the phenomenon.
- When the battery is alternately shorted to resistances R1 and R2, an equal amount of heat is released in them at the same time. Determine the internal resistance of the battery.
useful power- - [Ya.N.Luginsky, M.S.Fezi Zhilinskaya, Yu.S.Kabirov. English-Russian dictionary of electrical engineering and power engineering, Moscow, 1999] useful power Power (of a machine, equipment, power unit or other technical device)… …
Net power- Useful capacity – power (machine, equipment, power unit or other technical device) given by the device in a certain form and for a certain purpose; equal to total power minus costs... ... Economic and mathematical dictionary
useful power- 3.10 net power: Effective power in kilowatts, obtained on a test bench on the crankshaft end or measured by the method according to GOST R 41.85. Source … Dictionary-reference book of terms of normative and technical documentation
useful power- naudingoji galia statusas T sritis Standartizacija ir metrologija apibrėžtis Galia, susijusi su tam tikros sistemos, įrenginio, aparato ar įtaiso atliekamu naudingu darbu. atitikmenys: engl. net power; useful power vok. Abgabeleistung, f;… … Penkiakalbis aiškinamasis metrologijos terminų žodynas
useful power- naudingoji galia statusas T sritis fizika atitikmenys: engl. net power; useful power vok. Abgabeleistung, f; Nutzabgabe, f; Nutzleistung, f rus. useful power, f pranc. puissance utile, f … Fizikos terminų žodynas
The power that can be obtained at the motor shaft; same as Effective power... Great Soviet Encyclopedia
Net power- – power supplied by a device in a specific form and for a specific purpose. ST IEC 50(151) 78 ... Commercial power generation. Dictionary-reference book
useful pump power- Power supplied by the pump to the supplied liquid medium and determined by the relationship where Q pump flow, m3/s; P pump pressure, Pa; QM pump mass flow, kg/s; LP useful specific work of the pump, J/kg; NP net pump power, W. [GOST... ... Technical Translator's Guide
useful power (in motor vehicles)- net power Power, expressed in kilowatts, obtained on a test bench at the end of the crankshaft or its equivalent and measured in accordance with the power measurement method established in GOST R 41.24. [GOST R 41.49 2003] ... Technical Translator's Guide
useful power in watts- - [A.S. Goldberg. English-Russian energy dictionary. 2006] Energy topics in general EN watts out ... Technical Translator's Guide
The power developed by the current source in the entire circuit is called full power.
It is determined by the formula
where P rev is the total power developed by the current source in the entire circuit, W;
E-uh. d.s. source, in;
I is the magnitude of the current in the circuit, a.
In general, an electrical circuit consists of an external section (load) with resistance R and internal section with resistance R0(resistance of the current source).
Replacing the value of e in the expression for total power. d.s. through the voltages on the sections of the circuit, we get
Magnitude UI corresponds to the power developed on the external section of the circuit (load), and is called useful power P floor =UI.
Magnitude U o I corresponds to the power uselessly spent inside the source, It is called loss power P o =U o I.
Thus, the total power is equal to the sum of the useful power and the loss power P ob =P floor +P 0.
The ratio of useful power to the total power developed by the source is called efficiency, abbreviated as efficiency, and is denoted by η.
From the definition it follows
Under any conditions, efficiency η ≤ 1.
If we express the power in terms of the current and resistance of the circuit sections, we get
Thus, efficiency depends on the relationship between the internal resistance of the source and the resistance of the consumer.
Typically, electrical efficiency is expressed as a percentage.
For practical electrical engineering, two questions are of particular interest:
1. Condition for obtaining the greatest useful power
2. Condition for obtaining the highest efficiency.
Condition for obtaining the greatest useful power (power in 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 maximum power is 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 is short circuit mode (short circuit) . Almost all the power will be developed at the internal resistance of the current source. This mode is dangerous for the current source and also for the entire circuit.
If we increase the load resistance, the current in the circuit will decrease, and the power on the load will also decrease. If the load resistance is very high, there will be no current in the circuit at all. This resistance is called infinitely large. If the circuit is open, its resistance is infinitely large. This mode is called idle mode.
Thus, in modes close to a short circuit and no-load, the useful power is small in the first case due to the low voltage, and in the second due to the low current.
Condition for obtaining the highest efficiency
The efficiency factor (efficiency) is 100% at idle (in this case, no useful power is released, but at the same time, the source power is not consumed).
As the load current increases, efficiency decreases according to a linear law.
In short-circuit mode, the efficiency is zero (there is no useful power, and the power developed by the source is completely consumed within it).
Summarizing the above, we can draw conclusions.
The condition for obtaining maximum useful power (R = R 0) and the condition for obtaining maximum efficiency (R = ∞) do not coincide. Moreover, when receiving 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 wasteful expenditure of large powers. Therefore, for electrical stations and substations, the operating modes of generators, transformers, and rectifiers are calculated so as to ensure high efficiency (90% or more).
The situation is different in weak current technology. Let's take, for example, a telephone. When speaking in front of a microphone, an electrical signal with a power of about 2 mW is created in the device’s circuitry. Obviously, in order to obtain the greatest communication range, it is necessary to transmit as much power as possible into the line, and this requires 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 case where a coordinated mode is not ensured when the generator and load are directly connected, measures are used to match their resistances.
Consider a closed unbranched circuit consisting of a current source and a resistor.
Let us apply the law of conservation of energy to the entire circuit:
Since , and for a closed circuit points 1 and 2 coincide, the power of electrical forces in a closed circuit is zero. This is equivalent to the statement about the potentiality of the direct current electric field, 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 the circuit is called 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):
The role of electrical forces in a circuit. In the external circuit, at the load R, electric forces do positive work, and when moving a charge inside a current source, they do negative work of the same magnitude. In the external circuit, heat is released due to the work of the electric field. The work given in the external circuit is “returned” by the electric field 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.
Dependence of total, useful power and efficiency on load resistance R .
These dependencies are obtained from formulas (1 – 2) and Ohm’s law for the complete chain:
You can see the graphs of these dependencies in the figure.
The total power decreases monotonically with increasing , because the current in the circuit decreases. Maximum apparent power is released at , i.e. at short circuit. The current source does the maximum work per unit of time, but all of it goes to heating the source itself. The maximum apparent power is
The useful power has a maximum at (which you can verify by taking the derivative of function (5) and equating it to zero). Substituting into expression (5), we find the maximum useful power.
