When solving problems involving a mixed connection of conductors, you must try to transform the circuit and replace parallel and series connected conductors with equivalent conductors.
In the above example, it should be borne in mind that the first and second conductors cannot be considered connected in series, since there is a branch at the point of their connection. For the same reason, conductors 1–3 and 4–5 cannot be considered connected in series.
Conductor 1 and conductor 2,3 are connected in series. They can also be replaced by one equivalent conductor, the resistance of which is equal to the sum of the resistances of conductors 1 and 2.3. Having found this resistance, we again draw the transformed circuit. In this circuit, conductor 1,2,3 is connected in parallel with conductor 5. The resistance of parallel-connected conductors can also be calculated using a well-known formula and replaced with one conductor with an equivalent resistance of 1,2,3,5.
For example, if we were dealing with four conductors connected as shown in the diagram, the problem would be solved in an elementary way. Pairs of conductors 1,2 and 3,4 are connected in series. They can be replaced with equivalent conductors. These equivalent conductors are connected in parallel and can also be easily replaced by one common conductor. (If the resistance of the conductors were equal to 10 ohms each, then the total resistance of the circuit would also be equal to 10 ohms).
Let a current of force I 0 flow into point A. At this point the current branches. Some of it flows through the upper part of the chain, part through the lower. It may happen that the current that flows through the upper and lower sections is the same.
In problems involving the calculation of electrical circuits, it is useful to draw an analogy between electric current and water flow in pipes. Let's try to mentally carry out such a replacement in the problem under consideration.
For simplicity, let pipes 1, 2, 3, 4 be the same in cross-section and length. Equal currents flow through two parallel branches. Then the pipes converge into one pipe. It is obvious that the current flowing in is equal to the current flowing out. If you put an isthmus connecting two pipelines, then into this isthmus, due to the equality of pressures on both sides, water will not flow in either direction, no matter what the isthmus is. This isthmus can well be excluded from consideration of the process.
The same is true in electrical circuits. If it turns out that the potentials of points C and D are equal to each other, then there will be no current through conductor 5.
Thus, when we reach a fundamentally non-transformable electrical circuit, we must try to find points with equal potentials in this circuit. If this can be done, then any conductor connecting these points can be excluded from the circuit. Also, points with equal potentials can be connected to each other by any conductor, including one with zero resistance.
In this case, the potentials of points C and D will be equal if the resistances of conductors 1–4 are equal.
The resistances of conductors 1 and 3, 2 and 4 may be equal. All the same, the current strengths in the upper and lower branches will be equal to each other. The voltage drops on conductors 1 and 3, 2 and 4 will also be equal to each other, so there will be no current in the circuit of resistor 5. Because of this, resistor 5, regardless of its resistance, can be discarded from consideration.
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However, it may turn out that the potentials of points C and D are not equal to each other. Then the flow of currents I 1 and I 3 should be considered further. Let's assume that the current I 1 > I 3. I 1 reaches point C and branches further. Part of the current goes through resistor 2, and part through resistor 5. Currents I 4 and I 3 converge at point D. These currents go further through resistor 4, so current I 5 is equal to the sum of currents I 4 and I 3. Current I 5 merges with current I 2 and forms a current equal to the original current I 0 .
Thus we conclude the following.
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Next, you need to select closed contours in the chain. To do this, take an arbitrary point and begin moving along the chain so as to return to this point. When walking around, you must stick to one direction. The number of circuits must be such that all circuit elements can be bypassed.
If there are no current sources in the circuit, then the sum of the voltage drops is zero. Let's go around elements 1–5–3, moving clockwise.
The resulting system of equations can be solved for unknown quantities.
1. What is the time it takes for a current of 5 A to pass through a conductor if, at a voltage at its ends of 120 V, an amount of heat equal to 540 kJ is released in the conductor? (Give your answer in seconds.)
2. In an electric heater with constant resistance of the spiral through which it flows D.C., during t amount of heat released Q. If the current strength and time t doubled, then how many times will the amount of heat released in the heater increase?
3. Resistor 1 with an electrical resistance of 3 Ohms and resistor 2 with an electrical resistance of 6 Ohms are connected in series in a DC circuit. What is the ratio of the amount of heat released by resistor 1 to the amount of heat released by resistor 2 in the same time?
4. The figure shows a graph of the dependence of the current in an incandescent lamp on the voltage at its terminals. What is the current power in the lamp at a voltage of 30 V? (Give your answer in watts.)
5.
The student assembled the electrical circuit shown in the figure. What energy will be released in the external part of the circuit when current flows for 10 minutes? (Express your answer in kJ. The necessary data is indicated in the diagram. Consider the ammeter ideal.)
6. A capacitor with a capacity of 1 μF is connected to a current source with an emf of 2 V. How much work was done by the source when charging the capacitor? (Give your answer in µJ.)
7. A capacitor with a capacity of 1 μF is connected to a current source with an emf of 2 V. How much heat will be released in the circuit during the charging of the capacitor? (Give your answer in µJ.) Neglect radiation effects.
8.
A capacitor with a capacity of 1 μF is connected to an ideal current source with an emf of 3 V once through a resistor 
and the second time - through a resistor 
How many times more is the heat generated by the resistor in the second case compared to the first? Neglect radiation.
9.
To a current source with EMF 4 V and internal resistance 
connected the load resistor. What should it be equal for the source efficiency to be 50%? (Give your answer in ohms.)
10.
In the electrical circuit, the diagram of which is shown in the figure, the measuring instruments are ideal, the voltmeter shows a voltage value of 8 V, and the ammeter shows a current value of 2 A. How much heat will be released in the resistor in 1 second? (Give your answer in joules.)
11. The room is illuminated by four identical bulbs connected in parallel. Electricity consumption per hour is Q. What should be the number of parallel-connected light bulbs so that the electricity consumption per hour is equal to 2 Q?
12. An electric kettle with a power of 2.2 kW is designed to be connected to an electrical network with a voltage of 220 V. Determine the current strength in the heating element of the kettle when it is operating in such a network. Give your answer in amperes.
13. On the body of the electric roaster there is an inscription: “220 V, 660 W.” Find the current consumed by the roaster. (Give your answer in amperes.)
14. On the base of an electric incandescent lamp it is written: “220 V, 60 W.” Two such lamps are connected in parallel and connected to a voltage of 127 V. What power will be released in these two lamps with this connection method? (Give your answer in watts, rounded to the nearest whole number.) When solving the problem, assume that the resistance of the lamp does not depend on the voltage applied to it.
15. On the base of an electric incandescent lamp it is written: “220 V, 100 W.” Three such lamps are connected in parallel and connected to a voltage of 127 V. What power will be released in these three lamps with this connection method? (Give your answer in watts, rounded to the nearest whole number.) When solving the problem, assume that the resistance of the lamp does not depend on the voltage applied to it.
16. There are two round conductors in the school laboratory. The resistivity of the first conductor is 2 times greater than the resistivity of the second conductor. The length of the first conductor is 2 times the length of the second. When these conductors are connected to the same constant voltage sources at the same time intervals, the amount of heat released in the second conductor is 4 times greater than in the first. What is the ratio of the radius of the second conductor to the radius of the first conductor?
17. There are two round conductors in the school laboratory. The resistivity of the first conductor is 2 times greater than the resistivity of the second conductor. The length of the first conductor is 2 times the length of the second. When these conductors are connected to the same constant voltage sources at the same time intervals, the amount of heat released in the second conductor is 4 times less than in the first. What is the ratio of the radius of the first conductor to the radius of the second conductor?
18.
R 1, included in the electrical circuit, the diagram of which is shown in the figure? (Answer in watts.) R 1 = 3 Ohm, R 2 = 2 Ohm, R
19.
How much power is released in the resistor R 2, included in the electrical circuit, the diagram of which is shown in the figure? (Answer in watts.) R 1 = 3 Ohm, R 2 = 2 Ohm, R 3 = 1 Ohm, source emf 5 V, internal resistance of the source is negligible.
20.
R= 16 Ohm, and the voltage between the points A And B equal to 8 V? Give your answer in watts.
21.
What power is released in the section of the circuit, the diagram of which is shown in the figure, if R= 27 Ohm, and the voltage between the points A And B equal to 9 V? Give your answer in watts.
22. I= 6 A. What is the current strength shown by the ammeter? (Give your answer in amperes.) Neglect the resistance of the ammeter.
23.
A resistor with resistance is connected to a current source with EMF and internal resistance. If you connect this resistor to a current source with EMF 
and internal resistance 
then how many times will the power released in this resistor increase?
24.
I U on the lamp. Such a lamp is connected to a constant voltage source of 2 V. How much work will it do? electricity into a lamp filament in 5 seconds? Express your answer in J.
25.
The graph shows the experimentally obtained dependence of the current strength I, flowing through an incandescent lamp, from voltage U on the lamp. Such a lamp is connected to a constant voltage source of 4 V. How much work will the electric current do in the filament of the lamp in 10 seconds? Express your answer in J.
26. A direct current flows through a section of the circuit (see figure) I= 4 A. What current will be shown by an ideal ammeter connected to this circuit if the resistance of each resistor r= 1 Ohm? Express your answer in amperes.
27. A point positive charge of 2 µC is placed between two extended plates, uniformly charged with opposite charges. The modulus of the electric field strength created by a positively charged plate is 10 3 kV/m, and the field created by a negatively charged plate is 2 times greater. Determine the magnitude of the electric force that will act on the indicated point charge.
28. A point positive charge of 2 µC is placed between two extended plates, uniformly charged positive charges. The modulus of the electric field strength created by one plate is equal to 10 3 kV/m, and the fields created second plate, 2 times larger. Determine the magnitude of the electric force that will act on the indicated point charge. Give your answer in newtons.
29.
WITH, resistor resistance R and key K. The capacitor is charged to voltage U= 20 V. The charge on the capacitor plates is q= 10 –6 Cl. How much heat will be released in the resistor after switch K is closed? Express your answer in mJ.
30.
The figure shows a diagram of an electrical circuit consisting of a capacitor with a capacity WITH, resistor resistance R and key K. Capacitance of the capacitor C= 1 µF and it is charged to voltage U= 10 V. How much heat will be released in the resistor after closing key K? Express your answer in mJ.
31. The fuse of the electricity meter in a residential network with a voltage of 220 V is equipped with the inscription: “6 A”. What is the maximum total power of electrical appliances that can be connected to the network at the same time without the fuse melting? (Give your answer in watts)
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 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, the efficiency reaches its 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 it 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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5. Set the external circuit resistance 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 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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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.
Let's consider the energy relationships in a closed DC circuit. In Fig. 106, a closed DC circuit was presented, powered by an element e. d.s. Ш and with internal resistance, we denote the external resistance of the circuit by R. The total power released in the circuit will be the sum of the powers released in the external and internal parts of the circuit:
W = l1R-rriR№ = ]i(R-:-Rll),
or, since by formula (For) § 164 I (R-(- R0) - £, then
Thus, the total power released in the circuit is expressed by the product of the current strength and e. d.s. element. This power is released due to any third-party energy sources; such energy sources can be, for example, chemical reactions occurring in the element.
Consequently, in a direct current circuit, external forces develop positive power 1Sh.
6 S. Frisch to A. Tiyoreva
is closed by an external resistance R-, we determine the dependence on R of the following "values: the total Power W released in the circuit, the power Wa released in the external part of the circuit, and the efficiency % which is numerically equal to the ratio of the power released in the external part of the circuit , to all power.
The current strength I in value is expressed according to Ohm’s law by the relation:
It reaches its greatest value at R = 0; in this case the current is called short circuit current, its strength is equal to:
As the external resistance increases, the current decreases, tending asymptotically to zero with an infinite increase in external resistance (see Fig. 108).
The total power released in the circuit will be:
It reaches its greatest value at short circuit current (R = 0):
Rice. 108. Dependence of current strength.„, Ш3
from external resistance. Wmax-^g
As R increases, the power decreases, tending asymptotically to zero as R increases indefinitely.
The power released in the external part of the circuit is equal to:
At short circuit current R = 0, whence the power released in the external part of the circuit is equal to zero. Wa reaches its greatest value at R = R(I, i.e. when the external resistance is equal to the internal one. In this case
i.e. equal to a quarter of the power during a short circuit.
To make sure that the maximum power Wa is obtained at R=Rt>, let’s take the derivative of Wa with respect to external resistance:
1- -(R*-R*) dR (R + Ro)4
According to the maximum condition, the first derivative must be equal to zero:
Where does R = Ra.
We can verify that under this condition we obtain a maximum and not a minimum for Wa by determining the sign of the second derivative.
With an infinite increase in external resistance, the power released in the external circuit tends to zero.
We determine the efficiency by the ratio of the power Wa released in the external part of the circuit to the total power W:
For R = 0 we have -rj = 0; with increasing R, efficiency t) increases, tending to the value i] =;l with an unlimited increase in R, however, the power released in the external circuit tends to zero, therefore the condition for maximum efficiency with not interesting from a practical point of view.
In Fig. 109 curve / gives the dependence of the power Wa released in the external part of the circuit on the resistance of the external part of the circuit R-, curve 2 "gives the dependence on R of the total power W; finally, curve 3 gives the variation of the efficiency sch from the same external resistance R. As can be seen, "] increases with increasing R.
The most interesting, from a practical point of view, power Wa, released in the external part of the circuit, first increases, and then, having reached a maximum at R = R(), begins to decrease.
At R = R0, when Wa has a maximum, =
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