|
Arabic
Bengali
Chinese
English
French
German
Hebrew
Hindi
Italian
Japanese
Korean
Malay
Polish
Portuguese
Spanish
Turkish
Ukrainian
Vietnamese|
ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING Calculation of complex and branched chains. Encyclopedia of radio electronics and electrical engineering
Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur If two resistors are connected in series (Fig. 6, a), then the same current I flows through them. The voltage drops across the resistors will be: U1 = I R1 and U2 = I R2. The total voltage drop will be U = U1 + U2 = I(R1 + R2). In brackets is the total resistance R = R1 + R2. Thus, when resistors are connected in series, their resistances add up.
Let us turn to the parallel connection (Fig. 6b). Here, the common voltage for both resistors is U, and the total current I branches into currents I1 = U/R1 and I2 = U/R2, with I = I1 + I2. Let's use Ohm's law and express the currents in terms of voltage and resistance in the last formula: U/R = U/R1 + U/R2. Reducing U, we get 1/R = 1/R1 + 1/R2. When resistors are connected in parallel, values are added that are inverse to the resistances - conductivity. It is curious to note that with a series connection, the total resistance is greater than the largest of the summed, and with a parallel connection, it is less than the smallest. The easiest way to deal with the same resistors: connecting N pieces in series, we get the same amount of resistance, and connecting in parallel, the same amount less. The formula for calculating the resistance when resistors are connected in parallel does not cause much enthusiasm; for this case, a very convenient nomogram was invented a long time ago (Fig. 7).
We set aside on a piece of paper in a cell vertically the value of R1 and at any distance on the side - R2. The scale does not matter, one cell can correspond to 10 Ohm or 100 kOhm, it is only important that it be the same. We draw lines along the ruler from the top of one segment to the base of another (dashed in Fig. 7), and the height of their intersection point gives the value of R on the same scale. Using the formulas for parallel and series connection of resistances, it is possible to advance quite far in the calculation of complex circuits consisting only of passive elements. As an abstract example, consider the circuit in Fig. 8a, which is somewhat reminiscent of an avalanche of decay products during the intrusion of a cosmic particle into the Earth's atmosphere. It is required to find the resistance between the upper terminal and the common wire.
Let's start simplifying the circuit by calculating the total resistance of R4, R5 and R6, R7 connected in parallel (Fig. 8, b). Then the calculated values of R4-5 and R6-7 are added to R2 and R3, respectively (serial connections). It turns out a very simple scheme of Fig. 8, c. Having now calculated the total resistance of the lower resistors connected in parallel, we obtain the circuit of Fig. 8, d, in which the calculated value of R2-7 can only be added to R1 (Fig. 8, e) to get the answer. Currents and voltages are found using the simplest Ohm's law for a circuit section, "unwinding" the circuits in the opposite direction. We apply voltage U to the upper output. Dividing it by the total resistance of the circuit, we obtain the total current I (Fig. 8, e). Resistors R1 and resistor R2-7 equivalent to the rest of the circuit form a voltage divider (Fig. 8d), in which U2-7= I R2-7. We obtain currents I1 and I2 by dividing the resulting voltage by the resistances of the corresponding branches (Fig. 8, c), etc. The process is long, but not complicated. For training, calculate in your mind the total resistance of the circuit, if all the resistors are the same, and also, what proportion of the total voltage will be allocated to R7? (Answer: 1,75R, U/7). The method is not applicable if the circuit has transverse (bridge) connections between the branches or there are current or voltage sources in the branches. In this case, Kirchhoff's rules are used to calculate complex circuits. There are two of them: 1. The algebraic sum of the currents in each node is zero. 2. The sum of the voltage drops in each circuit is equal to the sum of the EMF. Recall that a node is a connection of three or more conductors, and a circuit is a closed circuit highlighted in the diagram. When using the Kirchhoff rules, it is necessary to indicate on the diagram the directions of the currents and the direction of bypassing the circuits. The current is considered positive if it flows into the node, and negative if it flows out of the node. If the current coincides with the direction of the circuit bypass, the corresponding voltage drop is considered positive, if the current through the source is directed from - to +, then the EMF is also positive. According to the first rule, no more than Y-1 equations should be composed, where Y is the number of nodes. The remaining equations are compiled according to the second rule, and for convenience, the simplest contours are chosen. The total number of equations corresponds to the number of branches or currents. You can solve equations in any way: substitution, adding and subtracting equations, making matrices, etc. Let us explain what has been said with simple examples. Let us calculate the balance condition of the Wheatstone bridge, the diagram of which with all the necessary notation is shown in Fig. 9.
First of all, note that the current I0 flowing into node A is equal to the current flowing out of node D, since no other conductors are connected to the bridge. When the bridge is in balance, the current I5 through the galvanometer RA is zero. Applying the first rule to points B and C, we obtain I1 = I3 and I2 = I4, and applying it to point A, we find I0 = I1 + I2. For the upper circuit (there is no EMF in it, and the current I5 and the voltage drop across the galvanometer are equal to zero), we have I1 R1 - I2 R2 = 0. Similarly, for the lower circuit I3 R3 - I4 R4 = 0. Replacing I3 with I1 and I4 to I2, then transferring the terms from I2 to the right side, we get I1 R1 = I2 R2, I1 R3 = I2 R4. It remains to divide one equality by another to obtain the well-known bridge balance condition: Kirchhoff's rules will have to be used in the case shown in Fig. 10, when two sources with different EMF and internal resistances work on a common load.
Suppose that all the values of the elements are known, it is necessary to find the current in the load and in each of the sources. We also assume, for definiteness, that we designated the source with a higher EMF as E1. There are two nodes in this circuit, therefore, according to the first rule, we will compose only one equation for node A: I1 + I2 = I3 (try, for fun, to make an equation for another node - nothing new will work). But we need three equations, according to the number of unknown currents. Let's choose simpler contours, so that each circuit includes one source, and write: for I - I1·r1 + I3·R = Е1; for II - I2 r2 + I3 R = E2. Now it remains to substitute the values \uXNUMXb\uXNUMXbof the EMF (in volts) and resistance (in ohms), solve three equations together and find out three currents (in amperes). A curious case is possible when a source with a lower EMF (E2) will not give current at all (a kind of bridge will turn out). Subtract the equation for circuit II from the equation for circuit I and set I2 = 0. We get I1 r1 = E1 - E2. This means that just such a voltage drops across the internal resistance of the first source that the voltage across the load is equal to E2. Naturally, under these conditions, there is no voltage drop across r2 and there is no current through the source. The current I1 = I3 flows into the load. If we now decrease E2 or increase R, the current I2 will flow in the opposite direction to that indicated (the solution for I2 will be negative), i.e., not from the source, but to the source (the battery in place of E2 will be charged). Question for self-test. The terminals of a 3336 battery (it consists of three identical cells connected in series) are short-circuited, and a voltmeter is attached to the middle cell. What will he show? Response. The voltage at the battery terminals is equal to zero according to the condition of the problem (the terminals are closed). The current in the circuit of elements is equal to the short circuit current: I = 1E/0r = E/r = Ikz. The voltage across each element is equal to its EMF minus the voltage drop across its internal resistance: U = E - XNUMX-g. Substituting the current into the expression for U, we get U \uXNUMXd E - E \uXNUMXd XNUMX. So, the voltmeter will not show any voltage. Author: V.Polyakov, Moscow
Air trap for insects
01.05.2024 The threat of space debris to the Earth's magnetic field
01.05.2024 Solidification of bulk substances
30.04.2024
▪ Cooler Master G Power Supplies 500, 600 and 700 W ▪ World's slowest pulsar discovered ▪ Tibet to build gravitational wave detector
▪ site section Power Amplifiers. Article selection ▪ article Combustion process and types of combustion. Fundamentals of safe life ▪ article Why are the Olympic Games participants depicted naked on ancient vases? Detailed answer ▪ article Garden currant. Legends, cultivation, methods of application ▪ article Increasing the hardness of rosin. Simple recipes and tips ▪ article Electroplating. Chemical Experience
Home page | Library | Articles | Website map | Site Reviews
www.diagram.com.ua |
See other articles Section 
Latest news of science and technology, new electronics:
All languages of this page