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ENCYCLOPEDIA OF RADIO ELECTRONICS AND ELECTRICAL ENGINEERING This complicated Ohm's law. Encyclopedia of radio electronics and electrical engineering
Encyclopedia of radio electronics and electrical engineering / Beginner radio amateur There is no doubt that everyone knows Ohm's law for the section of the circuit shown in Fig. 3a: U = IR, where U is the voltage drop across the section; I - current in the circuit; R is the resistance of this section of the circuit. It is a shame to make a mistake in Ohm's law, but if you have not memorized this formula yet, use fig. 3b. It is enough to close the desired value with your finger to get an answer on what to multiply or divide by. It is recommended to use the SI system of units, where voltage is expressed in volts, resistance is in ohms, and current is in amperes. However, when calculating radio circuits, it can be convenient to take the current in milliamps and the resistance in kiloohms - then the factors 10-3 and 103 will decrease and the voltage will still be in volts.
Let's express the current I = U/R. The dependence of current on voltage is directly proportional; on the graph l (U) it is displayed as a straight line (Fig. 3, c). This relationship is often called linear. So, we take a battery from a 4,5 V flashlight and connect a 1 ohm resistor and an ammeter connected in series to it (it is always connected in series with the load). Instead of the expected 4,5 A, we get much less! What's the matter, does Ohm's law really not work? We will have to investigate this phenomenon and connect a voltmeter in parallel with the resistor. It will show a voltage less than 4,5 V and equal to U = I R. Where does the rest of the voltage "fall"? On the internal resistance of the battery, which we did not take into account in the previous calculation. Here you need to use Ohm's law for a complete circuit: I \u4d E / (r + R), where E is the electromotive force of the battery (emf, it is indicated on the package, and not voltage at all); r - internal resistance. These two parameters fully characterize the current source. The scheme of the experiment and the order of turning on the instruments are shown in Fig. . XNUMX.
Let's see how the current and voltage at the load depend on its resistance R. The voltage at the load U = l R = ER/(r + R). If the load resistance is increased to infinity, the current will tend to zero, and the voltage will tend to EMF. Finding out the EMF is easy, you just need to connect a voltmeter (without load) to the battery terminals. It is assumed that the voltmeter is "good" - high-resistance, i.e., consuming a negligible current. If not, then a "bad" voltmeter will show a voltage that is less than the EMF by the value of Iv r where Iv is the current consumed by the voltmeter. Let us now direct the load resistance to zero, then the current in the circuit will be equal to the short circuit current Ikz \u4d E / r. Now the ammeter shown in Fig. XNUMX must be "good", i.e., having an exceptionally low intrinsic resistance ra. Otherwise, not Ikz will be measured, but a smaller current equal to E / (r + ra). It is possible to measure the short-circuit current with an ammeter only for the most low-power cells and batteries (then it is small, and a very short circuit of the terminals does not harm the battery). For many batteries, Ikz can reach hundreds and thousands of amperes - such a current will melt copper wires and iron nails and will certainly ruin your ammeter. Fortunately, it is not necessary to conduct such an experiment, and the internal resistance can be easily found by calculation. If you measure the EMF with a high-resistance voltmeter, and then the voltage U at a known load R, then from Ohm's law for a section of the circuit it is easy to find I \u1d U / R. You can also measure the current, then it is not even necessary to know the resistance. Now let's transform the formula of Ohm's law for the complete chain: r = E/I - R. Substituting I, we have r = R(E/U-XNUMX). The same calculation can be done graphically. For the complete circuit shown in Fig. 4, we plot the dependence of the current through the load on the voltage across it, provided that the resistance varies from 0 to infinity. When the resistance is 0, the current is maximum and equal to lK3, while the voltage is 0 - we get point a. Let's increase the resistance to infinity (turn it off) - the voltage will increase to E - we get the point b. Two points are enough to draw a straight line ab through them - it is called the load characteristic (thick line). Now turning on some resistance R, measuring the voltage U on it and calculating the current I, we get the point c. It is also easy to find it graphically by plotting l(U) in the same coordinates for a given resistance R, the same as in Fig. 3c (thin line in Fig. 5). The intersection of two straight lines gives the point c.
In the above calculation, we, in fact, found points b and c by measuring the EMF and voltage on the load. Drawing a straight line through them, we also find point a at the intersection with the vertical axis (Ikz), and hence the internal resistance r. Now let's try to answer the question, what power P is released in the load? As you know, Р = U·I. Volts multiplied by amps equals watts. If the current is measured in milliamps, and the voltage is in volts, then the power is obtained in milliwatts. Using this formula, it is easy to find the power dissipated by the resistors. For example, if a voltage of 1,2 V is applied to a 12 kΩ resistor, the current will be 10 mA, and the power dissipation will be 120 mW. Graphically, the power is equal to the area of a rectangle built on the coordinate axes and touching the vertex of the point c (it is shaded in Fig. 5). The load resistance can be chosen to be at a very interesting point d, where U = E/2 and I = lK3/2. Under these conditions, the load resistance is equal to the internal resistance of the source, i.e. R \uXNUMXd r, and the area of \uXNUMXb\uXNUMXbthe rectangle corresponding to the power P dissipated in the load will be maximum. Try to prove this position yourself for fun, either algebraically - by finding the maximum of the function, or by proving a geometric theorem. The condition R = r is called the matching condition, and the load is called matched. At the same time, the greatest power is released in it. Indeed, at high load resistances, the current drops, in the limit to zero, and the voltage cannot exceed the EMF. Consequently, the power in the load tends to zero. Another extreme case is less obvious, when the load resistance tends to zero Then the current increases to lK3, but the voltage U tends to zero, which means that the power in the load also drops. It should be noted that the power in this case is still dissipated, but not at all where it is needed - on the internal resistance of the source. It has been repeatedly observed that a short-circuited galvanic cell heats up, while quickly depleting its capacity. The last question for today's discussion is what is the efficiency of the circuit shown in fig. 4? By definition, efficiency is equal to the ratio of the power dissipated in the load to the total power consumed in the circuit. The latter is equal to E 1, and efficiency = U l/E l = U/E. This shows that the efficiency is close to unity only at high load resistances, when working with low currents, when U is almost equal to E, and the voltage drop across the internal resistance of the source is small. When matching efficiency = 0,5 (50%) and half of the total power is spent inside the source, and the other half - in the load. In modes close to a short circuit, the efficiency is very small. This is one of the reasons why it is more profitable to discharge galvanic cells with a small current. And now another "homework". You have been brought to the island, night is falling, the next boat flight has been delayed and it needs to give a light signal. Among the expedition equipment, you found a flashlight with a half-discharged battery, a multimeter and three light bulbs: 12 Vx0,1 A, 6 Vx0,2 A and 3 Vx0,4 A. Measurements of the battery parameters showed its EMF 12 V and short circuit current 0,4 A. Which light bulb to choose so that the light is as bright as possible? (Note that the circuit of the lantern corresponds to Fig. 4, only the switch is not shown.). Author: V.Polyakov, Moscow
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