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Figura 01-01

    Note that the A-B terminals represent the two output terminals of the source. In case it is a stack or battery, we have access to the A-B terminals, where terminal A would be the so-called positive pole battery and terminal B, it would be the negative pole. Therefore, what appears inside the orange rectangle it is the internal part of the cell or battery, which we do not have access to. So we can calculate the voltage that appears on these terminals. This voltage will be the potential difference (ddp) about resistance RL. If we don't have any load (RL = ∞) connected to the battery, the current circulating by the circuit will be null. So we won't have a voltage drop on Ri . Then, the ddp coincides with the efm (here represented by the letter V) of the battery. In practice, we can use a digital multimeter to measure this value, because in general, digital multimeters have an internal strength of the order of 10 megaohms, representing a despicable charge.



    2.   Types of Sources

      2.1   Independent Sources

    In this item we will study how voltage sources and independent current sources behave, analyzing Its main characteristics as suppliers of electrical energy for electrical circuits. Let's start studying the voltage sources and, later, the current sources.

        2.1.1   Voltage Sources

    Voltage sources are sources that ideally provide constant tension in their output terminals, regardless of the load. In practice, however, its output voltage is a function of the load, that is, the higher the load, the lower the output voltage. This is due to its internal resistance, as already explained in the previous item. As examples, we can cite common batteries, alkaline and rechargeable, commonly used in flashlights, portable radios, MP3, MP4, etc...

    In popular language, when batteries or batteries are worn out, we say that they are "unloaded". This is due to the wear and tear battery, and this causes a considerable increase in your internal resistance. So, when we want it to feed a load, the current circulating through the circuit is sufficient so that virtually all of the efm (electromotive force) of the battery or batteries, is reduced in the output due to the drop of tension that occurs in the internal resistance the same.

    If it is ordinary batteries, we must get rid of them, because there is no way to recover them. However, if it is batteries or rechargeable batteries, there is a way to recover them by simply using a "charger". This appliance makes it circulate current by the internal part of the cell battery or battery, in the opposite direction, causing a reduction in the internal resistance and with this we say, after a certain time of recharge, that the battery is charged. This will allow them to be used normally, as if they were new. In general, we can repeat this process from 500 to 1000 times, depending on the type of battery or batteries and its manufacturer.

    In addition to cell batteries or batteries, today we have as sources of voltage electro/electronic appliances, from the simplest like the cell phone chargers, to other very complex and specific types.

    On our website we will represent an independent voltage source as shown in Figure 01-02. Note that the font is represented in orange.

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Figura 01-02

        2.1.2.   Sources of Current

    Sources of current are sources that ideally provide a constant current, regardless of the load. In practice, these sources are possible using active circuits, that is, electronic devices such as transistors, integrated circuits, etc.... Passive circuits can be used as long as the value of the load is much lower than the internal resistance of the source. Thus, we can build a source of current, for example, putting up a resistance, of say 1000 ohms in series with a voltage source of 12 V. If the load is approximately 1% of the resistance value of 1000 ohms, that is, 10 ohms, so we have a current source of approximately 1.2 mA, current that will circulate by the load of 10 ohms.

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Figura 01-03

    In the Figura 01-03 we see an example of the current source and the symbol used to represent it in a basic circuit. Notice that the current I will circulate by the resistance R and on this we'll have a voltage drop equal to VR = R I.


        2.2   Dependents Sources


    Dependent sources, which may be of tension or current, are those that their value depends of other factors connected to the circuit, such as a voltage or a current on some component belonging to the circuit. The electric model of a transistor it has dependent sources.

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Figura 01-04

    In the Figure 01-04 we see an example of current source and voltage dependent on a basic circuit. The current source depends on the value of the output current I2 and the constant g12, where this is a feature of the device and its value can be tabled, calculated, etc....

    The voltage source depends on the input voltage V1 of the circuit as well as the constant g21. This is just an example.

    On this site, dependent sources will be represented by the color light blue.


    3.   Sources Association

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Figure 01-05
    We see in the Figure 01-05 on the side, the combination of voltage sources and current sources. When we have two sources of voltage in series we can replace them with a single source of value equal to the sum of the two. This is valid when the polarity of the sources points to the same side. This is extensive for any amount of voltage sources.

    In the case of sources of currents in parallel, it is worth the same principle. If the arrows are oriented in the same direction, we can replace them with a single current source of value equal to the sum of the values of each current source that composes the circuit.

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Figure 01-06
    In the Figura 01-06 we see the combination of voltage sources and current sources when we have opposite polarities. In this case, we must replace the sources with one whose value will be the subtract from their values. For more than two sources we must add them algebraically.

    Pay attention to the fact that the orientation (direction) of the single source, seen in the above circuit, assumes that   V1 > V2   and that   I1 > I2.

    Attention
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    4.   Explosion Sources Techniques

    In many circuits the sources associations present themselves in a very complex way for your resolution. One of the most employed techniques in these situations is the use of the so-called "source explosion".

    What does this technique consist of?   Look at the circuit below.

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Figure 01-07
    We see in the Figure 01-07 that we have two sources of current and a source of voltage. Note that the 2-ohm resistor that is in parallel with the V voltage source, It can be eliminated without prejudice to the solution of the problem. This is because we know the value of the source of voltage and the value of the resistor. So we know the electrical current that goes through this resistor.

    The same goes for the 8 ohms resistor in series with the 2 A current source. By removing it from the circuit there will be no changes in the value of the current source, it will remain 2 A. Therefore, the new circuit can be seen in the Figure 01-08.

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Figure 01-08
    With the removal of the resistor of 8 ohms we have a current source of 2 A which comes out of the "ground" and reaches the node e1. This way, if we turn this source into two current sources of 2 A, one coming to the node e2 and one coming out of the node e2, we don't alter the equations of the nodes. On the knot e3 the voltage is V.

    So if we put two sources of voltage, one in series with the resistor of 4 ohms and another in series with the resistor of 3 ohms, we also do not change the mesh equations.


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Figure 01-09
    The Figure 01-09 show the transformation of the circuit. Of course the current sources that are between the nodes e1 and e2 can be summed algebraically.

    With this we obtain a single source of current which value is 1 A, leaving the node e2 and coming to the node e1. The Figure 01-10 show the new transformation of the circuit.

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Figure 01-10
    So we can simplify the circuit and make your solution a lot easier. By applying techniques that we will study later, we will easily find the solution to the problem. In the link below, we present one of the techniques used to solve this type of circuit.



    5. Millman's Theorem

    Millman's Theorem allows us to reduce any number of voltage sources in parallel into a single source. See as an example the circuit shown in Figure 01-11.

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Figure 01-11

    This transformation will allow us to calculate the electric current that flows through the charge RL or the voltage between its terminals without applying methods such as Kirchhoff, law of Knots, Norton's theorem and others. Therefore, we must understand how to apply this theorem. Let's summarize it in three steps.


    Step 1

    We must convert all voltage sources into current sources using the source transformation   (access Source Transformation Method   clicking here! ). See what the resulting circuit looks like in Figure 01-12.


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Figure 01-12

    Step 2

    After executing the source transformation, we must add the values ​​of the current sources that resulted from the transformation. Carefully observe the polarity of the sources and the calculation of the resulting equivalent resistance. See what the resulting circuit looks like in Figure 01-13.


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Figure 01-13

    Step 3

    After the two steps above were performed, an equivalent current source was created in parallel with the equivalent resistance of the circuit. Now, we apply the source transformation again to obtain an equivalent voltage source in series with the equivalent resistance. Okay, we have the transformed circuit. See the result in Figure 01-14.


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Figure 01-14

    Now, simply by applying Ohm's law to the circuit it is possible to calculate the electric current in the charge RL, as well as the potential difference across it , Vab .