In the same way we solve for V 1, let's solve for V 2. In this case, the equivalent resistance that the circuit offers to V 2 Note that with the short circuit, resistor R 2 is now in parallel with R 1. For this circuit we must develop an equation that relates the current I 2 to the voltage source V 2. In the Figure 14-03 we present the circuit with the short-circuited voltage source V 1. To calculate K 2, let's proceed in the same way. Note that K 1, in the equation above, is this equation, but without the presence of Using the basic equation, already presented in the chapter 10 - law of Ohm, and after some algebraic Therefore, to calculate I 1, we will apply a current divider to the b node. Keep in mind that by dividing V 1 by R eq, we are calculating the current that flows through the source and by the R 1 resistor, arriving at the b node. Note that the parallel of R 2 and R 3 is in series with R 1, then: Note that with the short circuit, resistance R 2įor the calculation of I 1, we must calculate the equivalent resistance of the circuit. For this circuit we must develop an equation that relates the current I 1 to the voltage source V 1. In the Figure 14-02 we present the circuit with the short-circuited voltage source V 2. It should be noted that in this case, the values of I 1 and I 2 So, to get the final value of the current I, we simply add I 1 with I 2, or: Of I that corresponds to the voltage source V 2. Subsequently, we shorted V 1 and we calculated the fraction The fraction of I that corresponds to the voltage source V 1. To compute K 1, we will short-circuit V 2, so we will be calculating Therefore, to compute I we must first calculate K 1 and K 2. If there were more sources, we would have a number of parcels equal to the number of sources in the circuit. Of course, this equation, with two parcels, is for the particular case presented here as an example, since we have only two voltage sources. Then, by the Superposition theorem the current I will be given by: We want to calculate the current I that runs through the resistor R 2. The circuit has two voltage sources, V 1 and V 2. To understand how this method works, let's look at a simple circuit shown in the Figure 14-01. Also we must define in which variable we are interested. Thus, in the Superposition theorem we obtain a number of equations according to the number of sources in the circuit. And to eliminate a current source we must remove it from the circuit, resulting in an open circuit. Let's remember that to eliminate a voltage source we must short-circuit it. To do so, we must eliminate the other sources of voltage or current, remaining only one source. This theorem states that we can calculate, for example, the current or voltage at a given point in the electric circuit, using a voltage or current source, at a time. Let us present another tool for solving electric circuits of very practical use for certain situations. In this case, only with computer programs or calculators of specific use. \(Δk\) is called the wavenumber spread of the wave packet, and it evidently plays a role similar to the difference in wavenumbers in the superposition of two sine waves - the larger the wavenumber spread, the smaller the physical size of the wave packet.Although Kirchhoff's law is a very useful tool, in practice it proves to be a bit cumbersome, since when we have an electrical circuit with many meshes it can result in a rather complex mathematical system to find its solution. If \(Δk\) is changed to 2, so that wavenumbers in the range 2 ≤ k ≤ 6 contribute significantly, the wavepacket becomes narrower, as is shown in figures 1.11 and 1.12. e., for \(3 ≤ k ≤ 5\) in this case, contribute significantly to the sum. The quantity \(Δk\) controls the distribution of the sine waves being superimposed - only those waves with a wavenumber k within approximately Δk of the central wavenumber k 0 of the wave packet, i. 12: Representation of the distribution of wavenumbers and amplitudes of 20 superimposed sine waves with maximum at k 0 = 4 and half-width Δ k = 2.
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