This lesson deals, first, with the electrical impedance of an inductive coil.
In a second chapter, we will examine the electrical power in alternating current.
Finally, the last part is devoted to the analysis of magnetic circuits.
1. - IMPEDANCE
We will start this new theory by examining the inductances.
An inductor, or coil, is a winding consisting of many turns of a conductive wire, generally of small section.
These inductances are obstacles to the passage of the current. If the current is continuous, it is said that they offer resistance ; if it is alternative, then it is called reactance and, more specifically in the case of inductors, inductive reactance.
Resistance is a greatness we already know. This is the ohmic resistance of the driver. Its symbol is obviously R and it is expressed in ohms.
The reactance, which appears only when the coil is traversed by an alternating current, is a function of a characteristic clean of the coil, called coefficient of self-induction and symbolized by the letter L, and the frequency of the alternating current. Like resistance, it is expressed in ohms.
When the value of the resistance is small compared to that of the reactance, it is neglected. In this case, the inductance is shown schematically as shown in Figure 1-a.
If we want instead to consider the value of the resistor R, the equivalent diagram of the inductor is that shown in Figure 1-b.
The symbol of a resistive inductance makes it possible to consider the real inductance as consisting of an inductance without ohmic resistance (L) associated in series with a pure resistor (R).
The actual terminals of the resistive inductance are A and B (Figure 1-b). The voltage V, applied to the terminals A and B, is distributed across L and R which form a voltage divider. The voltages VL and VR are thus obtained at the terminals of each of these elements.
One could thus think that between the ends A and B of the inductance, it is necessary to apply a voltage V whose effective value is the sum of the effective values of VL and VR.
In fact, this is not so because the voltages VL and VR do not have the same phase with respect to the current as shown in Figure 2.
Two periods (or cycles) of the current are represented in figure 2-a, the second is represented by a more accentuated line. The voltage VR, necessary for the resistive part, is represented in Figure 2-b, during two periods as well.
It appears that VR is in phase with the current I. The voltage VL is represented in Figure 2-c.
The bold line indicates a complete period (or cycle). This voltage VL is shifted by a quarter of a period with respect to VR or I.
It starts a quarter of period before VR and also ends a quarter of period before VR.
Thus, the voltage VL appearing across the coil (inductor) is out of phase by a quarter of a period with respect to the voltage VR.
As a result of this phase shift, at the moment when one of the two voltages reaches the maximum value, the other reaches the value zero and vice versa ; in fact, by comparing Figures 2-b and 2-c, it can be seen, for example, that at time 0 seconds correspond the voltages VR zero and VL maximum ; on the other hand, at the time of 0.05 seconds, there corresponds a zero voltage VL and a maximum voltage VR.
At time 0 seconds, the voltage V applied between the ends A and B of the coil is equal to the maximum value of VL since VR is zero at this time. On the other hand, at time 0.05 seconds, V is equal to the maximum value of VR since VL is zero.
The maximum value of V is therefore not equal to the sum of the maximum values of voltages VL and VR because these two maximum value are reached at different times.
To calculate the voltage V, it is necessary to use the vector representation, which makes it possible to highlight the phase difference between VL and VR.
Figure 3 represents vectorially the voltages VR, VL and current I. These representations are the same as those already seen for the ohmic circuit and for the inductive circuit.
The elements L and R being in series, they are traversed by the same current I. We will therefore take the vector representing this current as a reference vector.
We know that in pure resistance, current and voltage are in phase. The two vectors representing them will therefore be on the same axis (Figure 3-a).
We have just seen that in a pure inductance, the voltage VL is out of phase with a period front ahead of VR. As VR is in phase with I, we can also say that VL is ahead on IR. A quarter period corresponding to 90°, the vector representation of VL with respect to I will be that of Figure 3-b.
We can finally represent on the same graph the different values represented in Figures 3-a and 3-b. We thus obtain Figure 3-c.
To find the vector representing the voltage V, it is necessary to carry out the vector sum of the two vectors representing VR and VL. This takes into account the maximum values of VR and VL and the phase difference between these two voltages.
Figure 4-a shows the sum of the two vectors VL and VR. We notice that the direction, the direction and the length of the vectors are identical to those of Figure 3-c. The phase shift is always 90°. It is enough to join the ends O and A of the two vectors to obtain the one representing the tension V.
To calculate the value of V, it is enough to take into account that one centimeter represents 10 volts.
(If you have forgotten this theorem of Pythagoras, click on the link opposite). Pythagore's theorem.
The vector VL (2 cm long) represents a voltage of 20 volts, while the vector VR (3 cm long) represents a voltage of 30 volts.
By measuring in the figure the length of the vector V, we see that it is 3,6 cm and that, consequently, the voltage V is 36 volts.
We can conclude that the voltage V to be applied across the coil to be traversed by a current of 1 A, has the maximum value of 36 volts.
It is also possible to calculate this value by applying the Pythagorean theorem to the right triangle formed by the three vectors.
Let's briefly recall this theorem using Figure 4-b.
The sides AB and AC are the sides of the right angle ; BC, which is the longest, is the hypotenuse.
According to Pythagoras' theorem, by summing the squares of the lengths of the sides of the right angle, we obtain the square of the length of the hypotenuse.
In the case of Figure 4-b, we have :
42 + 32 = 16 + 9 = 25 = 52
This theorem can be applied to the diagram of figure 4-a :
302 + 202 = 900 + 400 = 1 300
This number 1 300, is the square of the maximum value of V.
The square root of 1 300 is 36,055. We find the same value as that measured previously (36 volts).
We know that the maximum value of an AC voltage is equal to 1.41 times its Effective Value.
We can therefore say that the effective value of the voltage that must be applied to the ends of a coil is equal to the square root of the number obtained by adding the squares of the rms values of the voltages required for the inductive part and for the ohmic part of this coil.
We have already said that the resistive part and the inductive part constitute an obstacle to the passage of current in the coil. This obstacle is called electrical impedance and is indicated symbolically by the letter Z.
Electrical impedance is measured in ohms, such as resistance and reactance.
This impedance is expressed by the following relation :
XL is the reactance of the inductive part. To determine this relationship, one must start from the relationship :V2 = VR2+
VL2 (application of the Pythagorean theorem).
Thus, the OHM law applies to a resistive coil, just as it applied to pure resistance and inductance.
The value of the voltage required for a given alternating current to flow through a coil is obtained by multiplying this current by the impedance it has.
It remains for us to see the phase difference between this voltage and the current.
(To make it easier to read, we report the same diagram (figure 4) :
In figure 4-a, to find the vector V, we drew the vector VR horizontally, then the vector VL vertically, at the end of VR.
We can also adopt the procedure shown in Figure 5-a, that is to say, draw horizontally the vector VR and vertically the vector VL. By then connecting the points O and A, we obtain the vector V in every way identical to that of figure 4-a.
The vector V can also be deduced directly from Figure 3-c, which is indicated in Figure 5-b.
To locate the point A, it is sufficient to draw two parallels to the vectors VL and VR ; they are indicated in dashed lines in this figure 5-b. This new method has the advantage of highlighting the phase difference existing between V and I.
This phase difference corresponds to the angle "phi" (
is the lowercase letter that corresponds to the letter
which is a capital letter of the Greek alphabet). We note that this angle
is inferior to 90°.
This means that a resistive coil has characteristics intermediate between those of a pure resistance (
= 90°).
The phase shift angle is therefore proportional to the ratio of the reactance to the resistance.
Figure 5-c represents this phase shift in the case where the reactance is much greater than the resistance. In this case, the angle
is close to 90°.
In Figure 5-d, it is the opposite case which is represented ; angle
is close to 0° because the reactance is very small compared to the value of the resistance.
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"phi" (
which is a capital letter of the Greek alphabet). We note that this angle 
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