Created it, 05/10/15
Update it, 05/11/19
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ELECTRIC IMPEDANCE “1st PART”
This milked lesson, initially, of the electric impedance of an inductive reel.
In a second chapter, we will examine the electric output in AC current.
Lastly, the last part is devoted to the analysis of the magnetic circuits.
We will begin this new theory by examining inductances.
An inductance, or winds, is a rolling up made up of many whorls of a discussion thread, generally of weak section.
These inductances are obstacles with the passage of the current. If the current considered is continuous, it is said that they offer a resistance; if it is alternate, one speaks then about reactance and, more precisely in the case of inductances, about inductive reactance.
Resistance is a size which we know already. It is the ohmic resistance of the driver. Its symbol is obviously R and it is expressed in ohms.
Reactance, which appears only when the reel is traversed by a AC current, is related to a particular characteristic of the reel, called coefficient of self induction and symbolized by the letter L, and of the frequency of the AC current. Like resistance, it is expressed in ohms.
When the value of resistance is low compared to that of the reactance, it is neglected. In this case, inductance is represented schematically as shown in the figure 1-a.
If one wants on the contrary to consider the value of resistance R, the equivalent diagram of inductance is that represented figure 1-b.
The symbol of a resistive inductance makes it possible to regard real inductance as consisting of an inductance without ohmic resistance (L) associated in series with a pure resistance (R).
The real terminals of resistive inductance are A and B (figure 1-b). The tension V, applied at the boundaries A and B, is distributed at the boundaries of L and R which form a tension divider. One thus obtains at the boundaries of each one of these elements tensions VL and VR.
One could thus think that between ends A and B of inductance, it is necessary to apply a tension V whose effective value is the sum of the effective values of VL and VR.
In fact, it is not thus because tensions VL and VR do not have the same phase compared to the current as shown in the figure 2.
Two periods (or cycles) of the current are represented figure 2-a, second is represented by a feature more accentuated. Tension VR, necessary for the resistive part, is represented figure 2-b, during two periods also.
It appears that VR is in phase with current I. tension VL is represented figure 2-c.
The fatty feature indicates one period (or a cycle) complete. This tension VL is out of phase a quarter of period compared to VR or with I.
It begins a quarter from period before VR and also finishes a quarter of period before VR.
Thus, tension VL appearing at the boundaries of the reel (inductance) is out of phase in advance of a quarter of period compared to tension VR.
In consequence of this dephasing, at the moment when one of the two tensions reaches the maximum value, the other reached zero value and conversely; indeed, by comparing the figures 2-b and 2-c, one realizes, for example, that to time 0 second null tensions VR and maximum VL correspond; on the contrary, to times of 0,05 second a null tension VL and a maximum tension VR correspond.
In the time 0 second, the tension V applied between ends A and B of the reel is equal to the maximum value of VL since VR is null at this moment. On the other hand, in the times 0,05 second, V is equal to the maximum value of VR since VL is null.
The maximum value of V is thus not equal to the sum of the maximum values of tensions VL and VR because these two maximum values are reached at different moments.
To calculate the tension V, it is necessary to have recourse to the vectorial representation, which makes it possible to highlight dephasing between VL and VR.
Figure 3 represents tensions VR vectoriellement, VL and current I. These representations are the same ones as those already seen for the ohmic circuit and the inductive circuit.
The elements L and R being in series, they are traversed by same current I. We will thus take the vector representing this current like vector of reference.
We know that in a pure resistance, running and tension are in phase. The two vectors the representative will be thus on the same axis (figure 3-a).
We have just seen that in a pure inductance, tension VL is out of phase of a face of period in advance on VR. As VR is in phase with I, one can also say that VL has a lead over IR. A quarter of period corresponding to 90°, the vectorial representation of VL compared to I will be that of the figure 3-b.
We can finally represent on the same graph the various values represented figure 3-a and 3-b. We obtain the figure 3-c thus.
To find the vector representing the tension V, it is necessary to carry out the vectorial sum of the two vectors representing VR and VL. One thus takes account of the maximum values of VR and VL and of dephasing between these two tensions.
The figure 4-a, indicates the sum of two vectors VL and VR. We notice that the direction, the direction and the length of the vectors are identical to those of the figure 3-c. Dephasing is always equal to 90°. It is enough to join together the ends O and A of the two vectors to obtain that representing the tension V.
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To calculate the value of V, it is enough to hold account owing to the fact that one centimetre represents 10 volts.
(If you forgot this theorem of Pythagore, click on the bond opposite). Theorem of Pythagore.
Vector VL (2 cm length) represents a tension of 20 volts, while vector VR (3 cm length) represents a tension of 30 volts.
By measuring on the figure the length of the vector V, we see that it is 3, 6 cm and that, consequently, the tension V is worth 36 volts.
We can conclude that the tension V to be applied at the boundaries of the reel so that it is traversed by a current of 1 A, to the maximum value of 36 volts.
It is also possible to calculate this value by applying the theorem of Pythagore to the right-angled triangle formed by the three vectors.
Let us make a short recall of this theorem using the figure 4-b.
The sides AB and AC are the sides of the right angle ; BC, which is longest, is the hypotenuse.
According to the theorem of Pythagore, by making the sum of the squares lengths on the sides of the right angle, one obtains the square length of the hypotenuse.
In the case of the figure 4-b, one a :
This theorem can be applied to the diagram of the figure 4-a
This number 1 300, is thus the square of the maximum value of V.
The square root of 1 300 is 36,055. We find the same value well as that measured previously (36 volts).
We know that the maximum value of an alternating voltage is equal to 1,41 times its value effective.
We can thus say that the effective value of the tension which one must apply at the ends of a reel is equal to the square root of the number obtained by making the sum of the squares of the effective values of the tensions necessary for the inductive part and the ohmic part of this reel.
We already said that the resistive part and the inductive part constitute an obstacle with the passage of the current in the reel. This obstacle is called electric impedance and one symbolically indicates it by letter Z.
The electric impedance is measured in ohms, like resistance and the reactance.
This impedance is expressed by the following relation :
XL is the reactance of the inductive part. To determine this relation, it is necessary to leave the relation : V2 = VR2 + VL2 (application of the theorem of Pythagore).
Thus, the law of OHM applies to a resistive reel, just like it applied to a resistance and a pure inductance.
The value of the tension necessary so that a given AC current crosses a reel obtains by multiplying this current by the impedance which it has.
It remains us to see existing dephasing between this tension and the current.
(To facilitate the reading of it, we defer the same diagram to knowing figure 4) :
On the figure 4-a, to find the vector V, we drew vector VR horizontally, then vector VL vertically, at the end of VR.
We can also adopt the procedure indicated figure 5-a, i.e. to draw horizontally vector VR vertically then vector VL. By connecting then points O and A, one obtains the vector V in any point identical to that of the figure 4-a.
The vector V can also result directly from the figure 3-c, which is indicated figure 5-b.
To locate point A, it is enough to trace two parallels with vectors VL and VR ; they are indicated in dotted on this figure 5-b. This new method has the advantage of highlighting existing dephasing between V and I.
This dephasing
corresponds to the angle
“phi” (
is the small letter which corresponds to the letter 
which is a capital letter of the Greek alphabet). We note that this angle
is lower than 90°.
That means that
resistive reel has intermediate characteristics between those of a pure
resistance (
= 90°).
The angle of dephasing is thus proportional to the report/ratio of the reactance on resistance.
The figure 5-c represents this dephasing if the
reactance is much higher than resistance. In this case, the angle
is close to 90°.
On the figure 5-d, in fact the contrary case is
represented ; the angle
is close to 0° because the reactance is
very weak compared to the value of resistance.
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