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  1. Until now we have seen the components liabilities RLC (Coil-Condensing resistance) and have drawn the conclusion that:
  2. The behavior that has the resistance to the passage of the current is always the same, gives just as he is CC or CA, in addition, is independent of the frequency of CA
  3. The coil behaves as a pure driver in DC and becomes but “resistivo” when increasing the frequency in a circuit of CA

The condenser behaves as a circuit opened in CC and becomes but “driver” when increasing the frequency in a circuit of CA.

Shelp of another form, when a tension is applied to the resistance, the current that happens through the same is proportional to the value of the tension at every moment. In L and C that does not happen, because due to the processes of the magnetic fields in some (Coils) and electric fields in others (Condensing) it is going away to fulfill the one of before, but is going to exist a phase angle of the tension with respect to the current in the tips of each.

In addition, the frequency, that does not modify the conditions in pure a resistivo circuit, is going to influence in the behavior of L and C and, therefore, in the passage of the current by the circuit.

 sum of vectors

End. The coordinates where the vector arrives (part of shoot with an arrow)

In the image superior, to add to but b, the origin of b at the end of a takes and the vector is created a+b from the origin of a at the end of b


series-circuit-rlc If before, two natural values, 3 and 4, gave 7, now, if the out of phase vectors are 180º, the result would be zero. THERE IS A GREAT DIFFERENCE, NO?

In the condenser, we observed that tension V (cs) is delayed 90 º with respect to the current, and as in the case of the condenser, the impedance is going to depend on the frequency and the capacity of the condenser, of the form:

The element w represents the angular velocity, and is equal to 2 * ∏ * f


Mnemonic rule

In the condenser, now the one is I that is advanced, indicating that the current is advanced 90º with respect to the tension

We consider that the representation of each vector is different, being the resistance real values and the impedances of L and C imaginary values, which they imagine in the vertical axis, accompanied of imaginary value j.





It is necessary to consider the angles of phase angles because it repels on the behavior of the circuits.

circuit-rlcWe will see three cases.

value zr


 Case 1: We have a circuit RLC with the values of r = 20 Ω, 40 Ls = mHr and 40 Cs = µF. The frequency of the source is 100 Hertz. In this case, we have the following impedances:

 At the time of calculating the total impedance, the real numbers and the imaginary ones are added:

  Zt = 20 - 14.67 j Ω Case 2. Resonant circuit:


Circuit RLC in series with a small resistance and a capacitive impedance (Zc) that is equal to the Inductive impedance Zl for a certain frequency.  In this case, the module of the resulting vector will be the minor of the possible ones and this way, we have the major of the possible currents that they happen through the circuit. So that Zl is equal to Zc, it must fulfill that:


Giving a result of:

 Where fr is the frequency of resonance of the circuit for which, the intensity is maximum.

This is very important in radio communications because it allows us to tune a concrete frequency of the phantom of radio frequencies

  1. Activity. In a circuit RLC, we have it coil has a value of 20 mF and 400 condenser μF. To calculate
  2. The frequency of resonance of the circuit
  3. The values of Zl and Zc. To represent those values in the vectorial diagram
If the frequency of the generator is 400 to do the 2 KHz and condenser change μF, what value must have L now so that the circuit is resonant?

Case 3

 We have in an industrial ship many refrigerating machines with compressors of great windings (much inductive impedance). In this case, there is little resistance and much winding with hardly capacitive elements. In this situation, in the coils we have the out of phase current and 90 tension º. At the time of handling the powers, we have for the tension and current, the signals has this form: v (t) = Vo sen (wt)

As the circuit is inductive, the current is going to have a phase angle with respect to the tension. In this case, the current will be: , where Φ Is the phase angle angle that is between the tension and the current.


 The power is the product of the current by the tension, therefore:


By trigonometry sen (wt - θ) = 1/2 is demonstrated that to sen (wt) * (cos θ - cos (2wt - θ). Therefore, the equation we have left:

inductive power

 We have seen that the peak voltage is √2 of the effective value. We can put the previous expression in the form of effective values V and I, of the form:

 In the power we have a real value (V*I*cosØ) and cos a reactive power V* I * (2wt- Ø). The first power is a real and constant value, but second it is a power that “enters and leaves the house”.

This power is not desirable for the companies since it is circulating around cables if to produce an effective work in the load.

We will not enter but in this aspect but the following thing it would be to make a compensation of loads to diminish this effect. In these three cases we have seen the importance that has the correct treatment of the impedances and its vectorial representation is going to us to facilitate much its handling.

it forms complex binary polestar and of nunero


Forms to represent the impedance

 Until now we have seen that to represent an impedance, specific his real part and its imaginary one (accompanied by I number j), but there is another form that, to solve problems, is going to us to come very well. In the image of the left, if we know the coordinates the point, b, we can represent it, agreeing the value of a, with the real part x and the b with the imaginary one and, but that point can also be represented if distance r is known and the angle α For it we must apply to the theorem of Pitágoras and a little trigonometry.


Pitágoras shelp, and shelp well, that in a triangle rectangle, the Hypotenuse to the squared one is equal to the sum of the squared ones of the legs. In the figure superior, the hypotenuse is r and the legs are x to and. Therefore:

We already have what is worth the module of r. Now we are going to see as we calculated the angle. For it, we must know what is the sine and the cosine of an angle. We are going to do some nonformal definitions but that are going to serve to us.

We suppose that we put a lantern in the high part of the vector and another lantern to the right of the vector.

 The cosine of an angle (cos α) is the part of shade that vector r projects on the horizontal axis, giving rise to projection X.

The sine α would be the part of shade that vector r projects on the vertical axis, generating the shade and in the drawing

 Therefore, if we know the sine and cosine, we will be able to know the projections x and, that evidently, never will be but great that the same vector. Mathematically:

Those equalities we can divide one to them among another one, giving rise a

Don't Know:

If we are dividing two polar numbers, the modules are divided and the arguments are reduced

  1.  Activities. To conduct the following operations:
  2.  (3 +3j) + (6 - 5 J)
  3.  (5 - 4j) * (2 + 3j)
  4.  3˪45º * 5 ˪27 º 

5˪65º + 6˪ 120 º

Calculation of the current in a circuit RLC.

In order to know the current that happens through that circuit we must apply the law of Ohm: . This formula says to us that the current is obtained dividing the modules of the tension and the impedance and the module are going away to obtain reducing the angles of the tension and the impedance. We have therefore that: φ

triangle of powers

Factor of Power


We have seen that in the circuits RLC, the normal thing it is that the tension and the current are out of phase a certain angle. That phase angle brings about that “Energy that goes and comes by the network” without producing effective work. we go. This gives rise to the denominated triangle of the powers, where two new concepts are introduced, the reactive power (the one of the load) and the power pretends (the product of V * I without considering the phase angle angle). We see this in the graph of the left.  On the one hand we have P that represents the active power of the system.  Soon two reactive powers exist (one of the condenser and another one of the coil) that gives rise total to the reactive power (Ql - Qc) and finally, the power pretends S is obtained to multiply V by I and it occurs in Volt-Ampere.

 Problem circuit rlc We will see all this in an example. To calculate the impedances, tension and powers of the following circuit if the frequency of the generator is 50 Hertz

  The Power pretends is V*I = 220 * 19.99 = 4397.8 GOES

PROBLEMS. Aside from the activities of the book, to do the following:

problem rlIf the frequency of the generator is 10 Hertz, to calculate:
The Impedance of the coil
The total impedance of the circuit

 The current that leaves the battery

Activity 1.  To represent in the form of vectors the impedances of a circuit RLC if R = 40 Ω, Xl = 90 and Xc = 50 Ω. To calculate the active power and reactivates.

Fasorial diagram of the tension and current in the circuit

Activity 3. We have a refrigerating machine with a reactive component of 220 jΩ and one resistance of 75 Ω. If the frequency of the network is of 50 Hertz, to calculate the condenser that we must put so that to 50 the reactive component lowers jΩ.


electrical motors. etc

Overload of the condensers of compensation of the factor of power, etc  



In the loads we can find two cases. First that is linear, as a pure resistance, giving rise to a current wave that “follows” the form that has the tension, therefore, the current in that load is not distorted. Example of this type of load is the resistance of the heating engineers, the incandescent or motor lamps of induction of constant speed. Nevertheless we have another type of loads where the current varies desproporcionalmente to the tension during the stress cycles. These are the loads nonlinear and the form of the current contains distortions. These distortions are signals whose frequency is multiple of the out of phase main signal. To that group of signals, whose superposition gives rise to the signal of the real current, it is called overtones to them. (To see figure superior) We have examples of this type of load in shippers of batteries, electronic reactancias, regulating of power, etc it is not necessary to forget that the circuits are designed to work with certain frequencies and the presence of overtones gives rise to bad operation of the same one, reason why is important to eradicate them.

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