FUNDAMENTAL PRINCIPLES OF THE AC VOLTAGE
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.

• It is necessary to understand the concept of vectors as mathematical unit that goes to represent a magnitude in the circuits. It is defined by:
• Module, that makes reference to the length of the vector
• Direction (phase angle), that comes dice by the angle that forms the vector with the horizontal axis.
• Origin. The coordinates from which it divides the vector 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 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 circuit of the figure we have what happens with the tension in each component when it passes the current to him i, that is the same for all. We will see one by one.
• In the resistance, the signal that offers to us is just as the signal of the current, with its corresponding value given by Vr = i * R. We did not appreciate phase angle •  In the coil, the tension (vl) is advanced 90 degrees. When the current one happens through zero, the tension already is in the maximum value. The Impedance (the opposition to the passage of the AC voltage) that presents depends on a factor of the winding (Coefficient of self-induction L) and of the frequency, so that: 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

• We can remember the phase angles that we have in a circuit RLC if we remembered word ELICE. In the image one is what represents each letter.
• In L, we have the tension and before letter I (Intensity), which indicates to us that the 90 tension is advanced º

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. We will see three cases.   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: 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. 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  :

• When we used both. Following the operation that we do, he is better to use an expression of the impedance or another one. For example, when we make sum and subtractions, the best thing is to use the binary one, because it is on the other, to add the real parts on the one hand and the imaginary ones but we are multiplying or dividing, the best thing is to use the polar forms. In the polar form:
• If two numbers are multiplying, the modules multiply and the arguments are added

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 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. 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

• 1º Calculation of the impedances.
• The Impedance of the resistance is the same, therefore 20 Ohms • The impedance of the coil is obtained from: • The impedance of the condenser is:
• The total impedance is 20 + (25,13-24,48) = 20 + 0,64 j Ohms • The angle that forms the imaginary part and the real one calculates as • The module of the impedance is
• The module of the current comes given by the law from Ohm:
• I = 10.99 V/Z = 220/20.01 = A
• The current in vectorial form has form 10.99 ∠-1,85
• The Effective power is P = V*I*cosφ = cos 220 * 19.99 * 1.85 = 4395 W
• The Reactive power Q = V*I*senφ = 220 * 19.99 * sen 1.85 = 141.97 BAR (volts reactive amperes)
The Power pretends is V*I = 220 * 19.99 = 4397.8 GOES

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

• Activity 2º Tenemos a circuit RLC with the values of R = 40 Ω, 20 Ls = mF and C = 400 microfarads in a circuit series fed by a generator 220V 50 Hertz. To calculate
• The impedance of each element
• The total impedance
• The diagram of vectors of each impedance and the total
• The current in polar form
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. Overtones:

• Overtones, that soon we will say what is, have been between us since the first generator became of AC, already does but of 100 years, although in those times, these were not of much importance within the mains. The overtones are frequencies multiple of the base frequency, where the amplitude is decreasing to increases the frequency. For the mains, the S-values 100, 150, 200, etc. In the same generator of CA, owing to the fact that the magnetic fields in the stator and rotor perfectly are not distributed in the space, we have some small overtones of 1 0 2% of the total. They are not very important but already we are generating them. The overtones usually are produced by
•  exchanged power supplies
• Electronic stabilizers

electrical motors. etc

• The overtones produce:
• Overheat of the transforming
• Overload of the neutral drivers
• Unexpected firings in the interrupting differentials of the houses
• Noise and possible damages in electronic circuits

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