Continuing, let's see how behave circuits that contain only resistors
and inductors.
For alternating current (AC), let's simplify using the phasor concept. So we will use sources in the
complex form and impedances as well. This causes that, in impedances, the real part is a resistorand the imaginary part a reactance.
In this way, everything happens as if they were associations of resistors and so we can use all the theorems previously learned for direct current.
Then, law of Ohm, theorem of superposition, method nodal, Thévenin and Norton theorem, etc... all are valid.
Let us take as an example a very simple circuit, as can be seen in the Figure 54-01, where we have a sinusoidal source,
a resistor and a series inductor.
Figure 54-01
Let's assume that we are working on the frequency used in Brazil,
that is, f = 60 Hz. So we can calculate the frequency value
angular ω.
We know that ω = 2 π f = 377 rad/s.
We can now calculate the reactance that the inductor of 53.32 mH offers for the circulation of an alternating current of 60 Hz. Let us recall the equation that allows to calculate the reactance of an inductor.
XL = ω L
By making the numerical substitution of the values of the components we find:
XL = 20.1 Ω
With this value we can write the impedance in the rectangular form and in the polar form of the whole circuit, or:
Z = 10 + j 20.1 Ω ⇒ Z = 22.45 ∠ +63.55° Ω
Applying the Ohm law, we can easily determine the electric current
that circulates around the circuit. Like this:
I = V / Z = 220 ∠ 0° / 22.45 ∠ +63.55° A
Performing the calculation, we find:
I = 9.80 ∠ -63.55° A
Pay attention to the fact that the angle - 63.55° means that the current is lagging behind the voltage applied to the circuit. As expected, we know that the inductor delays the current in relation to the voltage. When this happens (delayed current), we say that we have an inductive circuit.
With the value of the current we can calculate the values of VR and VL.
VR = R I = 10 x 9.80 ∠ -63.55° = 98 ∠ -63.55° V
Note that because resistances do not cause lags between voltage and current,
VR is absolutely in phase with I. The same does not occur with the
inductive reactance, since we know that the inductor delay the current in relation to the voltage. Like this:
VL = XL I = 20.10 ∠ +90 x 9.80 ∠ -63.55° V
Performing the calculation, we find:
VL = 197 ∠ +26.45° V
In the Figura 54-02 we show in a graph the situation of all the quantities calculated above.
Note that the electric current I, is delayed 63.55° in relation to the source voltage,
V. E delayed of 63.55° + 26.45° = 90° in relation to the voltage on the inductor,
VL.On the other hand, the voltage on the resistor is in phase with the electric current. Note that if we add phasorially the voltage on the inductor and on the resistor, we must obtain the voltage of the source, that is:
V = √ (|VL|2 + |VR|2) = √ (1972 + 982) = 220 V
Figura 54-02
To conclude, let's calculate the power factor of the circuit. As we know that the current is
63.55° delayed in relation to the voltage, then the power factor is:
PF = cos 63.55° = 0.45 inductive or delayed
In the same way that we draw your attention to the case of the capacitor, here too, increasing the value of the inductor inductance, its inductive reactance increases. This causes the angle of lag between current and voltage to increase as well, tending the PF to
zero. In other words: in
alternating current, the higher the value of the inductor the more influence it has on the circuit. Because inductive reactance depends on two variables, L and ω, this means that if we keep the inductance with a fixed value and increase
considerably the value of ω, the result will be the same as above. It is
feature will be explored when we study filters.
And if we do the inverse, that is, we decrease the inductance or the angular frequency, we get the opposite result. In other words, the inductor will have little influence on the circuit and the
power factor tends to one. The previously stated is valid for an RL SERIES circuit.
Let's form a parallel circuit with the same components used in the previous item.
The current on the resistor will not suffer a lag. Current in the inductor will be delayed
of 90° in relation to the voltage. Then, the current that will be supplied by the voltage source
will be the phasor sum of the two previous ones.
Figure 54-03
See in the Figure 54-03 how the circuit with the components in parallel was.
We already know that ω = 2 π f = 377 rad/s, because we are assuming
f = 60 Hz.
We also know the inductor reactance of 53.32 mH, that is,
XL = 20.10 Ω.
What we must calculate now is the equivalent impedance of the resistor in parallel with the inductor. As previously stated,
we can use the same principles studied for DC to calculate the equivalent impedance. Thus, we assume the reactance as a
resistor and perform the calculation as if they were two resistors in parallel. Just do not forget that the reactance will be a complex number.
Then we can write:
Zeq = R ( jXL ) / (R + jXL )
Zeq = 10 j20.10 / (10 + j20.10)
Notice that, in the equation below, we transform + j201 into 201 ∠ +90° and we also
transform 10 + j20.10 into 22.45 ∠ +63.55°. Then
Zeq = 201 ∠ +90° / 22.45 ∠ +63.55°
By placing in polar, numerator and denominator format, it is very easy to calculate. Therefore, for the equivalent impedance,
after the calculation, we find:
Zeq = 8,95 ∠ +26,45° = 8 + j4 Ω
Note that when we put the components in parallel, current continue delay of
26.45°
in relation to tension. Looking at the equivalent impedance in rectangular form we find that
represents an impedance with two components in series. An 8 ohm resistor and an inductor with
4 ohms reactance. In short: a 8 ohms resistor in series with an inductor of 10.61 mH, will behave electrically as the originally presented circuit.
Since we know the angle of the impedance, we can calculate the power factor, or:
PF = cos 26.45° = 0.90
Thus, when we put the components in parallel, the power factor increased by
a factor 2 (for this particular case).
Let's calculate the currents that circulate in the circuit.
IR = V / R = 220 ∠ 0° / 10 = 22 ∠ 0° A
IL = V /XL = 220 ∠0° / 20.10 ∠+90° = 10.95 ∠ -90° A
I = V / Zeq = 220 ∠ 0° / 8.95 ∠ +26.45° A
Performing the calculation, we find:
I = 24.58 ∠ -26.45° A
Another way to calculate I is to calculate the phasor sum of
IR with IL.
In the upper part of the figure below, we graphically represent the equivalent impedance in its rectangular and polar form. How is an inductive impedance the angle of
26.45° is positive.
In the bottom part of the figure, we show the phasors of the currents. Notice that
IL is delayed 90° in relation to IR.
This current (IR) is in phase with the voltage V.
Figura 54-04
See the Figura 54-04, that IR and IL they are out of time
of 90° or, as is commonly said, are in quadrature. To calculate the
module of I it is no more obvious that we use the Pythagorean theorem. So
we write that |I| = √(IR2 + IL2).
This means that |I| = √(222 + 10.952) = 24.58 A.
To find the angle we must compute the arcotangent of the quotient
IL / IR. Doing the calculation we find 26.45°.
Thus
I = 24.58 ∠ -26.45° A
So the calculated final result is exactly the same when we apply the Ohm law
to find I.
