In the chapter referring to RC and RL circuits in direct current, we saw that to solve the problem we had to resort to differential equations.
For alternating current, let's simplify using the phasor concept. So we will use fonts in the
complex form and impedances as well. This causes that, in impedances, the real part
be a resistance and the imaginary part a reactance. In this way, everything happens as if they were associations of resistances 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.
Here our focus will be only on circuits containing resistors and
capacitors. We will start with a very simple circuit, as can be seen
in the Figure 53-01, where we have a sinusoidal source feeding a series circuit formed by a resistor and a capacitor.
Figure 53-01<
Let's assume that we are working on a frequency like that used in Brazil,
that is, f = 60 Hz. So we can calculate the angular frequency
value ω.
We know that ω = 2 π f = 377 rad/s.
We can now calculate the reactance that the capacitor of 265 µF
offers to an alternating current of 60 Hz. Let us recall the equation that allows to calculate the reactance of a capacitor.
XC = 1 / (ω C)
By making the numerical substitution of the values of the components, we find:
XC = 10 Ω
With this value we can write the impedance of the circuit
in rectangular form and in polar form, or:
Z = 10 - j 10 Ω ⇒ Z = 10 √2 ∠- 45° Ω
Applying the Ohm law, we can easily determine the electric current
that circulates around the circuit. Like this:
I = V / Z = 220 ∠ 0° / 10 √ 2 ∠ - 45° A
Performing the calculation we find:
I = 15.56 ∠ + 45° A
Pay attention to the fact that the angle + 45° means that the current is advanced in relation to the voltage. Fact already expected. When this happens (advanced current), we say that we have a capacitive circuit.
With the value of the current we can calculate the values of VR and VC.
VR = R I = 10 15.56 ∠ +45° = 155.6 ∠ + 45° V
As we know resistors do not cause lags, VR it is
absolutely in phase with I. The same does not occur with VC,
because we know that the capacitor causes a current lag. Like this:
VC = XC I = 10 ∠- 90 x 15.56 ∠+45° V
Performing the calculation, we find:
VC = 155.6 ∠- 45° volts
In the Figure 53-02, we show with phasors all calculated quantities.
Notice the current I advanced
45° in relation to the applied voltage, V. Note that the voltage across
the resistor,
VR, has the same phase of the electric current.
On the other hand, we have the voltage on the capacitor, VC, delay
of 45° with respect to the voltage source V and delayed
45° + 45° = 90° in relation to current
I. Notice how all quantities calculated analytically, "fit" perfectly in graphical analysis.
Figure 53-02
To conclude, let's calculate the power factor of the circuit. As we know that the current is
45° advanced in relation to the voltage, then the power factor is:
PF = cos 45° = 0.71 capacitive or advanced
We would like to draw your attention to the following fact: increasing the value of the capacitance
of the capacitor, its capacitive reactance decreases. This causes the angle of lag between
current and voltage also decreases, tending the power factor for the unit. In other words: in
alternating current, the higher the capacitor value, the less influence it has
in the circuit. Since the capacitive reactance depends on two variables, C and ω
this means that if we keep the capacitance with a fixed value, but 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 opposite, that is, we decrease the capacitance or the angular frequency,
we get the inverse result. In other words: the capacitor will have great influence on the circuit and the
power factor tends to zero. The above is valid for an RC 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. The current in the capacitor will
be advanced by
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 53-03
See in the Figure 53-03 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 capacitor reactance of 265 µF, that is,
XC = 10 Ω.
What we must calculate now is the equivalent impedance of the resistor in parallel
with the capacitor. 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 is not a real number. Then we can write:
Zeq = R jXC / (R + jXC )
Zeq = 10 (-j10) / (10 - j10)
In the equation below, notice that we have transformed - j10 into 10 ∠ -90° and also
10 - j10 into 10 √2 ∠ -45°. Then:
Zeq = 100 ∠ -90° / 10 √2 ∠ -45°
By placing in polar, numerator and denominator format, it is very easy to calculate. Therefore, for the equivalent impedance we find:
Zeq = 5 √2 ∠ -45° = 5 - j5 Ω
Note that by placing the components in parallel, the current continue advanced
of 45°
in relation to tension. Looking at the equivalent impedance in rectangular form, we find that
represents an impedance with two components in series: a resistor of 5 ohms and
a capacitor with a reactance of 5 ohms. In short: a
5 ohms in series with a capacitor of 530 µF, will have a behavior electrical
as the originally presented circuit.
We already know that the power factor is 0.71 advanced. So to finish
let's calculate the currents in the circuit.
IR = V / R = 220 ∠ 0° / 10 = 22 ∠ 0° A
IC = V / XC = 220 ∠ 0° / 10 ∠ -90° = 22 ∠ +90° A
I = V / Zeq = 220 ∠ 0° / 5√2 ∠-45° = 31.11 ∠+45° A
Another way of calculating I is to calculate the phasor sum of
IR and IC.
Figure 53-04
See in the Figure 53-04, that IR and IC are out of phase with
90°, or as they are commonly referred to as quadrature. To compute the I module, it is no more obvious that we use the Pythagorean theorem. So
we write that |I| = √(|IR|2 + |IC|2).
This means that |I| = √(222 + 222) = 31.11 ampère. Which is the result we found earlier.
To find the angle, note that the two sides have the same measure (22), so they form a
square and I is the diagonal of the square. However, we know that the diagonal of a square
forms an angle of 45° with the base. Therefore, the final result is exactly the same when we apply the law of Ohm to find I, that is
In practice, capacitors have a certain loss. This loss can be represented by a resistor in parallel or series. Thus, at a given frequency
ωo, have a
equivalent series and parallel representation. When passing from one representation to another, the value
of the components is changed, mainly the resistor. This property is used to modify the
load impedance level. Then, considering a frequency ωo the circuit can be represented as shown in Figure 53-05.
Figure 53-05
We have that the quality factor Q is the same in both representations, and the impedance modulus and
of admittance is given by
The quality factor Q being the same in both representations is defined by eq. 53-01.
eq. 53-01
Using the above equations and doing |Y (j ωo)| = 1 / |Z (j ωo)|, after some algebraic manipulations we obtain the following relations:
eq. 53-02
eq. 53-03
eq. 53-04
eq. 53-05
Important Note
When the quality factor Q is high, that is, Q > 10, the value of the capacitor almost does not change during
the transformation and we can approximate its value by
CS ≈ Cp and vice versa. However, be aware that the same does not happen in the
case of resistors, as their values vary greatly.
