Theory Circuit RLC in AC

This chapter consists of the items below. If you want to go straight to an item, Click on it.

1 - Introduction Click here!

2 - RLC Series Circuit Click here!

2.1 - Impedance Diagram Click here!

2.2 - Fasorial Diagram of Voltages Click here!

3 - RLC Paralell Circuit Click here!

3.1 - Fasorial Currents Diagram Click here!

After studying separately circuits RC and RL let's study how circuits containing the three elements behave simultaneously.

2. RLC Series Circuit

Let us first recall that the capacitive reactance is represented, in complex form, as - j X_C and the inductive reactance as + j X_L. Therefore, if these components are in series, the equivalent reactance will be the algebraic sum of the two reactances.

In this case, we easily perceive that if X_L > X_C, let's get a circuit with inductive predominance. Otherwise, that is, X_C > X_L, then we will have a circuit with capacitive predominance.

Taking this into account, the equivalent impedance of an RLC series can be written as:

eq. 55-01

In the next figure we see an RLC series circuit. The values of the supplied components allow determining whether the circuit has inductive or capacitive predominance. Note that the current I is the same in all components.

Also we must pay attention to the fact that the voltage on the resistor will be the only one that will be in phase with the current I. As the voltage on the capacitor will be 90° delayed in relation to the current and the voltage on the inductor will be 90° advanced in relation to the same current, we conclude that the voltages on the capacitor and on the inductor will be out of phase by 180°.

Writing the equivalent impedance value in rectangular form according to eq. 55-01 we have:

Z_eq = 9.95 + j (30 - 15) = 9.95 + j15 Ω

And in polar form:

Z_eq = 18 ∠ +56.31° Ω

Now that we know the equivalent impedance value, we can calculate the value of the current flowing through the circuit. Like this:

_eq

Performing the calculation we find:

I = 5 ∠ -56.31° A

Pay attention to the fact that the angle - 56.31° means that the current is delayed in relation to the voltage V. As we know, delayed current means that the circuit has inductive predominance.

With the value of the current we can calculate the values of V_R , V_L and V_L.

Note that because resistances do not cause lags, V_R is absolutely in phase with I. The same does not happen with the inductive and capacitive reactance.

V_L = X_L I = 30 ∠ +90 5 ∠ -56.31° V

Notice that we have transformed j 30 into 30 ∠ 90°. Performing the calculation, we find:

V_L = 150 ∠ +33.69° V

For the calculation of the voltage on the capacitor we have:

Notice that we have transformed -j 15 into 15 ∠ -90°. Performing the calculation, we find:

V_C = 75 ∠ -146.31° V

As we know that the current is 56.31° delayed in relation to the voltage, then we can calculate the power factor, or:

PF = cos 56.31° = 0.55 inductive or delayed

We find a delayed power factor because the circuit has an inductive predominance. Hence we conclude that if we had a circuit with capacitive predominance the power factor would be advanced.

It is possible to realize in a series RLC circuit that if the inductive reactance is exactly equal to the capacitive reactance they cancel out, and the equivalent impedance will simply be the value of the resistor and with this we obtain a unit power factor. When that happens the circuit is said to be in resonance. We will study about it in chapter 58. If interested on that subject click here!.

Reflexion moment!

"Question:" From what is said above, we easily realize that we can correct the power factor by installing a series capacitor with an RL circuit. In addition, we know that to correct the power factor industries install capacitors in parallel in your electrical installations. So why do not industries use this method (series capacitor) to correct the power factor?

2.1 Impedance Diagram

See the Figure 55-02 for the representation of the impedances involved in the problem. On the left side (of the figure), pointing upwards (green color) we have the inductive reactance. Pointing down (Purple Color) we have the capacitive reactance. These two quantities are on the imaginary axis. And in the real plane we have the resistor R.

On the right side (of the figure), on the vertical (imaginary) axis we have the result X_L - X_C, pointing up then X_L > X_C. Otherwise, it would point down. Note that to find the equivalent impedance we must add in the form phasor the magnitudes R and (X_L - X_C). The module will be given by:

eq. 55-02

Notice that we only employ the Pythagorean theorem. And to compute the angle φ it is enough to find the arc-tangent of the quotient between the imaginary and real part, that is:

eq. 55-03

Thus, with this data we can easily write the equivalent impedance on polar form:

eq. 55-04

2.2. Fasorial Diagram of Voltages

The Figure 55-03 shows the representation of the voltages involved in the problem. We take as reference the voltage V (horizontally, φ = 0°).

Note that current I is delayed 56.31° in relation to V and V_R in phase with I. On the other hand, we see V_L advanced 90° and V_C delayed 90° with respect to I.

Thus, V_L and V_C are out of phase with 180° between each other, as discussed above.

3. RLC Paralell 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 inductor will be delayed by 90° in relation to the voltage and the current in the capacitor will be advanced by 90° with respect to the same voltage. Then, the current that will be supplied by the voltage source will be the phasor sum of the three currents that circulate through the components. See the circuit show in the Figura 55-04.

For the sake of didactics we will initially compute the equivalent impedance that the three components in parallel present to the voltage source.

As we know, we can use the same principles studied for DC to calculate the equivalent impedance.

We will assume the reactances as resistors and perform the calculation as if they were three resistors in parallel. But it should not be forgotten that reactances are complex numbers. So, for clarity, let's first calculate the parallel of the reactances.

X_eq = X_L X_C / (X_L + X_C )

Let's replace the variables by their numerical values and perform the calculation.

X_eq = j20 (-j10) / (j20 - j10) = - j20 Ω

Now, by calculating the parallel between R and X_eq let's get Z_eq.

Z_eq = R X_eq / (R + X_eq)

Let's replace the variables by their numerical values and perform the calculation.

Z_eq = 10 (- j20) / (10 - j20)

By placing in polar, numerator and denominator format, it is very easy performing calculate. Therefore, for the equivalent impedance we find:

Z_eq = 8.94 ∠ -26.57° = 8 - j4 Ω

For the values of the supplied components in this problem, we find that the equivalent impedance has capacitive predominance. We can also reach to this conclusion by analyzing the angle φ, which is negative.

Since we know the angle of the impedance, we can calculate the power factor, or:

PF = cos (-26.57°) = 0.89 advanced or capacitive

Let's calculate the currents I_R, I_L and I_C.

I = V / Z_eq = 220 ∠ 0° / 8.94 ∠ -26.57° A

Performing the calculation, we find:

I = 24.60 ∠+26.57° A

3.1. Fasorial Currents Diagram

In order to calculate the value of I we must add phasorally the currents I_R, I_L and I_C.

See the Figure 55-05 on the left, as I_R is in phase with V. I_L is delayed 90° and I_C advanced of 90° in relation to V. As I_L and I_C are out of phase with 180° between each other, we just need to subtract them and get the result module. So we have as a result |I_L - I_C| = |11 - 22| = 11 A . In this way we obtain two phasors that form an angle of 90° with each other.

And as I_C > I_L the resulting phasor points up as shown in the Figure 55-05, on the right. To calculate the module of I we can use the Pythagorean theorem. So we write that:

|I| = √(|I_R|² + |I_L - I_C|²)

Substituting the numerical values and performing the calculation, we find:

|I| = √(22² + 11²) = 24.60 A

To find the angle we must compute the arcotangent of the quotient I_L / I_R.

φ = tg^-1 (11 / 22) = 26.57°

And so we can write the final value of the current, or:

I = 24.60 ∠+26.57° A

So the end result is exactly the same when we apply the Ohm law to find I.

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