# The Series RLC Circuit

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16.16The Series RLC Circuit

In this Physics tutorial, you will learn:

• Why the maximum voltage in a series RLC circuit is not the arithmetic sum of maximum voltages in each component?
• How can we use the help of phasors to understand how a series RLC circuit works?
• What is the phase relationship between current and voltage across the resistor in a series RLC circuit?
• The same for the inductor and capacitor in series RLC circuits
• What is the impedance of an electrical circuit? What is its unit?
• What is phase constant? How can we calculate it?
• What is resonance? What does the phase constant tells us about resonance in a RLC circuit?
• What are the effective (rms) values of current and voltage in an alternating circuit? How to calculate them?

## Introduction

In the previous tutorial, we discussed about some general features of RLC circuits such as the induced current and voltage or energy in such circuits. In addition, we gave the definition of inductive and capacitive reactance, which were analogue to the resistance provided by resistors.

In this tutorial, we will extend the study of RLC circuits in the interaction between components, especially in regard to the total resistance of the circuit. In addition, we will provide the equations of current amplitude and other quantities not discussed so far.

## Recap on the Series RLC Circuit

As explained in the previous tutorial, a series RLC circuit contains all three elements - a resistor, an inductor and a capacitor - connected in series in the same conducting wire. This means the electromotive force produced by the source produces three separate voltages across each component, as shown in the figure. We denote the maximum voltages across each component as ΔVR(max), ΔVL(max) and ΔVC(max) respectively.

In the previous tutorial, we have also explained that the current is in phase with voltage only at the resistor; the current in the inductor leads the voltage by one quarter of a cycle (π/2) while in the capacitor, current is behind the voltage by a quarter of a cycle (π/2). Since there is the same current flowing throughout the circuit (i.e. the current is in phase in all components), the maximum voltage produced by the source will not be the arithmetic sum of the three maximum individual voltages, i.e.

∆Vsource(max) ≠ ∆VR(max) + ∆VL(max) + ∆VC(max)

This is because voltages are not in phase with each other. However, when considering the instantaneous voltages ΔV(t) for each component, we have

∆V(t)source = ∆Vr (t) + ∆VL (t) + ∆Vc (t)

The last equation is always true and you can consider the loop rule to convince yourself about this.

As for the current flowing in the circuit at any instant, we can write

i(t) = imax ∙ sin⁡(ωd ∙ t-φ)

In this tutorial we will explain how to find the current amplitude imax and the phase constant φ, and investigate how they depend on the driving angular frequency ωd. For this, we will use the phasor diagrams, which - as explained in the previous tutorial - represent a simplified version of current and voltage representation, especially in the actual context, where these values change continuously with time.

## The Current Amplitude

First, let's express the current in a series RLC circuit through a phasor diagram like the one shown below. We can draw the three phasors of voltage for the above position of the current phasor. Thus, since current and voltage across the resistor are in phase, the phasor arrow of the resistive voltage will be collinear with that of current.

On the other hand, the current in the capacitor leads the voltage by π/2 (a quarter of a cycle, or rotation). Therefore, the capacitive voltage phasor is displaced by π/2 radians anticlockwise to the current phasor because capacitive voltage is quarter a cycle behind the current.

Finally, since the current is behind by π/2 to the voltage (it lags voltage by quarter of a cycle), the inductive voltage phasor is displaced by π/2 clockwise to the current phasor.

The following figure shows all four phasors discussed above. The projections of each voltage phasor in the vertical axis give the instantaneous values of the corresponding voltages. They are not shown in the diagram to avoid making it too much crowded.

Since ΔVC(max) and ΔVL(max) have opposite directions, it is better to subtract them, just as we do when subtracting two vectors. Then, we find the resultant of (ΔVL(max) - ΔVC(max)) and ΔVR(max), which represents the net maximum voltage Vnet(max) by applying the rules of vectors addition. Giving that at any instant the phasors obey the rule

εsource = ∆VR + ∆VL + ∆Vc

we obtain for the amplitudes of the above quantities (when phasors are taken as vectors):

εsource(max) = ∆Vnet(max) = ∆VR(max) + ∆VL(max) + ∆VC(max)

Let's use the rules of vector addition to find the net maximum voltage in terms of the other three voltages. Using the notation (ΔVL(max) - ΔVC(max)) instead of their separate notation, we obtain for the net voltage:

εsource(max)2 = ∆VR(max)2 + (ΔVL(max) - ΔVC(max) )2

The last equation is obtained by applying the Pythagorean Theorem. Using the Ohm's Law for each component, we obtain

εsource(max)2 = (imax ∙ R)2 + (imax ∙ XL - imax ∙ Xc)2

where XL and XC are the inductive and capacitive reactances in the circuit respectively.

Rearranging the last equation for the maximum current, we obtain

imax = εsource(max)/R2 + (XL-Xc )2

The expression R2 + (XL-Xc )2 is known as impedance Z of the RLC circuit for the given driving angular frequency ωd. It represents the total opposition a RLC circuit presents to current flow. The unit of impedance is Ohm, Ω. Hence, we have

Z = √R2 + (XL-Xc )2

Thus, we can write

imax = εsource(max)/Z

If we substitute the reactances XL and XC with their corresponding expressions found in the previous tutorial, we obtain

imax =εsource(max)/R2 + (ωd ∙ L-1/ωd ∙ C)2

### Example 1

The voltage in the series RLC circuit shown in the figure oscillates according the expression ε(t) = 150 sin (120π ∙ t). The values of resistance, inductance and capacitance of the corresponding circuit elements are 20Ω, 50mH and 0.4mF respectively. Calculate:

1. The frequency in the circuit
2. The impedance in the circuit
3. The maximum current flowing in the circuit

### Solution 1

1. Since the equation of voltage in a RLC circuit has the form
ε(t) = εmax ∙ sin⁡(ωd ∙ t)
It is clear that εmax = 150V and ωd = 120π rad/s. Also, we have
ωd = 2 ∙ π ∙ f = 120π
2f = 120
f = 60Hz
2. From the equation of impedance Z, we have (giving that L = 50mH = 0.05 H and C = 0.4mF = 0.0004 F):
Z = √R2 + (XL-Xc )2
= √R2 + (ωd ∙ L-1/ωd ∙ C)2
= √202 + (120 ∙ 0.05 - 1/120 ∙ 0.0004)2
= √202 + (6 - 20.8)2
= √202 + (-14.8)2
= √400 + 219
= √619
= 24.9Ω
3. The maximum current flowing through the circuit is given by:
imax = εmax/Z
= 150 V/24.9Ω
= 6.02A

## The Phase Constant

We have explained earlier that the only voltage in phase with current is the resistive voltage. This means for resistive voltage the phase constant is zero (φ = 0). As for the other two voltages, the phase constant is -π/2 for ΔVC and + π/2 for ΔVL.

However, we are more interested about the phase constant of the total voltage, than for individual voltages. In this regard, the general phase constant φ will be the angle formed by the resistive voltage and the total voltage phasors, as shown in the figure. Applying the trigonometry rules, we have:

= ΔVL(max) - ΔVC(max)/ΔVR(max)
= imax ∙ XL-imax ∙ Xc/imax ∙ R
= XL-Xc/R

This formula obtained above is very important, as we don't have to know the amplitudes of current and potential difference in a RLC circuit to calculate the initial phase. It is enough knowing the values of resistance and the two reactances for this.

The following cases are present when considering the last equation of phase constant φ in a series RLC circuit:

1. If XL > XC, the circuit is more inductive than capacitive. The phase constant φ is positive, so the phasor εmax is ahead of the current phasor imax.
2. If XL < XC, the circuit is more capacitive than inductive. The phase constant φ is negative, so the phasor εmax is behind the current phasor imax.
3. If XL < XC, we say the circuit is in resonance, for which we will discuss in the next paragraph. The phase constant is zero, so the current and voltage rotate together.

As special cases, in purely inductive circuits (XL ≠ 0 and XC = R = 0), we have the maximum value possible of phase constant (φ = π/2); in purely capacitive circuits (XC ≠ 0 and XL = R = 0), the phase constant is minimum (φ = π/2); while in purely resistive circuits (R ≠ 0 and XL = XC = 0) the phase constant is zero because φ = 0 and therefore, tan φ = 0.

### Example 2

The equation of voltage in a series RLC circuit is

ε(t) = 40 ∙ sin⁡(100π ∙ t)

and the values of resistance, inductance and capacitance of the corresponding circuit elements (resistor, inductor and capacitor) are 10Ω, 40mH and 80μF respectively.

1. What kind of circuit is it (more resistive, more inductive or in resonance)?
2. What is the phase constant in radians and degrees?
3. What is the maximum current in the circuit?
4. What is the voltage in the circuit at t = 0.506 s after the switch turns on? ### Solution 2

First, let's write some useful clues. Thus, from the equation of voltage, we see that

εmax = 40 V
ωd = 2π ∙ f = 100π rad/s = > f = 50 Hz
R = 10 Ω
L = 40 mH = 4 × 10-2 H
C = 80 μF = 8 × 10-5 F

1. To know what kind of circuit is it, we have to find the two reactances. Thus,
XL = ωd ∙ L
= (100π rad/s) ∙ (4 × 10-2 H)
= 4π Ω
= 12.56Ω
and
Xc = 1/ωd ∙ C
= 1/(100π rad/s) ∙ (8 × 10-5 F)
= 39.81 Ω
Thus, since XC > XL and R ≠ 0, the circuit is more capacitive than inductive.
2. The phase constant is calculated by
tan⁡φ = XL-Xc/R
= 12.56 Ω - 39.81 Ω/10 Ω
= -2.725
Therefore, the phase constant φ is
φ = tan-1 (-2.725)
= -69.80
The negative sign means the phasor εmax is behind the current phasor imax by 69.80.
3. The maximum current in the circuits is
imax = εmax/Z
= εmax/R2 + (XL-Xc )2
= 40 V/(10 Ω)2 + (12.56 Ω-39.81 Ω)2
= 40 V/29 Ω
= 1.38 A
4. The voltage in the circuit at t = 0.506 s is
ΔV(t) = ε(t) = 40 ∙ sin (100π ∙ t)
Thus,
ΔV(0.506) = 40 ∙ sin (100π ∙ 0.506)
= 40 ∙ sin (50.6π)
= 40 ∙ sin (0.6π)
= 40V ∙ 0.951
= 38.04 V

## Resonance in a Series RLC Circuit

As stated earlier, resonance is a state of a RLC circuit in which the current and net emf are in phase. This occurs when XL = XC. The graph of voltage and current versus time in a situation involving resonance for a complete cycle is shown below. The amplitudes may be different from those shown in the above graph; they depend on the corresponding values. This means the current graph may be above the voltage graph or they may have the same height, no matter which graph is higher. The graph above is only for illustration purpose.

We will show below the corresponding phasor diagram as well. In absence of resonance, the graphs of current and voltage do not have their peak at the same time because either current is ahead of voltage (when the phase constant is negative, i.e. when the circuit is more capacitive than inductive), or it is behind the voltage (when the phase constant is positive, i.e. when the circuit is more inductive than capacitive). In such cases, the voltage and current phasors are not collinear, as discussed earlier.

Giving that during resonance XL = XC, we have after substitutions:

ωd ∙ L = 1/ωd ∙ C
ωd2 = 1/L ∙ C
ωd = √1/L ∙ C = 1/L ∙ C

This is the known equation obtained earlier for natural angular frequency of free oscillations. In this case, the oscillations are forced (driven) by the power source. This makes possible the generation of long-lasting (sustainable) oscillations, for which we can apply the approach used in free oscillations, as they do not fade with time.

As for the frequency of RLC circuit oscillations, we have (giving that ωd = 2π ∙ f):

2π ∙ f = 1/L ∙ C

Hence,

f = 1/2π ∙ √L ∙ C

## Effective Values of Alternating Current and Voltage

From all discussed so far about the current and voltage in an alternating circuit, we can point out two important features:

1. Both the current and voltage oscillate in a sinusoidal fashion
2. Current and voltage can be either positive or negative

If we are asked to find other related quantities in an AC circuit such as power or energy, we cannot use the maximum values of current and voltage as this gives a considerable error; all values would be higher than actually they are. We cannot consider other known methods to find the average values such as the arithmetic mean, or approximations of the sinusoids to obtain a series or rectangles for example. Therefore, the only available method remains the root mean square method, similar to that discussed in the kinetic theory of gases, in which we calculated the rms speed of an ideal gas molecule.

Giving that

i(t) = imax ∙ sin⁡(ωd ∙ t)

we obtain for the rms current

irms = √[imax ∙ sin⁡(ωd ∙ t) ]2ave
= √i2max ∙ sin2d ∙ t)ave

Since from trigonometry it is known that

sin2d ∙ t)ave = 1/2

we can write

irms = √i2max1/2
irms = imax ∙ √1/2
= imax1/r = √2
= imax/2
≈ 0.707 ∙ imax

Thus,

iave = irms = imax/2

We have used this value earlier in some exercises but now you know what does it mean.

Likewise, for the average voltage, we have

∆Vave = ∆Vrms = irms ∙ R
= imax/2 ∙ R
= ∆Vmax/2

### Example 3

A series RLC circuit is shown in the figure. #### Calculate:

1. The resonance frequency of the circuit
2. The root mean square of current when the circuit is in resonance
3. The maximum voltages across the resistor, inductor and capacitor when the circuit is in resonance

### Solution 3

We have

εmax = 250 V
R = 5 Ω
L = 5 mH = 5 × 10-3 H
C = 4 μF = 4 × 10-6 F

1. The resonance frequency of the circuit is obtained when XL - XC = 0 or when XL = XC. Thus, we have
XL = Xc
ωd ∙ L = 1/ωd ∙ C
ωd2 = 1/L ∙ C
2π ∙ f = 1/L ∙ C
f = 1/2π ∙ √L ∙ C
= 1/2 ∙ 3.14 ∙ √(5 × 10-3 ) ∙ (4 ∙ 10-6 )
= 1/6.28 ∙ √2 ∙ 10-8
= 1/6.28 ∙ 1.414 × 10-4
= 1126 Hz
2. First, let's calculate the impedance in the circuit. If the circuit is in resonance, we have
Z = √R2 + (XL - Xc )2
= √R2 + 02
= √R2
= R
= 5 Ω
The maximum current in the circuit therefore is
imax = εmax/Z
= 250 V/5 Ω
= 50 A
Hence, the rms current in the circuit is
irms = imax/2
= 50 A/1.414
= 35.4 A
3. The maximum voltages in each component of the circuit are
ΔVR(max) = imax ∙ R
= (50 A) ∙ (5Ω)
= 250 V
ΔVL(max) = imax ∙ XL
= imax ∙ 2πf ∙ L
= (50 A) ∙ (2 ∙ 3.14 ∙ 1126 ∙ 0.005 Ω)
= 1768 V
ΔVC(max) = imax ∙ Xc
= imax1/2πf ∙ C
= (50 A) ∙ (1/2 ∙ 3.14 ∙ 1126 ∙ (4 × 10-6 ) )
= 1768 V

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