Introduction to RLC Circuits

The "envelope" that surrounds the graph of damped oscillations represents a decreasing exponential function of the form x(t) = A₀ ∙ e-γ ∙ t where A₀ is the initial amplitude and γ is a constant. Despite the amplitude decreases with time, this is still a simple harmonic motion because the other quantities such as period and frequency remain unchanged. This kind of SHM has the equation

On the other hand, in sustainable SHM, the amplitude of oscillations does not change with time. The envelope shows a horizontal function of the type x(t) = A₀."

In the next paragraph, we will see that a RLC circuit without a sustainable energy supply from outside, represents a system of damped oscillations because the energy of system decreases with time. This occurs because part of the total energy of the system converts into thermal energy and is therefore dissipated in the form of heat from the resistor to the environment. This brings a continuous decrease in the total energy of the system.

Equation of the Damped Oscillations in a RLC Circuit

In the previous tutorial, we have explained that the total electromagnetic energy in a LC circuit is

When a resistor is added in this circuit, the total electromagnetic energy will decrease at a rate of

because some of the energy in the system turns into thermal energy of resistor and is dissipated in the environment in the form of heat energy. The negative sign in the equation means that the total energy of the system decreases.

Differentiating the above equation with time, we obtain

Since i = dQ/dt, and di/dt = d² Q/dt² we obtain

Simplifying both sides of the last equation by dQ/dt, we obtain

This is the differential equation for the damped oscillations in a RLC circuit.

The charge decay in such a circuit is calculated through an expression, which is a combination of exponential and sinusoidal equation, as occurs in all types of damped oscillations. Thus, the charge left in a RLC circuit after a given time t of operation is found by:

where

is the angular frequency of damped oscillations and

is the angular frequency of undamped oscillations.

In addition, you can see that the amplitude also contains an exponential decaying term e^{-R ∙ t/2L}. This means the amplitude of every successive oscillation is smaller than the previous one as the power of Euler's Number e here is negative.

Example 1

The potential difference between the plates of a 5μF capacitor connected in an alternating 50Hz RLC circuit shown in the figure is 20V and the resistance of resistor in the circuit has a value of 0.25Ω.

Calculate:

1. The initial charge stored in the capacitor
2. The charge stored in the circuit 0.4 s after the switch turns on

Take the initial phase as zero.

Solution 1

1. The initial charge stored in the capacitor is
   Q₀ = C ∙ ∆V
   = (5 × 10^-6 F) ∙ (20V)
   = 10^-4 C
2. First we must find the inductance and then calculate the charge stored in the circuit at the given time. Thus, since the circuit has a frequency of 50Hz, we have
   ω = 2π ∙ f
   = 2π ∙ 50Hz
   = 100π rad/s
   Hence, giving that
   ω = 1/√L ∙ C
   or
   ω² = 1/LC
   we obtain for the inductance L of the inductor:
   L = 1/ω² ∙ C
   = 1/(100π)² ∙ (10^-4 )
   = 1/10⁴ ∙ π² ∙ 10^-4
   = 1/π²
   = 1/(3.14)²
   = 0.1 H
   The charge stored in the capacitor after 20s therefore is
   Q(t) = Q₀ ∙ e^{-R ∙ t/2L} ∙ cos⁡(ω' ∙ t + φ)
   Q(t) = Q₀ ∙ e^{-R ∙ t/2L} ∙ cos√ω² - (R/2L)² ∙ t
   Q(20) = (10^-4 ) ∙ e^{-0.25 ∙ 0.4/2 ∙ 0.1} ∙ cos√(100π)² - (0.25/2 ∙ 0.1)² ∙ 0.4
   Q(20) = (10^-4 ) ∙ e^-5 ∙ cos(√(100π)²-(1.2)² ∙ 0.4)
   = (10^-4 ) ∙ e^-5 ∙ cos(100π ∙ 0.4)
   = (10^-4 ) ∙ e^-5 ∙ cos(40π)
   = (10^-4 ) ∙ e^-5 ∙ 1
   = 10^-4 ∙ 6.7 × 10^-2 = 6.7 × 10^-6 C

From the above results, we draw the following conclusions:

For small values of resistance, ω' ≈ ω, so we can neglect the (R/2L)2 factor and focus only on the value of ω.

We can use a similar approach when calculating the energy decay in a RLC circuit operating without an external energy supply. For this, we observe what happens to the electric energy in the capacitor. Since

we can write this expression as a function of time, i.e.

This means the energy of the electric field in a RLC circuit oscillates in a cos2 fashion while the amplitude decreases exponentially with time.

Example 2

A 50Hz RLC circuit contains a 10Ω resistor, a 0.4H inductor and a 2nF capacitor connected in series. The capacitor initially stores a charge of 5μC. Calculate:

1. The initial electric energy stored in the capacitor plates
2. The energy left in the capacitor plates 2 s after the switch turns OFF.

Take the initial phase equal to zero.

Solution 2

Clues:

R = 10 Ω
L = 0.4 H
C = 2nF = 2 × 10^-9 F
Q₀ = 5μC = 5 × 10^-6 C
f = 50 Hz
a) W_{e initial} = ?
b) W(2) = ?

1. The initial electric energy stored in the capacitor plates is
   W_{e initial} = Q²₀/2C
   = (5 × 10^-6 C)²/2 ∙ (2 × 10^-9 F)
   = 6.25 × 10^-3 J
2. The energy left in the capacitor plates after 2 s of supply interruption is
   W_e (t) = Q²₀/2C ∙ e ^{- R ∙ t/L} ∙ cos²⁡(ω' ∙ t + φ)
   W_e (t) = Q²₀/2C ∙ e ^{- R ∙ t/L} ∙ cos²⁡(√ω²-(R/2L)² ∙ t)
   W_e (2) = (6.25 × 10^-3 J) ∙ e^{-10 ∙ 2/0.4} ∙ cos²√(2 ∙ π ∙ 50)²-(10/2 ∙ 0.4)² ∙ 2
   = (6.25 × 10^-3 J) ∙ e^-50 ∙ cos²(√98596 - 156.25 ∙ 2)
   = (6.25 × 10^-3 J) ∙ e^-50 ∙ cos²(√98439.75 ∙ 2)
   = (6.25 × 10^-3 J) ∙ e^-50 ∙ cos²(627.5 rad)
   = (6.25 × 10^-3 J) ∙ (1.93 × 10^-22 ) ∙ (0.6833)²
   = 5.63 × 10^-25 J
   This value is very small, close to zero. Hence, we say the flow of energy through a RLC circuit, practically stops immediately after the switch turns off. Again, here you can see how necessary a sustainable external source such as a battery or an AC power supply is, in order to maintain constant the electricity flow through a RLC circuit.

Forced Oscillations. Alternating Current and Emf in a RLC Circuit caused by Forced Oscillations

As stated in the previous paragraph, a RLC circuit needs an external sustainable source of emf to make it operate for a long time at steady values. This is made possible by connecting the circuit to an AC power supply, which on the other hand is supplied by an AC generator. An AC generator consists on a current-carrying loop placed inside an external magnetic field, as discussed in the tutorial 16.3 "Magnetic Force on a Current-Carrying Wire. Ampere's Force." Such a system produces forced oscillations in the circuit, which makes it operate for a long time with constant periodic values.

A simplified version of an AC generator is shown in the figure below. A conducting loop is placed inside an external magnetic field. The slip rings are used to create the contact with multiple coils. Each ring is connected to one end of the loop and to the rest of the circuit through metal brushes. As a result, the rings slip against the metal brush and are free to rotate. This brings an induced current I in the circuit.

As the conducting loop of the generator is forced to rotate in the external magnetic field B, an emf is induced in the loop. The equation of this induced emf is

where ε_max is the amplitude of the induced emf in the generator (usually the amplitude is the initial induced emf), while ωd is called the "driving angular frequency" because it drives an induced current in the circuit, the equation of which is given by

where i_max is the amplitude of the driven current in the circuit. The induced current may not be in phase with the induced emf, so the inclusion in the formula of the initial phase φ is necessary.

In addition, we may express the driven angular frequency as

where f_d is the driven frequency of the induced current.

Thus, in circuits with no or very small resistance, the induced current (and emf) oscillate at angular frequency

which is known as the "natural angular frequency". Such oscillations are known as free oscillations.

On the other hand, when a resistor is present is the circuit, the oscillations are known as damped (when no external source is present) or forced (when an external source is needed to keep the values of the induced current and emf constant). The following rule is true for the forced oscillations:

"The induced current and emf in a circuit always occur according the angular frequency of the forced oscillations, regardless the value of the natural angular frequency."

Resistive, Inductive and Capacitive Load

To make the understanding of a RLC circuit more digestible, we will discuss separately three simple circuits, each of them containing only an external emf and one of the three circuit components: resistor, capacitor or inductor, which produce a load in the circuit. Let's start with the circuit that produces the resistive load.

a) Resistive Load

Let's consider a circuit containing only an alternating emf source and a resistor as shown in the figure.

From the Law of Energy Conservation, we have

where ΔV_R is the potential difference across the resistor. We can write this equation as

When neglecting the resistance of conducting wire, we obtain

Moreover, we have for the current in the circuit:

Giving that in this type of circuit the current is in phase with the potential difference, we have φ = 0. The graph below shows one cycle of induced current and potential difference in an AC circuit containing a resistive load:

To make the graph plotting easier, we use arrows similar to vectors even though neither current nor voltage are vectors. These diagrams are known as phasor diagrams. The angle formed by the arrows and the horizontal (time) axis gives the ωt term. A phasor diagram pertaining the above graph is shown below:

Example 3

The voltage of an AC generator is given by the equation ΔV(t) = 90 sin ωt and the frequency of generator is 60 Hz. Calculate:

1. The maximum current produced by the generator if it is connected to a 30 Ω resistor
2. The potential difference in the circuit at t = 2.504 s.

Solution 3

1. From the equation of voltage, we notice than the maximum voltage (potential difference) in the circuit is 90 V. In addition, we obtain for the maximum current flowing through the circuit:
   i_max = ∆V_max/R
   = 90 V/30 Ω
   = 3A
2. The potential difference at t = 2.504 s (giving that ω = 2πf and f = 60 Hz), is
   ΔV(2.504) = 90 ∙ sin (2π ∙ 60 ∙ 2.504)
   = 90 ∙ sin (2π ∙ 60 ∙ 2.504)
   = 90 ∙ sin 300.48π
   = 90 ∙ sin 0.48π
   = 90V ∙ 0.998
   = 89.82 V

b) Capacitive Load

Now, let's consider a circuit supplied by an AC source, which contains only a capacitor C as shown in the figure.

Using a similar approach as we did when dealing with the resistive load, we obtain for the potential difference at any instant across the capacitor:

From the definition of capacitance, we have for the charge stored in the capacitor at any instant t:

and for the current at any instant t:

The quantity

is called capacitive reactance of capacitor. It has the unit of resistance (Ohm).

From experiments, it results that current leads by one quarter of a cycle the voltage in such a circuit. If we replace the cos ωd ∙ t term with a phase-shifted sine expression of + π/2 rad, we obtain

Hence, we obtain for the current in the circuit:

In addition, we have for the maximum potential difference in the circuit

Since there is a shift in phase by one quarter of a period (π/2 = 2π/4 = T/4), the graphs of potential difference and current versus time for one complete cycle are:

The corresponding phasor diagram for this circuit is

The current is π/2 (a quarter of a cycle) in advantage to potential difference. Therefore, we say: "the current leads the voltage by π/2".

Remark! The capacitive reactance behaves as an AC resistance. As the frequency of current approaches zero, the capacitive reactance raises to infinity and as a result, the circuit behaves as a DC circuit. However, the current flow in this way (in one direction only) is prevented from the high resistance between the plates of capacitor (at the gap between the plates), which does not allow the current to flow between the plates and to close therefore the cycle.

Example 4

A circuit containing a 60 Hz AC power source that oscillates according the equation ΔVC(t) = 50 ∙ sin ωd ∙ t and a 20 μF capacitor as shown in the figure.

Calculate:

1. The capacitive reactance in the circuit
2. The maximum current flowing in the circuit
3. The value of voltage and current in the circuit at t = 1.406 s.

Solution 4

1. From the clues, it is clear that ΔV_C(max) = 50 V, C = 20μF = 2 × 10-5 F and f = 60 Hz. Thus, for capacitive reactance, we have
   X_c = 1/ω_d ∙ C
   = 1/2π ∙ f ∙ C
   = 1/2 ∙ 3.14 ∙ (60 Hz) ∙ (2 × 10^-5 F)
   = 132.7Ω
2. The maximum current flowing through the circuit is given by
   ∆V_C(max) = i_C(max) ∙ X_c
   Thus,
   i_c(max) = ∆V_c(max)/X_c
   = 50 V/132.7Ω
   = 0.38 A
3. The value of voltage in the circuit at t = 1.406 s is
   ∆V_c (t) = ∆V_C(max) ∙ sin ω_d ∙ t
   ∆V_c (1.406) = 50 ∙ sin (2π ∙ 60 ∙ 1.406)
   = 50 ∙ sin (168.72π)
   = 50 ∙ sin (0.72π)
   = 50V ∙ 0.77
   = 38.5 V
   As for the current at t = 1.406 s, we have
   i_c (t) = i_C(max) ∙ sin (ω_d ∙ t + π/2)
   i_c (1.406) = 0.38 ∙ sin (168.72π + π/2)
   = 0.38 ∙ sin (169.22π)
   = 0.38 ∙ sin (1.22π)
   = 0.38A ∙ (-0.64)
   = -0.24 A
   The negative sign shows direction; it means the current actually is flowing in the opposite direction to the initial current at t = 0.

c) Inductive Load

The reasoning is the same even when we have a circuit in which there is only an AC source and an inductor as shown in the figure.

The potential difference across the inductor is

From Faraday's Law, we have

Thus, combining the above equations, we obtain

Or

The current flowing at any instant in the circuit is obtained through integration techniques. Thus,

The quantity

is called inductive reactance and is measured in Ohms, similarly to capacitive reactance discussed in the previous paragraph.

Using the trigonometric identity

we obtain for the current flowing in a circuit containing an inductive load:

From this equation, we can see that for a purely inductive load, the current is delayed (is out of phase) by π/2 (a quarter of a cycle) to the potential difference. Therefore, we obtain the graph below:

Again, we use the phasor concept to simplify the understanding of the above graph. The phasor diagram that corresponds the above graph is shown below:

Example 5

A 0.2 H inductor is connected to an AC source of voltage ΔVC(t) = 40 ∙ sin(100 π ∙ t).

1. Calculate the inductive reactance of the circuit
2. Write the equation of current in the circuit as a function of time
3. Calculate the current and voltage in the circuit at t = 4.961 s.

Solution 5

1. From the clues, we have ΔV_L(max) = 40 V and L = 0.2 H. In addition, we have
   ω_d = 2πf = 100π
   so, f = 50 Hz. The inductive reactance in the circuit is
   X_L = ω_d ∙ L
   = 100π ∙ 0.2
   = 20π Ω
   = 20 ∙ 3.14 Ω
   = 62.8 Ω
2. Since the inductive reactance is like a resistance in the circuit, we obtain for the maximum current flowing through the circuit (applying the Ohm's Law):
   i_L(max) = ΔV_L(max)/X_L
   = 40V/62.8Ω
   = 0.64A
   Therefore, the equation of current in the circuit is
   i_L (t) = ∆V_L(max)/ω_d ∙ L ∙ sin⁡(ω_d ∙ t - π/2)
   = 40/100π ∙ 0.2 ∙ sin (100π ∙ t - π/2)
   = - 2/π ∙ sin (100π ∙ t - π/2)
   = 0.64 ∙ sin (100π ∙ t - π/2)
3. The current at t = 4.961s is
   i_L (4.961) = 0.64 ∙ sin (100π ∙ 4.961 - π/2)
   = 0.64 ∙ sin (100π ∙ 4.961 - π/2)
   = 0.64 ∙ sin (496.1π - π/2)
   = 0.64 ∙ sin (495.6π)
   = 0.64 ∙ sin (1.6π)
   = 0.64A ∙ (-0.95)
   = 0.61A
   and the voltage in the circuit at t = 4.961s is
   ∆V_L (t) = ∆V_L(max) ∙ sin⁡(ω_d ∙ t)
   ∆V_L (4.961) = ∆V_L(max) ∙ sin⁡(ω_d ∙ t)
   = 40 ∙ sin⁡(100π ∙ 4.961)
   = 40 ∙ sin⁡(496.1π)
   = 40 ∙ sin⁡(0.1π)
   = 40V ∙ 0.31
   = 12.4V

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