# Introduction to RLC Circuits

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16.15Introduction to RLC Circuits

In this Physics tutorial, you will learn:

• What is a RLC circuit?
• Why oscillations in a RLC circuit are damped?
• What is the equation of oscillations in a RLC circuit?
• What is angular frequency of damped oscillations? How does it differ from the angular frequency of undamped oscillations?
• How to find the energy in a damped RLC circuits at any instant?
• What are forced oscillations? What do they cause in an electric circuit?
• How to calculate the AC current and emf caused by forced oscillations in a circuit at any instant?
• How to cope with AC circuits containing a single resistor/capacitor/inductor?
• What are phasor diagrams? Why do we use them?

## Introduction

In the previous tutorial, we have discussed about LC circuits. What happens if we add a resistor in this circuit? Recall what we said in the previous tutorial about the resistance of LC circuits and how it affects the energy distribution in the circuit.

Do you think capacitors have resistance? What about inductors? Recall how an inductor behaves after a long time of circuit operation. What can you say about the resistance of inductor at the beginning of process?

Now, let's explain these questions by considering the most inclusive type of circuit - the RLC circuit. For simplicity, here we will consider only the series RLC circuit.

## What is a RLC Circuit? Damped Oscillations in a RLC Circuit

By definition, a RLC circuit is a circuit that contains at least a resistor, a capacitor and an inductor. The simplest RLC circuits are series RLC circuits, like the one shown in the figure below. As stated in the previous tutorial, when the resistance of circuit is considered, the total energy (electric plus magnetic) in the circuit does not remain constant but it rather decreases with time as some of this energy converts into thermal energy and is therefore dissipated in the form of heat by the resistor. As a result, the energy in the circuit will decrease after each cycle; the oscillations fade and becomes less visible until they disappear completely.

Due to this decrease in the electromagnetic energy of the system, all related quantities such as current, charge and potential difference will decrease too, as there is a continuous decrease in the amplitude of oscillations. We say the oscillations are damped, exactly as occurs in a real block-and-spring oscillation system, in which the amplitude of oscillations decreases after each cycle because of friction.

In tutorial 10.2 "Energy in Simple Harmonic Oscillations" we have provided a general overview on what damped oscillations are. Let's recall a passage from the core paragraph of that tutorial:

"If we make a system oscillate, the amplitude of oscillations would decrease until it becomes zero if energy is not provided continuously to the system. In this case, the oscillations fade away with time. Such oscillations are known as damped oscillations the amplitude of which decreases with time. Look at the graph below. ,/i> 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

x(t) = A0 ∙ e-γ ∙ t cos⁡(ω ∙ t + φ)

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

Wtot = WM + We = L ∙ i2/2 + Q2/2C

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

dWtot/dt = -i2 ∙ R

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

dWtot/dt = L ∙ i ∙ di/dt + Q/CdQ/dt = -i2 ∙ R

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

L ∙ dQ/dtd2 Q/dt2 + Q/CdQ/dt = -R ∙ (dQ/dt)2
L ∙ dQ/dtd2 Q/dt2 + R ∙ (dQ/dt)2 + Q/CdQ/dt = 0

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

L ∙ d2 Q/dt2 + R ∙ dQ/dt + 1/C ∙ Q = 0

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:

Q(t) = Q0 ∙ e-R ∙ t/2L ∙ cos⁡(ω' ∙ t + φ)

where

ω' = √ω2 - (R/2L)2

is the angular frequency of damped oscillations and

ω = 1/L ∙ C

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
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 ∙ cosω2 - (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.

• In absence of an external source, the charge in the circuits drops very fast (in only 0.4 s it has dropped about 15 times, from 10-4C to 6.7 × 10-6 C). This means that if the emf source is not active, the RLC circuit discharges almost immediately due to the electromagnetic-to-heat energy conversion, which takes place in the resistor.
• 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 = Q2/2C

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

We (t) = [Q(t)]2/2C
= [Q0 ∙ e-R ∙ t/2L ∙ cos⁡(ω' ∙ t + φ) ]2/2C
= Q20/2C ∙ e - R ∙ t/L ∙ cos2⁡(ω' ∙ t + φ)

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) We initial = ?
b) W(2) = ?

1. The initial electric energy stored in the capacitor plates is
We 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
We (t) = Q20/2C ∙ e - R ∙ t/L ∙ cos2⁡(ω' ∙ t + φ)
We (t) = Q20/2C ∙ e - R ∙ t/L ∙ cos2⁡(√ω2-(R/2L)2 ∙ t)
We (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

ε(t) = εmax ∙ sin ωd ∙ t

where εmax 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

i(t) = imax ∙ sin (ωd ∙ t - φ)

where imax 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

ωd = 2π ∙ fd

where fd 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

ω = 1/L ∙ C

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.

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

ε - ∆Vr = 0

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

∆Vr (t) = εmax ∙ sin ωd ∙ t

When neglecting the resistance of conducting wire, we obtain

ir (t) = ε(t)/R = εmax/R ∙ sin ωd ∙ t

Moreover, we have for the current in the circuit:

ir (t) = i(t) = imax ∙ sin⁡(ωd ∙ t-φ)

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:
imax = ∆Vmax/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

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:

∆Vc (t) = ∆VC(max) ∙ sin⁡ωd ∙ t

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

Qc (t) = C ∙ ∆Vc (t)
= C ∙ ∆VC(max) ∙ sin ωd ∙ t

and for the current at any instant t:

ic (t) = dQc/dt = ωd ∙ ∆VC(max) ∙ cos ωd ∙ t

The quantity

Xc = 1/ωd ∙ C

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

cos⁡ωd ∙ t = sin⁡(ωd ∙ t + π/2)

Hence, we obtain for the current in the circuit:

ic (t) = ∆VC(max)/Xc ∙ sin⁡(ωd ∙ t + π/2)

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

∆VC(max) = iC(max) ∙ Xc

Since there is a shift in phase by one quarter of a period (π/2 = /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 ΔVC(max) = 50 V, C = 20μF = 2 × 10-5 F and f = 60 Hz. Thus, for capacitive reactance, we have
Xc = 1/ωd ∙ 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
∆VC(max) = iC(max) ∙ Xc
Thus,
ic(max) = ∆Vc(max)/Xc
= 50 V/132.7Ω
= 0.38 A
3. The value of voltage in the circuit at t = 1.406 s is
∆Vc (t) = ∆VC(max) ∙ sin ωd ∙ t
∆Vc (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
ic (t) = iC(max) ∙ sin (ωd ∙ t + π/2)
ic (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.

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

∆VL (t) = ∆VL(max) ∙ sin ωd ∙ t

∆VL = L diL/dt

Thus, combining the above equations, we obtain

diL/dt = ∆VL/L = ∆VL(max)/L ∙ sin ωd ∙ t

Or

diL = ∆VL(max)/L ∙ sin (ωd ∙ t)dt

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

iL (t) = diL = ∆VL(max)/Lsin⁡(ωd ∙ t)dt
= ∆VL(max)/L ∙ (-1/ωd ) ∙ cos⁡(ωd ∙ t)
= -∆VL(max)/ωd ∙ L ∙ cos⁡(ωd ∙ t)

The quantity

XL = ωd ∙ L

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

Using the trigonometric identity

-cos (ωd ∙ t) = sin (ωd ∙ t - π/2)

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

iL (t) = ∆VL(max)/ωd ∙ L ∙ sin (ωd ∙ t - π/2)

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 ΔVL(max) = 40 V and L = 0.2 H. In addition, we have
ωd = 2πf = 100π
so, f = 50 Hz. The inductive reactance in the circuit is
XL = ωd ∙ 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):
iL(max) = ΔVL(max)/XL
= 40V/62.8Ω
= 0.64A
Therefore, the equation of current in the circuit is
iL (t) = ∆VL(max)/ωd ∙ L ∙ sin⁡(ωd ∙ 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
iL (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
∆VL (t) = ∆VL(max) ∙ sin⁡(ωd ∙ t)
∆VL (4.961) = ∆VL(max) ∙ sin⁡(ωd ∙ t)
= 40 ∙ sin⁡(100π ∙ 4.961)
= 40 ∙ sin⁡(496.1π)
= 40 ∙ sin⁡(0.1π)
= 40V ∙ 0.31
= 12.4V

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