Alternating Current. LC Circuits

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16.14Alternating Current. LC Circuits

In this Physics tutorial, you will learn:

• What are the main features of a LC circuit?
• What is AC current? How is it obtained?
• What happens to the energy in a LC circuit?
• What kind of function is more appropriate to describe the quantities involved in a LC circuit?
• What is the mechanical analogue of a LC circuit? Which are the corresponding quantities?
• How to calculate the angular frequency of LC oscillators?
• What is the differential equation of LC oscillators? What is its solution?
• How to calculate the potential difference in a LC circuit?

Introduction

Consider again the example in which a magnet moves towards or away from a coil. What happens to the direction of the current induced in the coil? Is it always the same?

What is frequency? What is its unit? (Recall the frequency of spring oscillations and that of circular motion to answer this question).

Choose an electric device you have at home. Look at the plate in which all operating values are shown. How many hertz (Hz) does the plate read?

Now do the same thing with another device. How many hertz does it read? What conclusion do you draw about the operating frequency of electric devices?

This tutorial is about alternating current, which is a type of current produced through induction. Then, we will explain how RC circuits operate, especially in regard to the resistance each component shows.

LC Oscillations

In the previous tutorials, we have discussed about the basic features of the electric and magnetic field, and methods how energy is stored in capacitors and inductors. Now, we will explore the methods, in which the energy stored in one place of the circuit can be transferred to another place in order to do any work. This is true for the energy transfer between different circuits as well. For example, the energy produced at a power plant is transferred through the mains electricity at our homes to run electric devices. Now, we will discuss how this is made possible through the energy transfer methods.

The current produced by cells and batteries is known as direct current (DC), as discussed in tutorial 16.2. This is because it is obtained through direct contact between battery and the conducting wire. However, most of energy in use today is transferred not as DC but rather, as alternating current (AC), which is a kind of electromagnetic oscillation propagating through the conductor in a sinusoidal way. Such oscillations are made possible through the induced current, which take place in LC circuits with inductance L and capacitance C.

So far, we have discussed about RC circuits, in which a capacitor is charged through a resistor and RL circuits, in which some of the energy produced by the source is stored in the magnetic field of an inductor. In both these kind of circuits, the current is not built immediately in the circuit but quantities such as current, charge or potential difference increase or decrease in an exponential fashion according a certain scale, represented through the time constant τ (τC for RC and τL for RL circuits).

LC circuits are different in this regard, i.e. the current and potential difference does not increase or decrease exponentially but in a sinusoidal fashion instead. The behavior of electric-related quantities here, is similar to all kinds of waves (especially electromagnetic ones) discussed in Sections 10 and 11 ("Simple Harmonic Motion" and "Waves"). The oscillations produced in the capacitor (resulting in the change of capacitor's electric field|) and in the inductor (resulting in the change in the inductor's magnetic field) are known as electromagnetic oscillations. Therefore, the energy is a LC circuit is always the sum of electric and magnetic energy stored in the capacitor and inductor respectively, i.e.

Wcircuit = We(capacitor) + Wm(inductor)

or

Wcircuit = Q2/2C + L ∙ i2/2

This situation is analogue to the mechanical energy stored in oscillating springs discussed in Section 5, which represents the sum of kinetic and potential energy of the system. If the mechanical energy is conserved (when no external factors interfere in the event), the kinetic energy is converted into potential and vice-versa; however, their sum (mechanical energy) is always constant.

We know from previous tutorials that a capacitor is charged and discharged in equal intervals (periods). When the capacitor is near the maximum value of charge it can store (near the full capacitance), the charge is built more slowly in it. This is analogue to when an object thrown upwards is near the highest position or when an oscillating spring is near the maximum deformation. When the capacitor is fully charged, the inductor does not store any energy in its magnetic field. As a result, the above equation becomes

Wcircuit = Qmax2/2C

On the other hand, when the capacitor is discharged, the entire energy of the circuit is stored in the magnetic field of the inductor. Hence, the equation representing the total energy in the circuit becomes

Wcircuit = L ∙ imax2/2

The inductor shows some resistance to the current flowing through it (from Lentz Law). Therefore, any change in the energy stored in the magnetic field of the inductor takes some time (period). Hence, any switch between the maximum and minimum values of energy stored in the capacitor or inductor occur in a periodical (and sinusoidal) fashion.

There is no direct source in such circuits, so we cannot assign a positive or negative direction to the current flow. As a result, the plates of capacitor are charged oppositely in equal time intervals. Look at the figure below.

In the stage 0, there is no current flowing in the circuit; this means no energy is stored in the magnetic field of inductor. The capacitor is fully charged, and all the energy of the system is only electric due to the charge stored in the capacitor plates.

In the stage 1, the capacitor is discharging; some current is flowing throughout the circuit and therefore, a part of the total energy is stored in the magnetic field of the inductor. On the other hand, the electric energy stored in the capacitor is decreasing.

In the stage 2, the capacitor is fully discharged and all the energy of the system is stored in the magnetic field of inductor. This means the energy in the system is entirely magnetic. In addition, the current flowing through the circuit in this stage has the maximum value.

In the stage 3, the capacitor's plates are charging at the opposite signs to the stage 2 (due to the inertia of current, i.e. the charges are accumulating in the opposite plates of the capacitor in respect to the stage 1). As for the current, its value is decreasing, so the energy stored in the magnetic field is less than in the stage 2.

In the stage 4, the capacitor is again fully charged and no current is flowing through the circuit. Therefore, all energy of the system is stored in the capacitor plates in the form of electric energy. In a certain sense, this stage is similar to the initial one (stage 0) except the direction of the magnetic field between the plates of capacitor, which is opposite to that in the stage 0.

In the stage 5, the capacitor is discharging and the current in the circuit is increasing. This situation is similar to stage 1, except the direction of current and electric field between the capacitor plates is opposite to those in stage 1.

The stage 6 is similar to 2. All the energy of system is due to the energy stored in the magnetic field of inductor; the capacitor has no charge stored in its plates. The current is flowing at maximum rate in the circuit. However, the direction of current is opposite to that in stage 2.

In the stage 7, the current in the circuit is decreasing and the charges are accumulating at the capacitor plates at opposite direction to stage 4. The energy in the system is the sum of electric and magnetic ones.

Then, the process start from the beginning, i.e. if we included in the figure an eighth stage, it would be identical to the stage 0. This means the process is periodical.

As you see, these eight situations represent a whole cycle, in which the current switches between zero and imax. In addition, the current changes direction during its flow through the circuit. The same thing can be said for the magnetic field of the inductor as well. You can see from the direction of magnetic field lines that the poles of the electromagnet obtained by means of the current flow through the inductor. Since the current changes direction, the magnetic field lines change direction as well.

Remark! In the above analysis, we have neglected the heating effect of the conducting wire. In fact, the maximum potential energy stored in the magnetic field of inductor is not equal to the maximum electric field stored in the capacitor because some of the energy of capacitor turns into heat energy of wire. Therefore, we can write

We(capacitor) ≥ Wm(inductor)

Obviously, the equality sign is used only when we neglect the resistance of conducting wire of the circuit.

The graph below shows the relationship between the values of electric and magnetic energy stored in the capacitor and inductor respectively at any instant, not only in the eight stages discussed above. (The resistance of conductor is considered as the two energies are not equal when they are at maximum).

Both graphs are sinusoidal. This means the corresponding graphs for the current I flowing through the inductor and for the potential difference ΔV between the plates of capacitor are sinusoidal as well. Since by definition current is the amount of charge in the unit time and giving that the circuit has a certain resistance (and therefore a potential difference ΔVR = i ∙ R between the terminals due to this resistance), we conclude that all the four above quantities - the current i flowing in the circuit, the potential difference ΔVC between the capacitor plates, the charge Q stored in the capacitor plates, and the potential difference ΔVR due to the resistance of the circuit - are all sinusoidal.

As stated earlier, the graph will not continue to exist in the actual shape infinitely; the initial energy in the circuit will be lower than before after each cycle as some of this energy converts to heat due to the resistance in the circuit. This means the amplitudes (the vertical displacements of the graph in respect to the horizontal axis) will be lower and lower until both graphs become linear and converge at the horizontal axis. Therefore, a sustainable source is required to keep the values of the four above quantities repeat themselves for a long time.

Example 1

The maximum current flowing through a LC circuit is 3A. The resistance of the circuit is 0.05 Ω while the time needed to the current to complete a whole cycle is 0.02 s. The inductance of the solenoid is 4H. From these data, calculate

1. The total energy stored in the circuit at the beginning of the process.
2. The initial potential difference across the capacitor plates if it stores 4μC of charge at this instant
3. The maximum charge stored in the capacitor plates after 5 cycles
4. The number of cycles after which the energy remained in the system will be half of the initial energy.

Hint! We will see later that the average value of current flowing in AC circuits is calculated by

Solution 1

1. The maximum initial energy of the system is calculated by finding the maximum energy stored in the inductor during the first cycle. Thus,
W0(max) = L ∙ imax2/2
= (4 H) ∙ (3 A)2/2
= 18 J
2. The maximum energy stored in the magnetic field of inductor during the first cycle is equal to electric energy stored in the inductor during the same cycle. Thus, since
W0(max) = (Q20)/2C
we obtain for the capacitance of capacitor:
C = Q20/2W0(max)
= 4 × 10-6 C)2/2 ∙ (18 J)
= 4.44 × 10-13 F
3. Based on the Joule's Law, the total energy in the circuit decreases by a value of
∆W = < i >2 ∙ R ∙ ∆t
after each cycle in respect to the previous value. Thus, the decrease in energy of the system after 5 cycles (N = 5) is
∆Wn = N ∙ < i >2 ∙ R ∙ ∆t
= 5 ∙ (imax/2)2 ∙ R ∙ ∆t
= 5 ∙ (3A/2)2 ∙ (0.05Ω) ∙ (0.02 s)
= 0.0225 J
Therefore, the energy remained in the system after 5 cycles is
W5 = W0(max) - ∆W5
= 18 J - 0.0225 J
= 17.9775 J
4. Now, we use the relation
Wn = W0(max) -N ∙ ∆W1
where N is the number of cycles and ΔW1 = < i > ∙ R is the decrease in the total energy in the system after each cycle. It represents the amount of energy converted into heat during each cycle due to the resistance of the circuit. Thus, since WN = 1/2 Wmax = 9J, we obtain
9 = 18-N ∙ (3A/2)2 ∙ (0.05Ω) ∙ (0.02 s)
N ∙ (0.0045 J) = 18J - 9J = 9J
N = 9J/0.0045 J
= 2000 cycles

Electrical to Mechanical Analogy between Two Oscillating Systems

As stated earlier, an oscillating block and spring system represents a good analogy when trying to understand the behavior of quantities involved in a LC circuit, as both systems oscillate in a sinusoidal fashion. Thus, there are two types of energy involved in the block and spring system: one is kinetic (due to the block movement) and the other is (elastic) potential (stored in the spring). Neglecting the friction in the block and spring system and resistance in the LC circuit, we have

Etot = KEblock + EPEspring

for the block and spring system and

Wtot = Wcapacitor + Winductor

for the LC circuit. The following table makes this point clearer.

From the above table, it is easy to deduce the following correspondences:

x ⇢ Q
k ⇢ 1/C
v ⇢ i
m ⇢ L

In Section 11 (more precisely in tutorial 11.2 "Energy in Simple Harmonic Motion"), we have seen that the general equation for the angular frequency ω of a SHM is

ω = /T

where T is the period of oscillation (i.e. the time needed to complete one oscillation). When applied for the mass and spring system, the equation of angular frequency becomes

ω = √k/m

where k is the spring constant and m is the mass of the attached object.

Substituting 1/C for k and L for m, we obtain for the angular frequency of a LC circuit:

ω = √1/C/L
= √1/L ∙ C
= 1/L ∙ C

Giving that

ω = 2π ∙ f

where f is the frequency of oscillations in Hertz, we obtain for the frequency of oscillations in a RL circuit (electrical frequency):

2π ∙ f = 1/L ∙ C

or

f = 1/2π ∙ √L ∙ C

The value of electrical frequency in US, Canada, Brazil, Colombia, some regions in Japan and some small countries in Central America is 60Hz, while in the rest of the world, this frequency is 50 Hz. This means the current makes 50 or 60 complete cycles in one second when flowing through an AC circuit.

Example 2

A LC circuit is operating somewhere in Europe. What is the capacitance of the capacitor if the inductance of the solenoid is 0.4 H?

Solution 2

Since the circuit is operating in Europe, the frequency of electricity must be 50 Hz. Therefore, giving that

f = 1/2π ∙ √L ∙ C

we obtain for the capacitance C after raising both parts of the equation in power two to remove the root:

f2 = 1/2 ∙ L ∙ C
C = 1/2 ∙ f2 ∙ L
= 1/4 ∙ 3.14 ∙ (50 Hz)2 ∙ (0.4 H)
= 7.96 × 10-5 F
= 79.6 μF

LC Oscillations - A Quantitative Approach

Before deriving the equations of the physical quantities involved in a LC oscillator, we must recall the concept of derivative as a rate of function's change. For example, in kinematics, the instantaneous velocity is the rate of position change in a very narrow interval, so we can write

v = dx/dt

We say, "velocity is the first derivative of position to the time". Likewise, acceleration is the first derivative of change in velocity to the time or the second derivative of the change in position to the time (derivative of derivative). We have

a = dv/dt = d2 x/dt2

and so on.

In addition, we will use again the block-and-spring and LC circuit analogy to describe the quantities involved in LC oscillation systems in a more comprehensive way. For, example, we can write for the total energy of a block-and-spring system of oscillations:

Etot = KEblock + EPEspring
= m ∙ v2/2 + k ∙ x2/2

If we ignore any friction, then the total energy is conserved and such oscillations are known as harmonic oscillations. Since the total energy does not change with time but remains constant instead, we can write

dEtot/dt = 0

or

dEtot/dt = d/dt (m ∙ v2/2 + k ∙ x2/2)
= m ∙ v ∙ dv/dt + k ∙ x ∙ dx/dt
= 0

Substituting v = dx/dt and dv/dt = d2x/dt2 as discussed above, we obtain

m ∙ d2 x/dt2 + k ∙ x = 0

This is the fundamental equation of block-and-spring oscillator, the general solution of which (as discussed in the tutorial 11.2), is

x(t) = xmax ∙ cos⁡(ωt + φ)

where xmax is the amplitude of oscillations (the maximum displacement from the equilibrium position, ω is the angular frequency, x(t) is the position of the block aa a given instant and φ is the initial phase of oscillations (not necessarily the block must be at the equilibrium position initially).

Similarly, if we substitute the above values with the corresponding ones for a LC system of oscillations given in the table of the previous paragraph, we obtain for the total energy stored in a LC system of oscillations

Wtot = WM + We
= L ∙ i2/2 + Q2/2C

Neglecting the resistance of the conductor, we obtain a harmonic LC oscillator in which the energy remains constant with time. Again, we can write

dWtot/dt = 0

or

dWtot/dt = d/dt (L ∙ i2/2 + Q2/2C)
= L ∙ i ∙ di/dt + Q/CdQ/dt
= 0

Again, substituting i = dQ/dt and di/dt = d2Q/dt2, we obtain

L ∙ d2 Q/dt2 + 1/C ∙ Q = 0

This is the fundamental equation of a LC oscillator, the general solution of which, is

Q(t) = Qmax ∙ cos⁡(ω ∙ t + φ)

where Q(t) gives the charge stored in the capacitor at a given instant, Qmax is the maximum charge variation in the capacitor, ω is the angular frequency of LC oscillations and φ is the initial phase of oscillations.

Taking the first derivative of the last equation, we obtain the equation of current in a LC circuit. Thus,

i(t) = dQ/dt = -ω ∙ Qmax ∙ sin⁡(ω ∙ t + φ)

where imax = ω ∙ t is the amplitude of the current in this kind of circuit. Hence, we can write the equation of current in a LC circuit:

i(t) = -imax ∙ sin⁡(ω ∙ t + φ)

Example 3

A current is flowing through a LC circuit according the sinusoidal function shown in the i vs t graph below.

Calculate:

1. The frequency of current in the circuit
2. The angular frequency
3. The initial phase
4. The equation of current in this circuit
5. The current in the circuit after 2 s
6. The current in the circuit after 3.079 s

Solution 3

1. From the graph, we can see that one cycle is completed in 0.02 s. This value represents the period T of oscillations in the LC circuit. Since frequency f is the inverse of period, we obtain
f = 1/T
= 1/0.02 s
= 50 Hz
2. The angular frequency ω of the LC circuit is
ω = 2 ∙ π ∙ f
= 2 ∙ 3.14 ∙ 50 Hz
3. From the graph we can see that the current at the beginning is zero, despite its amplitude imax is 0.5 A. Therefore, we obtain for the initial phase φ:
i(t) = -imax ∙ sin(ω ∙ t + φ)
i(0) = -imax ∙ sin(ω ∙ 0 + φ)
0 = -0.5 ∙ sin(314 ∙ 0 + φ)
0 = -0.5 ∙ sinφ
Thus, sin φ must be zero to fit the values. This means φ = 0 because sin 0 = 0.
4. The general equation of current in this LC circuit, therefore is
i(t) = -(0.5 A) ∙ sin⁡(100π ∙ t)
or
i(t) = -(0.5 A) ∙ sin⁡(314 ∙ t)
5. Substituting t = 2, we obtain for the current in the circuit after 2 s:
i(2) = -(0.5 A) ∙ sin(100 ∙ π ∙ 2)
= -(0.5 A) ∙ sin[100 ∙ (2π)]
= -(0.5 A) ∙ sin[100 ∙ 0]
= -(0.5 A) ∙ sin0
= 0
(We know that 2π rad = 3600, so sin (N ∙ 2π) = sin 2π = sin 0 = 0. Here we have N = 100.)
6. We must substitute t = 3.079 in the equation of current to find the current in this instant. Thus,
i(3.079) = -(0.5 A) ∙ sin (100 ∙ π ∙ 3.079)
= -(0.5 A) ∙ sin (307.9 ∙ π)
= -(0.5 A) ∙ sin (306 ∙ π + 1.9 ∙ π)
Since 306π is a multiple of 2π (306π = 153 ∙ 2π) and giving that sin 2π = 0, we write
i(3.079) = -(0.5 A) ∙ sin (1.9 ∙ π)
= -(0.5 A) ∙ (-0.309)
= 0.1545 A

Angular Frequency

We can check whether the equation

Q(t) = Qmax ∙ cos⁡(ω ∙ t + φ)

is a solution for the differential equation

L ∙ d2 Q/dt2 + 1/C ∙ Q = 0

The first derivative of equation

Q(t) = Qmax ∙ cos⁡(ω ∙ t + φ)

is

i(t) = dQ/dt = -ω ∙ Qmax ∙ sin⁡(ω ∙ t + φ)

while the second derivative of the same equation is

d2 Q/dt2 = -ω2 Qmax ∙ cos⁡(ω ∙ t + φ)

Hence, we can write

-L ∙ ω2 ∙ Qmax ∙ cos⁡(ω ∙ t + φ) + 1/C ∙ Qmax ∙ cos⁡(ω ∙ t + φ) = 0

Cancelling the term Qmax ∙ cos⁡(ω ∙ t + φ) from the above equation, we obtain

-L ∙ ω2 + 1/C = 0

Rearranging the last equation to isolate the angular frequency ω, we obtain

1/C = L ∙ ω2
ω2 = 1/L ∙ C
ω = 1/L ∙ C

This is the same equation for angular frequency obtained earlier through another method, so the approach is correct. The equation of Q(t) is a solution for the differential equation

L ∙ d2 Q/dt2 + 1/C ∙ Q = 0

As for the electrical and magnetic oscillations, we can write for the electrical energy stored in a LC circuit at a given time

We (t) = Q2/2C = Qmax2/2C ∙ cos2 (ω ∙ t + φ)

while for the magnetic energy stored in the magnetic field of inductor at a given time, we have

WM (t) = L ∙ i2/2 = L ∙ ω2 ∙ Qmax2/2 ∙ sin2 (ω ∙ t + φ)

Substituting ω = 1/L ∙ C, we obtain for the magnetic energy stored in a LC circuit

WM = L ∙ Qmax2/2L ∙ C ∙ sin2 (ω ∙ t + φ)
= Qmax2/2C ∙ sin2 (ω ∙ t + φ)

Example 4

A LC circuit is operating at standard values of mains electricity (240V, 50Hz). Calculate:

1. The current flowing in the circuit 1.753 s after the switch is turned ON if the resistance in the circuit is 30Ω
2. The energy stored in the magnetic field of inductor at this instant if its inductance is 0.8H
3. The maximum charge stored in the capacitor plates

Take the initial phase of oscillations as zero (φ = 0).

Solution 4

Clues:

ΔVmax = 240V
f = 50 Hz
φ = 0
R = 30Ω
L = 0.8H
i(1.753) = ?
Wm (1.753) = ?
Qmax = ?

1. The maximum current in the circuit is
imax = ∆Vmax/R
= 240 V/30Ω
= 8A
Thus, the amount of current flowing in the circuit at t = 1.753 s, is
i(t) = -imax ∙ sin (ω ∙ t + φ)
i(t) = -imax ∙ sin (2π ∙ f ∙ t + φ)
i(1.753) = -8 ∙ sin (2 ∙ π ∙ 50 ∙ 1.753 + 0)
i(1.753) = -8 ∙ sin (175.3 ∙ π)
i(1.753) = -8 ∙ sin (175.3 ∙ π)
i(1.753) = -8 ∙ sin (174 ∙ π + 1.3 ∙ π)
i(1.753) = -8 ∙ sin (1.3 ∙ π)
i(1.753) = -8 ∙ (-0.80)
= 6.4A
2. The energy stored in the inductor at the given instant (t = 1.753 s), is
WM = L ∙ i2/2
= (0.8 H) ∙ (6.4A)2/2
= 16.384 J
3. Given that the energy stored in the magnetic field of a LC circuit is
WM (t) = L ∙ i2/2 = L ∙ ω2 ∙ Qmax2/2 ∙ sin2 (ω ∙ t + φ)
we obtain for the maximum electric charge Qmax stored in the capacitor plates:
imax2 = ω2 ∙ Qmax2
or
Qmax = imax/ω
= imax/2 ∙ π ∙ f
= (8A)/2 ∙ 3.14 ∙ (50 Hz)
= 0.0255 C

Potential Difference in a LC Circuit

Since potential difference is proportional to current, we use the same approach as for the current to find the potential difference in a LC circuit. Furthermore, potential difference is in phase with the current because they are related to a positive constant (resistance) to each other. Hence, we have

i(t) = -imax ∙ sin(ω ∙ t + φ)

Multiplying both sides by R, we obtain

i(t) ∙ R = -imax ∙ R ∙ sin(ω ∙ t + φ)
∆V(t) = -∆Vmax ∙ sin(ω ∙ t + φ)

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