PHYS 5.4: AC circuits and electrical oscillations	PPLATO @
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1 Opening items

1.1 Module introduction

Perhaps before reading this module you should do a small electrical audit. Spend a minute or two wandering around your home looking at the electrical appliances which you find there. Most of them will have a small plate or sticker attached which tells you something about their electrical specifications. My toaster, which is quite a simple model, just says 240 V 1000 W 50 Hz. Whilst the back of my microwave – which incidentally was made in France – bears the more complicated legend

240 V ~ 50 Hz
fréq.: 2450 MHz
MW 1500 W/6,5 A

A good deal of this information concerns the sort of power supply that the appliance requires. In both the cases quoted the appliance requires a supply of alternating current (a.c.), with a root-mean-square voltage of 240 volts and a frequency of 50 hertz. Alternating currents are the central theme of this module. It seems clear that we must know something about them if we are to understand even the most basic information written on electrical appliances. (From 1995 the UK electricity supply was 230 volts.)

Section 2 begins by showing a mathematical way of describing alternating currents, and investigating their behaviour in various electronic circuit components. This involves building mathematical models of physical situations, a process that some students find daunting. To avoid confusion everything has been made as explicit as possible, so you may find some of the steps spelled out in rather a lot of detail. Other topics covered in Section 2 include impedance (the a.c. equivalent of resistance), and the construction of simple filter circuits that can be used to block signals in certain frequency ranges. Phasor methods are introduced and used in this discussion to determine the impedances (and other properties) of a variety of simple circuits.

Section 3 uses the results and concepts introduced in Section 2 to study the behaviour of charge, current and voltage in a variety of simple circuits. The circuits studied can exhibit various kinds of short–lived transient behaviour and, in some cases, sustained oscillations in which charges and currents flow back and forth repeatedly between different circuit components. In the most complicated cases this can even result in resonance when an alternating electrical supply excites an unusually large response from an appropriately designed circuit. This discussion of transients and oscillations also introduces the idea of a differential equation and some of the general principles that surround the use of such equations in particular physical contexts. However, no attempt is made to teach the mathematical techniques that are required to solve such equations in general.

Although most people have some understanding of direct current, it is alternating current which – in many ways – has the greatest bearing upon our lives. That is one of the reasons why this module is important since it deals with an aspect of electricity which impinges directly upon our everyday lives.

Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.

1.2 Fast track questions

Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 5.1Module summary and the Subsection 5.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 5.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.

1.3 Ready to study?

2 AC circuits

2.1 Describing alternating currents and voltages

You should already be familiar with the properties of a simple d.c. circuit of the kind shown in Figure 1. The box symbol represents a resistor, and the battery symbol represents a source of voltage (i.e. potential difference). The current I in the circuit will be such that the voltage drop V across the resistance R is equal to the voltage supplied by the battery. The voltage drop across the resistor can be measured using a voltmeter connected in parallel whilst the current flowing through the resistor can be measured using an ammeter connected in series. The resistance R can then be determined using Ohm’s law

$R = \dfrac VI$(1) i

In analysing a d.c. circuit such as that of Figure 1 we generally think of the battery as a source of constant voltage, so that the current in any part of the circuit is a steady flow of electric charge in a single direction; that, after all, is the meaning of d.c. – direct current. However, not all voltage sources are constant and neither are all currents direct. For example, the mains voltage supplied by standard wall sockets in the UK varies rapidly with time, the potential difference between the ‘live’ and ‘neutral’ terminals changing not just in magnitude but even in sign 50 times a second. Not surprisingly, the currents that such varying voltages produce in circuits will also vary with time, generally altering both their magnitude and direction in response to the variations in the voltage. A current that periodically reverses its direction in this way is called an alternating current (a.c.).

The simplest kind of alternating current is represented in Figure 2. The current changes smoothly and regularly with time in a manner that may be described by a sinusoidal function (i.e. a sine or a cosine function), just like a simple harmonic oscillator. Its peak value (or amplitude) is I₀, and the period over which it goes through one complete oscillatory cycle is T. In drawing Figure 2 we have chosen the time t = 0 to be a moment at which the current is zero, consequently, it is marginally easier to represent this particular current by means of a sine function rather than a cosine function so we can write

Note that we have chosen to represent the current by I(t) in order to emphasize that the value of I is changing with time and will generally depend on the value of t at which we choose to measure it.

Equation 2 is very similar in form to the equation used to describe the displacement of a simple harmonic oscillator from its equilibrium position. The quantity 2πt/T, is called the phase of the oscillator, and increases by 2π each time t increases by an amount T.

Since an increase of phase by 2π corresponds to a complete oscillation, the phase has many of the characteristics of an angle measured in radians and some authors prefer to treat it as an angle, in which case it may be called the phase angle.

It is possible to write the phase in Equation 2 in a couple of other ways. The frequency f of an oscillation is related to the period by f = 1/T, so it is always possible to replace 2πt/T by 2πft. More usefully we can introduce a related quantity called the angular frequency which is defined by ω = 2π/T so that the phase 2πt/T may be simply written as ωt.

In terms of frequency and angular frequency, the alternating current described by Equation 2 may therefore be written in the following equivalent ways.

Frequency and angular frequency are defined in such a way that either quantity may be measured in units of s⁻¹, thus the SI unit of each quantity is the hertz (Hz) since 1 Hz = 1 s⁻¹.

 

So far, our mathematical descriptions of alternating currents have all been limited to the case where I = 0 when t = 0. Obviously this need not be the case; it is quite possible for an alternating current to have any one of its allowed values at t = 0. Figure 3 indicates the general situation that can arise. The current I₁ is of the kind we have already discussed and is described by:

I₁(t) = I₀sin(ωt)(4a)

but the current I₂, which has the same angular frequency and amplitude as I₁, has some other value at t = 0. This more general current may be represented by the expression

I₂(t) = I₀sin(ωt + ϕ)(4b)

where the constant ϕ is called the phase constant. The phase constant is the value of the phase at t = 0 and determines the current at that time since I₂(0) = I₀sinϕ.

Given any two quantities that oscillate with the same angular frequency, such as I₁ and I₂, it is always possible to relate the oscillations of one to those of the other by describing the way in which the phase of one is related to that of the other. In the case of I₁ and I₂ we can describe this phase relationship by saying that I₂ phase_leadleads I₁ by ϕ or that I₁ phase_laglags I₂ by ϕ. Note that ϕ could have many values, though it is customary to limit it to the range −π < ϕ ≤ +π since adding or subtracting an integer multiple of 2π to the phase makes no difference to the value of the current.

The phasor representation

Another way of representing the alternating current of Figure 2 is in terms of a rotating phasor, as shown in Figure 4. A phasor is rather like the hand on a clock, though the phasors we will consider will all rotate in the anticlockwise direction. A phasor has a magnitude A and rotates at a fixed angular speed ω, so that the angle θ from the x–axis to the phasor increases with time. i The y–component of the phasor (i.e. its ‘projection’ y = Asinθ on to the y–axis) varies sinusoidally with θ, as indicated in Figure 4.

It follows that a graph showing the variation of this y–component with time will be identical to Figure 2 provided the following requirements are met:

1

The magnitude A of the phasor is equal to the amplitude (or peak value) I₀ of the current.

2

At any time t, the angle θ of the phasor is equal to the phase (generally, ωt + ϕ) of the current.

The fact that oscillating systems, such as alternating currents, can be represented by phasors is interesting but little more than that at present. However, later in the module you will see that the phasor representation becomes very useful when we have to work out the combined effect of several alternating currents that differ in phase and amplitude.

2.2 AC power and rms current

When dealing with direct current, the power dissipated by a current passing through a resistor may be calculated using the formula

P = I²R(5)

Resistors can become hot as a result of this power dissipation when current passes through them.

When dealing with alternating current the situation is not so simple, since the current changes continuously. If the current at any instant is given by I = I₀sin(ωt), then the instantaneous a.c. power dissipation is given by

P = I²R = I₀²Rsin²(ωt)

This instantaneous power is shown in Figure 5. Though it is useful to know the expression for the instantaneous power it is not really of much direct relevance when dealing with everyday a.c. devices such as light bulbs and electric heaters. The frequency of the mains supply (50 Hz in the UK) is such that the instantaneous power rises and falls 100 times every second, which is much too fast for any changes in heating effect to be seen or felt. When dealing with such everyday devices it is more useful to have an expression for the average a.c. power dissipated over a time that is much larger than the oscillation period of the current.

Over one complete period of oscillation T, the average value of an alternating current I = I₀sin(ωt) will be zero since every positive contribution to the average current will be counterbalanced by a negative contribution of equal magnitude. Nonetheless, an alternating current really does cause power dissipation in a resistor – you only have to switch on an electric fire to be sure of that. The reason for this is apparent from Equation 5: the instantaneous power is proportional to I², so the instantaneous power is positive whenever I is non-zero, irrespective of its sign.

Thus, the average power $\langle P\rangle$ i dissipated over one full period T will be equal to the mean of I² over that period. We may indicate this by using angular brackets to denote the mean and writing

$\langle P\rangle = \langle I^2R\sin^2(\omega t) = I^2R\langle\sin^2(\omega t)\rangle$ i

Now the mean square current on the right–hand side can be worked out straightforwardly using the techniques of integration developed elsewhere in FLAP, but there is a simpler method that uses the special properties of the function sin²(ωt). Since the angular brackets indicate an average over a full period of oscillation, the time at which we start timing is unimportant, so we could equally well write

$\langle P\rangle = \langle I^2R\sin^2(\omega t) = I^2R\langle\cos^2(\omega t)\rangle$ i

If we add the right–hand sides of these last two equations we find:

$2\langle P\rangle = \langle I^2R\sin^2(\omega t) + \langle I^2R\cos^2(\omega t) = I^2R\langle\sin^2(\omega t) + cos^2(\omega t)\rangle$

However sin²(ωt) + cos²(ωt) = 1, which is a constant, so its average value over a full period is the same as its instantaneous value, i.e. 1.

Thus,

$\langle\sin^2(\omega t) + cos^2(\omega t)\rangle = 1$

so$2\langle P\rangle = I_0^2R$

i.e.$\langle P\rangle = \dfrac12 I_0^2R$(6)

This amounts to saying that $\langle\sin^2(\omega t)\rangle = 1/2$, so the mean of I² over a full period is I₀²/2.

As you can see, when an alternating current of peak value I₀ passes through a resistance R, the power dissipation is the same as that which would occur if a direct current of steady value I₀/2 were to flow through the resistor. Because of the way it has been derived – by taking the square root of the mean of I² over a full period – this ‘effective average current’ is called the root_mean_square_r.m.s._currentroot–mean–square current or root_mean_square_r.m.s._currentrms current. The root-mean-square current is usually denoted I_rms so we can write:

If we had started this investigation of the power dissipated by a resistor connected to an a.c. supply by considering an alternating voltage V = V₀sin(ωt) across the resistor rather than an alternating current through the resistor, we could have used the d.c. expression P = V²/R to lead us to the a.c. results:

In this case V_rms is the root_mean_square_r.m.s._voltageroot–mean–square voltage. Finally, note that by multiplying together the right–hand sides of Equations 7a and 7b, and then taking the square root, it can be seen that the average power dissipated by the resistor is also given by:

2.3 Alternating current in resistors, capacitors and inductors

In the last subsection we examined the effect of an alternating current passing through a resistor. In this subsection we look at the origin of such a current. In particular we will answer the following question, ‘If an alternating voltage is applied across a resistor, what current will it cause to flow through the resistor?’ As you will see, in the case of a resistor the relationship between the voltage and the current is quite straightforward, but we will also examine two other electronic components, a capacitor and an inductor, in which the relationship is not so simple.

Alternating current in resistors

Figure 6a shows a simple a.c. circuit that supplies an alternating voltage V(t) = V₀sin(ωt) across a resistor.

Since Ohm’s law V = IR can be applied at any instant, we can expect that the varying voltage will produce a varying current given by

$I(t) = \dfrac{V(t)}{R} = \dfrac{V_0}{R}\sin(\omega t)$ i

so, if we define

$I_0 = \dfrac{V_0}{R}$(9)

then we can write

I(t) = I₀sin(ωt)(10)

A graph of this current, and the voltage that caused it, is shown in Figure 6b, along with the corresponding current and voltage phasors at an arbitrary time t. The peak current I₀ has been chosen for graphical convenience in Figure 6b; in practice it would be determined by the values of V₀ and R. The main point to notice about Figure 6b is that the current and the voltage rise and fall together, neither one leads or lags the other so there is no phase difference, and we can say that they are in phase. This is also clear from the phasors which rotate together, one on top of the other.

Alternating current in capacitors

A capacitor acts to store charge, and may generally be thought of as a pair of parallel conducting plates holding charges of equal magnitude but opposite sign. An alternating voltage V(t) = V₀sin(ωt) may be applied across a capacitor of fixed capacitance C by means of a circuit like that shown in Figure 7a.

No conduction current actually passes through the capacitor in such a circuit but there is an ebb and flow of charge from the capacitor as it is alternately charged and discharged; it is this flow that constitutes the current in the circuit. i

Predicting this current is a slightly indirect process but the general approach is the same as before; we start with a result that can be expected to apply at any particular instant. In the case of a capacitor, the formula that we need is that which relates the charge q stored on the capacitor to the potential difference V across it: q = CV. Assuming this is true at every instant we can write

q(t) = CV(t) = CV₀sin(ωt)(11)

Note that our (arbitrary) choice of direction for the arrow indicating the positive sense of voltages across the capacitor in Figure 7a means that the stored charge q(t) must be interpreted as the charge on the upper plate of the capacitor. Whenever V(t ) is negative (implying that the upper plate is at the lower potential), q(t) will also be negative.

Now, the current at any point in a circuit is determined by the rate of flow of charge at that point in the circuit. In Figure 7a the positive direction assigned to the current is such that a positive current will cause q(t) to increase. So, using the derivative notation of differential calculus, we find

$I(t) = \dfrac{dq(t)}{dt}$(12)

Combining this with Equation 11 gives

$I(t) = \dfrac{d[CV_0\sin(\omega t)]}{dt}$

This derivative can be easily evaluated by using the techniques of differentiation (or laboriously evaluated by drawing a graph of sin(ωt) and evaluating its gradient at many values of t). Using either method, the result will be

I(t) = ωCV₀cos(ωt)

so if we define I₀ = ωCV₀(13)

then we can write I(t) = I₀cos(ωt)(14)

which can be rewritten in terms of a sine function as

$I(t) = I_0\sin\left(\omega t + \dfrac{\pi}{2}\right)$(15)

Having expressed the current in terms of a sine function it is easy to see that in a circuit containing just a capacitor the current and the voltage are not in phase. The current leads the voltage by π/2. Figure 7b shows this relationship in graphical form, along with the associated phasors. As you can see the current phasor and the voltage phasor rotate at the same rate, but the current phasor is 90° ahead of the voltage phasor.

Alternating current in inductors

An inductor is essentially a coil of wire. When a current is made to flow through an inductor a magnetic field is produced in the coil.

When the current is changed, the changing magnetic flux in the coil causes an induced voltage in the coil (Faraday’s law) and the polarity of this voltage acts to oppose any change in the current.

Figure 8a shows a simple circuit that applies an alternating voltage V(t) = V₀sin(ωt) across an ideal inductor. i As the voltage changes the current will also tend to change. If the derivative dI/dt represents the rate of change of the current I(t) passing through the inductor then the size of the voltage drop across the inductor at any instant will be

$V(t) = L\dfrac{dI}{dt}$(16)

where the constant L is called the inductance of the inductor. You should be able to see from Equation 16 that the inductance can be expressed in units of V s A⁻¹; such a unit is called a henry (H), so 1 H = 1 V s A⁻¹. Widely used inductances vary from a few microhenry to hundreds of henry.

Since V(t) = V₀sin(ωt) Equation 16 gives

$\dfrac{dI}{dt} = \dfrac{V_0}{L}\sin(\omega t)$(17) i

Equation 17 is an example of a differential equation since it involves the derivative dI/dt of the quantity that interests us, the varying current I(t). The techniques needed to solve such equations are taught elsewhere in FLAP but you are not required to know them in order to study this module. Whenever such equations arise in this module we will simply quote their solutions and leave it to you to check by substitution that they are correct, should you wish to do so.

The solution to Equation 17, subject to the condition that there is no source of steady current in the circuit, is

$I(t) = -\dfrac{V_0}{\omega L}\cos(\omega t)$

so, if we define $I_0 = \dfrac{V_0}{\omega L}$(18)

then we can write I(t) = −I₀cos(ωt)

which we can rewrite in terms of a sine function as

$I(t) = I_0\sin\left(\omega t - \dfrac{\pi}{2}\right)$(19)

Having expressed the current in terms of a sine function we can again determine its phase relationship to the applied voltage V(t). In this case the current lags the voltage by π/2. Figure 8b shows this relationship in graphical form, along with the associated phasors. Note that the current phasor is 90° behind the voltage phasor.

Summing-up

As the above investigations show, when an alternating voltage V(t) is applied across a resistor, a capacitor or an inductor the resulting current has the same frequency as the applied voltage but is not necessarily in phase with it. In the case of the resistor the phases do coincide, but in the capacitor the current leads the voltage by π/2, and in the inductor the current lags the voltage by π/2. i

The peak value of the current, I₀, depends on the resistance, capacitance or inductance in each respective case, but in addition, it also depends on the frequency of the supply in the case of a capacitor or an inductor. We will say more about this relationship in the next subsection.

2.4 Resistance, reactance and impedance

The discussion of alternating currents in resistors in the last subsection shows no real surprises. In particular, if we measure the peak value V₀ of the voltage across a resistor and the peak value I₀ of the current that it causes to flow through the resistor (or the corresponding rms values), then we can easily determine the resistance R. It is given by

$R = \dfrac{V_0}{I_0} = \dfrac{V_{\rm rms}}{I_{\rm rms}}$

This resistance does not depend on the frequency of the a.c. supply and does not give rise to any phase difference between the voltage and the current.

In a similar way, we can use the quantity V₀/I₀ to define a sort of ‘effective a.c. resistance’ for capacitors and inductors. However, for these components the value of V₀/I₀ does depend on the frequency of the supply and there will be a phase difference between the voltage and the current. Since the response of capacitors and inductors depends on the nature of the supply they are said to be reactive components, and V₀/I₀ is called the reactance of a capacitor or an inductor, rather than its ‘resistance’. Nonetheless, the unit of reactance, like that of resistance, is still the ohm (Ω).

Using the symbols X_C and X_L to represent, respectively, the reactance of a capacitor and an inductor we see from Equations 13 and 18

I₀ = ωCV₀(Eqn 13)

$I_0 = \dfrac{V_0}{\omega L}$(Eqn 18)

(together with the relation ω = 2πf) that

These results, along with the relevant phase relationships, are summarized in Table 1. Graphs showing how resistance and reactance vary with angular frequency are shown in Figure 9.

Table 1 Resistance/reactance of various circuit components, together with the phase relationship they produce between voltage and current.

Component	Resistance/reactance	Phase relationship
resistor	R = V0/I0 (constant)	V and I in phase
capacitor	XC = 1/(ω C)	I leads V by π/2
inductor	XL = ω L	I lags V by π/2

We can summarize this in words by saying that the frequency does not affect the conduction through a resistor but the conduction through a capacitor increases with frequency while that through an inductor decreases with frequency. For direct current the reactance of a capacitor becomes infinite while that for an inductor falls to zero. At very high frequencies the inductor reactance rises without limit while that for the capacitor falls to zero. Inductors tend to block high frequency currents while capacitors block d.c. currents.

Resistance and reactance are both special cases of a more general a.c. phenomenon known as impedance. If, as in Figure 10, an alternating voltage V(t) = V₀sin(ωt) is applied to some general network of resistors, capacitors and inductors then the current I(t) that the network draws from the supply will have the general form I(t) = I₀sin(ωt + ϕ), where I₀ represents the peak value of the current and ϕ determines the phase relationship between the voltage and the current.

The values of I₀ and ϕ will depend on the resistances and reactances of the various components of the network, but in general terms we say that the impedance Z of the network is

If the network consisted of a single resistor, its impedance would simply be the resistance of that resistor and ϕ would be zero. Similarly, if the network consisted of a single capacitor or inductor then its impedance would equal the reactance of that capacitor or inductor and the value of ϕ would be +π/2 or −π/2, respectively.

In the next two subsections we will investigate the impedances and phase relationships that arise from more complicated networks.

It is important to note that the instantaneous power supplied to the network in Figure 10 is given by

P(t) = V(t)I(t) = V₀sin(ωt)I₀sin(ωt + ϕ)

In the case of a resistor (or even a network of resistors), where the phase difference ϕ = 0, we have already seen that the average a.c. power over a full period (T = 2π/ω) is $\langle P\rangle = V_0I_0/2 = V_{\rm rms}I_{\rm rms}$. However, for a network containing capacitors and inductors ϕ will not generally be zero and it may be shown that

It follows that both the instantaneous and average power dissipated when the network consists of a single capacitor or a single inductor is zero since ϕ = π/2 in these cases, and cos(π/2) = 0.

It is also true that for any network consisting only of pure inductors and capacitors, there is no power dissipation, since in each component the current and voltage are 90° out of phase. Power dissipation requires the presence of resistance in the circuit.

2.5 The series LCR circuit

Combining C and R

Figure 11 shows a capacitor and a resistor connected in series. When circuit components are connected in this way the same current I(t) flows through each of them, but the total instantaneous voltage V(t) across them is the sum of the instantaneous voltages V_R(t) and V_C(t) across the individual components. (To this extent they behave like d.c. circuits.) Since the voltage across the resistor is in phase with the current while that across the capacitor is not, it follows that the voltage across the capacitor cannot be in phase with the voltage across the resistor. This means that the total voltage across the resistor and capacitor, which is just the applied voltage V(t), will not be in phase with the voltage across either component. Nor therefore can the applied voltage be in phase with the current in the circuit.

In what follows we are going to determine the relationship between the applied voltage and the current in this circuit, which will involve finding expressions for the ratio of their peaks Z = V₀/I₀ and for the phase difference ϕ between them.

The easiest way to find the relationship between the peak voltages (and this is the key to finding the impedance) is to use phasor diagrams of the kind introduced in Subsection 2.1. Those needed in this case are also shown in Figure 11b. If the current is assumed to be of the form I(t) = I₀sin(ωt), then at t = 0 it can be represented by a horizontal phasor. i The voltage across the resistor V_R(t) is in phase with this current so it too can be represented by a horizontal phasor at t = 0. This voltage phasor is shown in Figure 11b, it has amplitude I₀R.

Now, remembering that the voltage across a capacitor lags the current through it by π/2 (since the current leads the voltage by π/2) we see that at t = 0, the voltage across the capacitor is represented by a phasor that points vertically downwards. This phasor is also shown in Figure 11b, and has amplitude I₀X_C, where X_C is the reactance of the capacitor. The phasor representing the total voltage across the resistor and capacitor at t = 0 is obtained by adding the other two phasors together as though they were vectors. Using Pythagoras’s theorem, the amplitude V₀ of this resultant phasor, which represents the total voltage across the circuit, is given by

$V_0^2 = I_0^2R^2 + I_0^2X_C^2\quad\text{i.e.}\quad V_0^2 = I_0^2(R^2 + X_C^2)$

so$V_0 = I_0\sqrt{\smash[b]{R^2 + X_C^2}}$

It follows that the impedance Z = V₀/I₀ of the series combination of a resistor and a capacitor is given by

$Z = \sqrt{\smash[b]{R^2 + X_C^2}}$(24)

It can also be seen from Figure 11 that the total voltage lags the voltage in the resistor by ϕ.

If we take the current phase as our reference zero then the voltage phase lag ϕ is given by

$\tan\phi = -\dfrac{X_C}{R}$

so$\phi = -\arctan\left(\dfrac{X_C}{R}\right)$(25)

Since the voltage in the resistor is in phase with the current in the circuit, and since we chose to represent the current by I(t) = I₀sin(ωt), it follows that we can represent the total voltage by V(t) = V₀sin(ωt + ϕ).

So, in a series RC circuit, the impedance is $\sqrt{\smash[b]{R^2 + X_C^2}}$, and the current leads the applied voltage by $\arctan\left(\dfrac{X_C}{R}\right)$, since ϕ is a negative quantity in this case. i

Combining L and R

The same thing can be done for a series combination of an inductor and a resistor – which is actually how a real inductor would behave.

The relevant circuit is shown in Figure 12. Once again it will be assumed that the current is of the form I(t) = I₀sin(ωt) and as before we will write the applied voltage as V(t) = V₀sin(ωt − ϕ). The challenge is to find the ratio V₀/I₀ and the value of ϕ that are appropriate to this case. Of course, V₀/I₀ can always be written as Z, so the first problem is to find the impedance Z of the series combination. We will use the voltage phasors at t = 0 to help solve this problem.

The current in each series component is the same, and the voltage across the resistor is in phase with that current, so at t = 0 the phasor for the voltage across the resistor is horizontal and points to the right. The voltage across an inductor leads the current through it, so the phasor for the voltage across the inductor points vertically upwards in this case. The phasor representing the total voltage is the (phasor) sum of these two, as indicated in Figure 12.

Following a similar argument to that used earlier, the impedance of the LR combination will be

$Z = \sqrt{\smash[b]{R^2 + X_L^2}}$(26)

In this case the current (which is in phase with the resistor voltage) lags the total voltage. So V(t) = V₀sin(ωt + ϕ) where in this case ϕ must be a positive quantity.

$\tan\phi = \dfrac{X_L}{R}$

or equivalently   $\phi = \arctan\left(\dfrac{X_L}{R}\right)$(27)

So, in a series LR circuit, the impedance is $\sqrt{\smash[b]{R^2 + X_L^2}}$ and the current lags the applied voltage.

Combining L, C and R

Figure 13 shows a resistor, a capacitor and an inductor combined in a series a.c. circuit. The current in each component is the same and so has the same phase. It is sensible to take this current as defining our phase zero, with the voltages across each component having phase angles with respect to the current phase. This time we must add together three voltage phasors at t = 0. Fortunately two of them point in opposite directions along the same line, so their resultant is easy to determine, and this can be combined with the third phasor in the usual way.

Thus$V_0^2 = I_0^2R^2 + I_0^2(X_L-X_C)^2$

i.e.$V_0^2 = I_0^2\left[R^2 + (X_L-X_C)^2\right]$

but the impedance Z is given by Z = V₀/I₀, hence

Substituting for the reactances (X_L = ωL and X_C = 1/(ωC)) gives us

$Z = \sqrt{R^2 + \left(\omega L -\dfrac{1}{\omega C}\right)^2}$(29)

It is also true that Z² = R² + X² where X is the total reactance of the circuit. So, for a series circuit

(impedance)² = (resistance)² + (reactance)²

The constant ϕ that determines the phase relationship between the current in the circuit and the supply voltage can also be deduced from the final phasor diagram in Figure 13.

The value of ϕ is given by

So, in a series LCR circuit, if X _C is greater than X_L then ϕ will be negative and the current in the circuit will lead the total voltage (as shown in Figure 13). However, if X_C is less than X _L, Equation 30 will yield a positive value for ϕ, this corresponds to the case in which the current lags the total voltage.

In this convention we are taking the conventional voltage in the resistor to define the zero phase condition, so I(t) = I₀sin(ωt) and the voltage applied to the whole circuit is then V(t) = V₀sin(ωt + ϕ), where V₀ = ZI₀ and ϕ is given by Equation 30.

2.6 The parallel LCR circuit

The basic principle that underlies the analysis of a parallel LCR circuit, of the kind shown in Figure 14 is that the instantaneous voltage V(t) is the same across each element but the current will generally differ from one component to another. When dealing with a series circuit, a phasor diagram was used to combine the voltages across currents through each component, but with a parallel circuit, phasors must be used to combine currents through each component. The total current being supplied to the combination of resistor, capacitor and inductor will therefore be the phasor sum of the individual currents in each component. The current in the resistor I_R(t) is in phase with the supply voltage V(t); the current in the capacitor I_C(t) leads V(t) by π/2, and the current in the inductor I_L(t) lags V(t) by π/2. i

Hence, if the supply voltage is given by V(t) = V₀sin(ωt) the currents in the resistor, capacitor and inductor at t = 0 can be represented by the phasors shown in Figure 14. These phasors may be added together (like vectors) giving a resultant phasor that represents the current I(t) drawn from the supply.

If we use Pythagoras’s theorem to evaluate the amplitude of the resultant phasor in Figure 14, we find that the peak value of the total current in the circuit is given by

$I_0 = \sqrt{V_0^2\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)^2 + \dfrac{V_0^2}{R^2}}$(31)

so$\dfrac{I_0}{V_0} = \sqrt{\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)^2 + \dfrac{1}{R^2}}$

but the impedance Z = V₀/I₀. Hence,

Substituting for the reactances (X_C = 1/(ωC) and X_L = ωL) we can rewrite this in the equivalent form

$\dfrac{1}{Z} = \sqrt{\left(\omega C-\dfrac{1}{\omega L}\right)^2 + \dfrac{1}{R^2}}$(33)

The constant ϕ that determines the phase relationship between the total current and the current through the resistor (and hence the voltage) may also be expressed in terms of the amplitudes of the three current phasors:

$\phi = \arctan\left[\left(\dfrac{V_0}{X_C}-\dfrac{V_0}{X_L}\right)\middle/\dfrac{V_0}{R}\right]$

Dividing both the numerator (the top) and the denominator (the bottom) of this expression by V₀ gives us

So, in a parallel LCR circuit, if 1/X_C is greater than 1/X_L, then ϕ will be positive and the current will lead the voltage, but if 1/X_C is less than 1/X_L, ϕ will be negative and the current will lag the voltage. In other words, if the voltage in the circuit is V(t) = V₀sin(ωt), the current will be given by I(t) = I₀sin(ωt + ϕ), where I₀ = V₀/Z and ϕ is given by Equation 34.

2.7 Combining series and parallel circuits

Using the rules that have been deduced above for calculating the impedance of components that are joined in series or in parallel (Equations 28 and 32, respectively)

$Z = \sqrt{\smash[b]{R^2 + (X_L-X_C)^2}}$(Eqn 28)

$\dfrac{1}{Z} = \sqrt{\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)^2 + \dfrac{1}{R^2}}$(Eqn 32)

it is now possible for us to work out the impedance of more complicated networks. For example, in the case of the circuit shown in Figure 15 a capacitor and an inductor are joined in parallel, and that combination is connected in series with a resistor.

According to Equation 32, the parallel part of the circuit is equivalent to a single reactance X given by

$\dfrac1X = \dfrac{1}{X_C}-\dfrac{1}{X_L}$   which implies   $X = \dfrac{X_LX_C}{X_L-X_C}$

and according to Equation 28, the total impedance Z of that reactance and the resistance with which it is in series is given by

Z² = R² + X²

Thus,$Z = \sqrt{R^2 + \left(\dfrac{X_LX_C}{X_L-X_C}\right)^2}$

The voltage is the same across the inductor and the capacitor, whilst the (phasor) sum of the currents through the combination of inductor and capacitor is the same as the current through the resistor.

2.8 Filter circuits

It is important to remember that the impedance of reactive circuit components such as capacitors and inductors depends on the (angular) frequency of the supply. This property can be put to good use in various contexts, such as the construction of filter circuits designed to block signals in certain frequency ranges while allowing others to pass. Two simple circuits of this kind R are shown in Figures 16 and 17.

In the case of Figure 16, an alternating supply of voltage V_in provides current to a capacitor and a resistor in series, and the varying voltage across the capacitor is used as an output voltage of rms value V_out that may supply some other circuit. The impedance of the series RC combination is $Z = \sqrt{\smash[b]{R^2 + X_C^2}}$ (Equation 12), so V_in is related to the rms current in the circuit by

$V_{\rm in} = I_{\rm rms}\sqrt{\smash[b]{R^2 + X_C^2}}$

butV_out = I_rmsX_C

so$V_{\rm out} = V_{\rm in}\dfrac{X_C}{\sqrt{\smash[b]{R^2+X_C^2}}}$

Now, this can be written in terms of angular frequency and capacitance as follows

$V_{\rm out} = V_{\rm in}\dfrac{1/\omega C}{\sqrt{\smash[b]{R^2+[1/(\omega C)]}}}$

A graph showing the rms value V_out as a function of the angular frequency ω of the supply is shown in Figure 16b.As you can see, if the input voltage has a low frequency the output voltage will have a fairly similar rms value to V_in, but if the input has a high frequency the output will have a relatively small rms value.

Thus, this particular circuit blocks high frequency signals and is therefore a crude example of a low–pass filter. You can easily imagine how useful such circuits might be in radios and other items of electrical equipment.

The next question invites you to find the formula that describes the graph in Figure 17 and hence to establish that the RC circuit shown there is an example of a high–pass filter.

3 Oscillations in electrical circuits

You have now seen how capacitors and inductors respond to sinusoidal currents and voltages in a.c. circuits, but to have a more complete understanding of their behaviour you also need to know how they respond when supplies are switched on or off, or new connections are made. Such changes generally result in transients – that is brief bursts of current and voltage whose behaviour is characteristic of the component (or network of components) involved. We will temporarily return to d.c. circuits to consider this point.

3.1 Transient currents in an RC circuit

Figure 18a shows a capacitor being charged through a resistor by a d.c. supply. When the switch is first closed, at t = 0, the potential difference across the capacitor is initially zero, but it will increase with time as charge accumulates in the capacitor. This increasing potential difference V_C(t) may be measured using a high impedance voltmeter (such as a cathode_rayscathode ray oscilloscope) and will change with time in the way indicated by Figure 18b.

Eventually, the charge stored in the capacitor will build up to a maximum value q₀ = V₀C, where V₀ is the voltage of the d.c. supply and C is the capacitance of the capacitor.

If the capacitor in Figure 18a is fully charged, and the battery is then replaced by a piece of wire the capacitor will discharge through the resistor. The stored charge will then decrease with time in the way shown in Figure 19.

Our aim now is to obtain the mathematical formulae that describe these curves.

Charging a capacitor

When charging the capacitor, the supplied voltage V₀ must equal the sum of the voltage drops across the resistor and the capacitor at any time t.

SoV₀ = V_R(t) + V_C(t) i

i.e.$V_0 = I(t)R + \dfrac{q(t)}{C}$

where I(t) represents the current through the resistor at time t.

Thus$I(t) = \dfrac{V_0}{R} - \dfrac{q(t)}{RC}$(35)

Now, the current I(t) is the rate of flow of charge that causes the stored charge q(t) to increase, so we can write I(t) = dq/dt. Equation 35 can therefore be rewritten as

$\dfrac{dq}{dt} = \dfrac{V_0}{R} - \dfrac{q(t)}{RC}$

i.e.$\dfrac{dq}{dt} = \dfrac{1}{RC}\left[V_0C - q(t)\right]$

Using the fact that q₀ = V₀C, and rearranging this last equation slightly gives us

$\dfrac{dq}{dt} = -\dfrac{1}{RC}\left[q(t) - q_0\right]$(36)

Now, R, C and q₀ are all constants, so what we have here is an equation which relates q(t), the stored charge at any time t, to dq/dt, the instantaneous value of its rate of change. This is an example of a first–order differential equation, similar to the one we met in Subsection 2.3. It is said to be ‘first-order’ because it does not contain the second derivative d²q/dt², nor any other derivative of q apart from the first derivative dq/dt. As before we will not work out the solution to the differential equation, we will just write it down and leave it to you to confirm that it is correct, should you wish to do so. In this case the general solution to the equation is

$q(t) = q_0\left[1-A\exp\left(-\dfrac{t}{RC}\right)\right]$(37) i

The arbitrary constant A that appears in this equation is a feature that arises in the general solution of every first–order differential equation. To determine its value in any particular case we must call on some appropriate item of additional information about the behaviour of q(t); some property that is not implicit in Equation 36. This additional information is called a boundary condition or an initial condition of the solution. In the case of the charging capacitor a suitable condition is the requirement that at t = 0, when charging commences, there is no stored charge in the capacitor.

In mathematical terms this means that q(0) = 0. Imposing this condition on Equation 37 (i.e. setting t = 0 and q(0) = 0) implies i

A = 1

So, the particular solution to the differential equation that is relevant to us is:

This is the mathematical description of the charging curve shown in Figure 18b.

Discharging a capacitor

When a charged capacitor is allowed to discharge through a resistor a changing current I(t) flows through the circuit until the stored charge has been reduced to zero. In this case, since the stored charge is decreasing with time we know that its rate of change dq/dt will be negative. It follows that the current in the circuit I(t) = dq/dt will also be a negative quantity, indicating that it flows anticlockwise, in opposition to the arrow shown in Figure 18a.

Now, in the absence of a battery V_R(t) + V_C(t) = 0

soV_R(t) = −V_C(t)

But,$V_R(t) = I(t)R = \dfrac{dq}{dt}R\quad\text{and}\quad V_C(t) = \dfrac{q(t)}{C}$

Hence$\dfrac{dq}{dt} = -\dfrac{q(t)}{RC}$

Once again we have a first–order differential equation that may be solved by standard methods discussed elsewhere in FLAP. This time the general solution is

$q(t) = A\exp\left(-\dfrac{t}{RC}\right)$(39)

where A is the arbitrary constant that we expect to arise in such a solution.

The particular solution that is relevant to our problem is found by imposing the initial condition that the capacitor was charged at time q₀ t = 0, so q(0) = q₀. Imposing this condition we see that in this case A = q₀, so

This equation shows the exponential decay expected from a discharging capacitor (Figure 19).

Now that you know how a capacitor charges and discharges, the next thing to look at is the transient currents which flow when a circuit containing a resistor and an inductor is switched on and off.

3.2 Transient currents in an LR circuit

Figure 20a shows a resistor and an inductor connected in series to a d.c. supply of voltage V₀. At any time after the switch is closed, the total voltage across the circuit is given by the sum of the V₀ voltages across the individual components, so

V₀ = V_R(t) + V_L(t)

Using Ohm’s law to find the voltage drop across the resistor and using Equation 16,

$V(t) = L\dfrac{dI}{dt}$(Eqn 16)

for that across the inductor we can rewrite this in terms of the current I(t):

$V_0 = I(t)R + L\dfrac{dI}{dt}$(41)

This can be rearranged to give

$\dfrac{dI}{dt} = -\dfrac RL\left(I(t)-\dfrac{V_0}{R}\right)$(42)

Now, when this circuit has been switched on for a sufficiently long time the current will reach a final steady value I(t) = I₀. When that happens dI/dt = 0, so it follows from Equation 40,

$q(t) = q_0\exp\left(-\dfrac{t}{RC}\right)$(Eqn 40)

that I₀ = V₀/R.

Thus, Equation 42 may be rewritten as

$\dfrac{dI}{dt} = -\dfrac RL\left[I(t)-I_0\right]$(43)

The particular solution that interests us is therefore

The transient current which flows in an inductor as the circuit is completed produces a curve similar to that in Figure 20b. Note that this curve for the growth of current in an inductor has the same overall V₀ shape as the curve for the growth of stored charge in a capacitor.

If the voltage supply is replaced by a piece of wire, then the current across the inductor will decay. In that case it follows from Equation 41that

$I(t)R = -L\dfrac{dI}{dt}$(45)

Notice that in this case the time required for the current to decay by a factor of 1/e is L/R, so increasing the resistance will speed–up the decay.

3.3 Oscillations in LC circuits

Having looked at the response of capacitors and inductors individually to transient currents and voltages, the next step is to look at how they behave when connected together, as shown in Figure 21.

In drawing the circuit, arbitrary ‘positive’ directions have been assigned to the current I(t) and the potential differences V_C(t) and V_L(t). i

At all times the total voltage drop around the whole circuit must be zero, so

V_C(t) + V_L(t) = 0

soV_C(t) = −V_L(t)

where$V_C(t) = \dfrac{q(t)}{C}\quad\text{and}\quad V_L(t) = L\dfrac{dI}{dt}$

Hence$L\dfrac{dI}{dt} = -\dfrac{q(t)}{C}$(47)

If the capacitor in Figure 21a initially has a stored charge q₀ as shown, then, when the switch is first closed, current will flow around the circuit in the anticlockwise direction, so I(t) = dq/dt will be negative and q(t) will decrease. It follows from Equation 47 that dI/dt will be negative, causing V_L(t) to be negative. This negative potential difference will counterbalance the positive (but decreasing) voltage V_C(t) across the capacitor.

At all times$I(t) = \dfrac{dq}{dt}$

so$\dfrac{dI}{dt} = \dfrac{d^2q}{dt^2}$

and consequently Equation 47 may be rewritten as

$\dfrac{d^2q}{dt^2} = -\dfrac{q(t)}{LC}$(48)

Now, this is a differential equation that contains a second derivative, but no higher derivatives, so it is called a second–order differential equation.

We can simplify Equation 48 a little, or at least make it more recognizable if we rewrite it in the following way:

The general solution to a second–order differential equation contains two arbitrary constants. If we call them A and ϕ we can write the general solution as

Equation 51 indicates that q(t) oscillates with angular frequency ω₀, increasing and decreasing in repetitive cycles between +A and −A, and having an initial value Acos(ϕ) when t = 0. In order to determine the arbitrary constants A and ϕ we will need two appropriate boundary conditions. The first of these is that at t = 0 the stored charge in the capacitor is q(0) = q₀. The second is that this is the maximum positive value of the charge. i

The second of these conditions implies that Acos(ϕ) = A, so ϕ = 0. The first condition then implies that Acos(ϕ) = Acos(0) = A = q₀. Thus, the particular solution that is of interest to us is

The oscillations of the stored charge described by this Equation 52 are shown in Figure 21b. Note that the period of the oscillations is

so the corresponding frequency is

This is usually called the natural frequency of the LC circuit, and ω₀ is called its natural angular frequency.

The charge oscillations described by Equation 52 cause the current in the circuit, I(t) = dq/dt, to oscillate and this causes associated oscillations in the voltages V_C(t) and V_L(t).

The oscillations described in general by Equation 51,

q(t) = Acos(ω₀t + ϕ)(Eqn 51)

and in particular by Equation 52,

$q(t) = q_0\cos(\omega_0t) = q_0\cos\left(\dfrac{t}{\sqrt{LC\os}}\right)$(Eqn 52)

are examples of simple harmonic oscillations.

The same sort of oscillations can be seen in the simple harmonic motion (SHM) of many mechanical systems, such as a mass on a spring or a pendulum swinging back and forth through a small angle. The capacitor in an LC circuit, with its ability to store charge, behaves to some extent like a spring, a ready source of energy. The inductor, constantly striving to prevent changes in the current, is to some extent like a mass, always reluctant to be accelerated.

As the charge in an LC circuit flows back and forth, the energy is ebbing and flowing between the capacitor and the inductor. At one moment in the cycle, all the energy is stored in the electric field of the capacitor. As this field decays, the energy is transferred to the inductor where it is stored in the magnetic field. As this field in turn decays, an electric field develops in the capacitor and the whole process continues its cycle. No power is dissipated in these interchanges because the voltages across the capacitor and the inductor always lead or lag the current by π/2. Thus in each case $\langle P\rangle = V_{\rm rms}I_{\rm rms}\cos\phi = 0$.

3.4 Damped oscillations in LCR circuits

The argument presented above showed that no energy is lost from an LC circuit, so the oscillations should continue for ever without any decrease in L amplitude or change of frequency. However, in reality, electromagnetic energy will be lost due to heating because real inductors always have some resistance and this will alter the phase relationship between the current and the voltage. This phenomenon of decaying oscillatory behaviour is called damping. i A more realistic circuit, containing a resistor as well as an inductor and a capacitor is shown in Figure 22. In this case the capacitor can be charged when the switch is set at A, and discharged through the inductor and the resistor when the switch is at B. i

The general solution to this second–order differential equation for values of R² which are less than 4L/C may be written:

As usual, boundary conditions (such as the values of q(t) and dq/dt = I(t) at some particular value of t) are required to determine the values of the arbitrary constants A and ϕ. However, irrespective of those values, provided A is not zero, it is possible to interpret Equation 56 as the mathematical description of a quantity that oscillates with angular frequency ω and with an ‘amplitude’ $A\exp\left(-\dfrac{Rt}{2L}\right)$ that decays exponentially with time.

This kind of oscillation is an example of a damped oscillation and is illustrated by the dashed line in Figure 23. Note that if we identify $1/\sqrt{LC\os}$ as the natural angular frequency ω₀ in the case where R = 0, then we can write

$\omega = \omega_0\sqrt{1-\dfrac{R^2C}{4L}}$(58)

showing that the angular frequency of a damped LCR circuit is lower than the natural frequency ω₀ of a pure (undamped) LC circuit.

The exponentially decaying damped oscillations that Equations 56 and 57 describe, those which arise for R² < 4L/C, are said to exhibit underdamping. Underdamping occurs with relatively small values of R.

If R² is very large compared with 4L/C then the capacitor takes a relatively long time to discharge and no oscillation takes place. In this case, the circuit is said to exhibit overdamping. In critical damping R² = 4L/C and the behaviour of q(t) lies between these two extremes, with no full oscillation but, in general, with a single overshot of the q = 0 condition. i

3.5 Driven oscillations in LCR circuits

Whether they do so rapidly or slowly, oscillations in a resistive circuit (i.e. one which has a resistance) will always decay with time. However, if a supply of alternating current is made part of such a circuit it can supply the electromagnetic power that is dissipated and sustain the oscillations indefinitely. The oscillations that occur under these circumstances are called driven oscillations or forced oscillations, since their (angular) frequency will generally be that of the alternating supply rather than any ‘natural’ (angular) frequency associated with the circuit itself.

A driven LCR circuit is shown in Figure 24a.

If the a.c. supply provides an alternating voltage V(t) = V₀sin(Ωt), then the sum of the voltages around the circuit must be zero and so the stored charge q(t) must satisfy the second–order differential equation

where Ω is the angular frequency of the supply.

This equation is harder to solve than those we dealt with earlier, so we will not bother to write down the general solution, we will simply round off our discussion of electrical oscillations by noting some of its main features.

The first thing to note is that after any initial transient behaviour has died away the solutions will take the form of simple harmonic oscillations described by

q(t) = Asin(Ωt + ϕ)

where$A = \dfrac{V_0}{\sqrt{\smash[b]{[(1/C)-L{\it\Omega}^2]^2+(R{\it\Omega})^2}}}$

and$\phi = \arctan\left(\dfrac{-R{\it\Omega}}{(1/C)-L{\it\Omega}^2}\right)$

Notice that the amplitude A of these driven oscillations depends on Ω, the angular frequency of the a.c. supply.

A graph showing the variation of A with Ω is given in Figure 25b. As you can see the amplitude in Figure 25b,

$A = \dfrac{V_0}{\sqrt{\smash[b]{[(1/C)-L{\it\Omega}^2]^2+(R{\it\Omega})^2}}}$

shows a pronounced maximum when Ω is close to the natural angular frequency $\omega_0 = 1/\sqrt{LC\os}$. This is the phenomenon of resonance, the existence of a driving frequency which provokes an exceptionally strong response from the circuit.

Since $\omega_0 = 1/\sqrt{LC\os}$ at resonance, or very nearly, we can rewrite the approximate resonance condition as

$\omega_0L= \dfrac{1}{\omega_0C}$(60)

that is, the reactances of the inductor and capacitor are equal

X_L = X_C(61)

and the phase   $\phi = \arctan\left(\dfrac{-R{\it\Omega}}{(1/C)-L{\it\Omega}^2}\right)$   becomes zero.

This means that the impedance of a series LCR circuit $Z = \sqrt{\smash[b]{R^2 + (X_L-X_C)^2}}$ has its minimum value, R, at resonance (see Figure 25a).

The rms current flowing in the circuit will be given by I_rms = V_rms/Z, so (for a constant rms supply voltage) the rms current will be a maximum at resonance in a series LCR circuit.

 

The voltage in a parallel LCR circuit (Figure 14) shows similar resonant behaviour. In that case

$\dfrac1Z = \sqrt{\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)^2+\dfrac{1}{R^2}}$

so, 1/Z is a minimum at resonance. Consequently Z has its maximum value R at resonance. The rms voltage across each component is the same in this case, and is related to the rms current in the circuit by V_rms = ZI_rms so (for a constant rms supply current) the rms voltage is a maximum at resonance in a parallel LCR circuit.

4 Closing items

4.1 Module summary

1

An alternating current is a flow of electric charge that reverses its direction periodically. It may be described by

I(t) = I₀sin(ωt + ϕ)(Eqn 3b)

where I₀ is the amplitude or peak value of the current, (ωt + ϕ) is its phase, and ϕ is its phase constant. ω is the angular frequency of the current, which also has a period T = π/ω and a frequency f = 1/T. The root-mean-square value of such a current is given by $I_{\rm rms} = I_0/\sqrt{2}$.

2

An alternating voltage V(t) = V₀sin(ωt) applied across a resistor, a capacitor or an inductor will cause an alternating current I(t) = I₀sin(ωt + ϕ). The current through the resistor will be in phase with the voltage, so ϕ = 0. The current through the capacitor will lead the voltage by ϕ = π/2, and that through the inductor will lag the voltage by ϕ = π/2.

3

The role of impedance in a.c. circuits is analogous to that of resistance in d.c. circuits. If a voltage V(t) = V₀sin(ωt) applied across a network of components causes an alternating current I(t) = I₀sin(ωt + ϕ), the impedance of the network is given by Z = V₀/I₀, and the average power dissipated in the network is

$\langle P\rangle = V_{\rm rms}I_{\rm rms}\cos\phi = I_{\rm rms}^2Z\cos\phi$

4

A capacitor of capacitance C has a capacitative reactance X_C = 1/(ωC) where ω is the angular frequency of the voltage supply. Similarly, an inductor of inductance L has an inductive reactance X_L = ωL. The impedance of a single capacitor or inductor is equal to its reactance, the impedance of a single resistor is equal to its resistance.

5

In a series LCR circuit the total impedance is

$Z = \sqrt{\smash[b]{R^2 + (X_L-X_C)^2}}$(Eqn 28)

and the voltage leads the current b

$\phi = \arctan\left(\dfrac{X_L-X_C}{R}\right)$(Eqn 30)

In a parallel LCR circuit the total impedance is

$\dfrac{1}{Z} = \sqrt{\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)^2 + \dfrac{1}{R^2}}$(Eqn 32)

and the current leads the voltage by

$\phi = \arctan\left[\left(\dfrac{1}{X_C}-\dfrac{1}{X_L}\right)\middle/\dfrac{1}{R}\right]$(Eqn 34)

6

Short lived variations in current and voltage called transients can arise when the conditions in a reactive circuit are suddenly disturbed. Such transients may be studied with the aid of an appropriate first-order differential equation, relating the instantaneous value of a quantity (such as the stored charge q(t)) to its instantaneous rate of change (dq/dt). Such an equation has a general solution which contains an arbitrary constant. The general solution may be particularized to a specific physical situation by using an appropriate boundary condition to determine the value of the arbitrary constant.

7

In an LC circuit consisting of an capacitor connected to an inductor, the stored charge will exhibit simple harmonic oscillations described by

q(t) = Acos(ω₀t + ϕ)

where A and ϕ are arbitrary constants with values that must be determined from appropriate boundary conditions and the natural angular frequency

$\omega_0 = \sqrt{\dfrac{1}{LC}}$(Eqn 50)

Such oscillations satisfy a characteristic second–order differential equation which may be written in the form

$\dfrac{d^2q}{dt^2} = -\omega^2 q(t)$(Eqn 49)

8

The addition of a series resistor to an LC circuit results in an LCR circuit in which, under appropriate conditions, the stored charge exhibits damped oscillations in which the ‘amplitude’ decays with time. The frequency of these oscillations is less than the natural frequency of the corresponding undamped LC circuit.

9

The further addition of an alternating voltage supply to the LCR circuit eventually results in driven oscillations of the stored charge that have the same angular frequency as the supply. The amplitude of such oscillations is a function of the supply frequency and is strongly peaked when the supply frequency is close to the natural frequency of the corresponding LC circuit. The ability of a supply of appropriate frequency to excite large amplitude oscillations in the stored charge (and the current) is an example of the phenomenon of resonance.

4.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of the module.

A2

Describe alternating currents and voltages in terms of sinusoidal functions.

A3

Characterize sinusoidally varying quantities in terms of their root-mean-square values and use those values in calculating power dissipation in simple a.c. circuits.

A4

Calculate the reactance of inductors and capacitors.

A5

Use phasor diagrams to determine current, voltage, impedance and phase in series and parallel LCR circuits, and in simple combinations of series and parallel LCR circuits.

A6

Describe the transient behaviour of stored charge (and consequently current) in RC and RL circuits, using appropriate first–order differential equations. (You are not expected to know how to find the general solutions to such equations.)

A7

Describe the simple, damped and driven oscillatory behaviour of stored charge (and consequently current) in LC, LCR and driven LCR circuits, using appropriate second–order differential equations. (You are not expected to know how to find the general solutions to such equations.)

A8

Describe the phenomenon of resonance and identify the conditions for resonance in circuits which display varying degrees of damping.

Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.

4.3 Exit test

Study comment Having completed this module, you should be able to answer the following questions each of which tests one or more of the Achievements.

FLAP material © copyright 1996 Open University