10-1

Unit 10

Alternating current

10.1 Introduction

10.2 Features of an A.C.

10.3 Phasor diagrams.

10.4 Behaviour of basic dipoles facing an A.C.

10.5 RLC series circuit. Impedance and phase lag.

10.6 Resonance on A.C. circuits.

10.7 Quality factor of a RLC series circuit.

10.8 RLC circuit as a filter.

10.9 Questions and problems

Objectives

• To know the main features of alternating current (A.C.) and their effect on

resistors, capacitors and inductors.

• Understand the phase lag between drop of potential and intensity of cur-

rent on A.C. circuits.

• Compute the rate between drop of potential and intensity of current in

RLC series dipoles.

• Define the impedance of a circuit.

• Analyze a RLC series circuit from the point of view of energy.

• To know the meaning of resonance on A.C. circuits

• To understand the working of an electric filter.

10.1 Introduction

Along 19th

century, the electricity was developed around of direct current (D.C.). This type of

electricity was useful for a lot of applications, on electric engines, heating, lighting, telegraph,

telephone, etc… One of the most important businessmen was Thomas Alva Edison, owner of Gen-

eral Electric Company. But when Edison tried to send the electric energy massively to a large dis-

tance, almost all the energy was lost by Joule heating along the transmission line.

In order to understand the problem of sending electric energy with direct current, let’s imag-

ine we want to send, for example, a power of 1000 W at a voltage of 100 V (an usual voltage), to a

little distance, for example 3 Km; the current along the line will be I=1000/100=10 A. The re-

sistance of a copper wire having a cross section of 50 mm2 is around 0,4 Ω/Km; as the current

must go and return, we need 6 Km of wire, with a total resistance of 0,4*6=2,4 Ω. The total power

lost by Joule heating on transmission line will be:

10-2

Lost Power=102*2,4=240 W ¡ 24% of sent power ¡

To solve this problem, the consumers of electricity should be placed near the generating fac-

tory and a big project to build a lot of factories in Niagara Falls (to profit its water jumping), was

presented.

But they were some scientists, engineers and businessmen supporters of alternating current

(A.C.) because of its easy generation (on before unit we saw that a loop turning inside a magnetic

field produces sinusoidal alternating current) and the low quantity of energy lost on the transmis-

sion line. We know that the transformer doesn’t work with direct current, but it works on alter-

nating current, and a voltage of 100 V can be easily converted in, for example, 10000 V. The lost

power sending 1000 W at 10000 V (intensity of current = 1000/10000=0,1 A) to a distance of 3 Km

with the same wire we have used for direct current will be:

Lost Power=(0,1)2*2,4=0,024 W only a 0,0024% of sent power

Nikola Tesla, a Croatian engineer, and George Westinghouse were two of the most active

supporters of A.C., and their disagreement with the wide use of D.C. produced the “War of cur-

rents” on last decade of 19th

century. They won this war and Niagara Falls was preserved.

From then, the electric energy can be sent at very large distances at very high voltages, even

100000 V, and only a few generating factories of electricity are needed. Of course, this voltage is

not useful at home (it would be very dangerous) and it must be reduced with a second transform-

er; this is the reason why we can found transformers along any town.

Besides its easy transportation, Fourier’s theorem states that any periodic function can be

written as an addition of sinusoidal functions, and then the results for alternating current can be

extrapolated to any type of periodic function, even digital signals.

10.2 Features of an a.c.

We saw on before unit that a sinusoidal alternating current can

be produced by turning a loop inside a uniform magnetic field. The

symbol for the A.C. generators on circuits is that on right, and the mathematical expression for

the difference of potential on terminals of generator can be written as:

)tcos(U)t(u um ϕω +=

this equation has be written as a cosine, but it could be also written as a sine, only adding 90º

(π/2 rad) to the phase:

)2

tsin(U)tcos(U)t(u umum

πϕωϕω ++=+=

The main features of this sinusoidal

function are:

• The amplitude, Um, is the maximum

value reached by the function

Figure 10-1). The units of amplitude are

the same that the magnitude we are

representing; in this case, volts.

-2

0

2

time (s)

D.d

.p.

(V)

Um

T

1

3

-1

-3

Figure 10-1. Amplitude and period of an

a.c.

10-3

• The period, T is the time taken for a whole cycle of function ( Figure 10-1). Its unit in

the I.S. is the second.

• The frequency, f, is the number of cycles by second. It’s the inverse of period: T

1f = . Its unit

is Hertz (Hz): 1 Hz = 1 s-1

.

• The angular frequency, angular speed or pulsatance, ω, is the angle (in radians) turned on a

second. Its unit is rad/s. We must remember that the radian is dimensionless, and then the

dimensions of angular frequency are T-1

. As a whole cycle equals 2π radians and the period, T,

is the time taken by a cycle: f2T

12 ππω ==

• The phase ut ϕω + , expressed in radians.

• The initial phase uϕ , is the magnitude

of phase on time t=0. For an easier

reading it is usual to give the magnitude

u in degrees instead radians, but when

it’s operated, it must be converted to

radians. The initial phase, really, locates

the sinusoidal wave along the X axis. On

right (Figure 10-2) the initial phase of

several sinusoidal functions can be

seen. It’s important to note that this

figure represents the function

)tcos(U)t(u um ϕω += ; if the sinus func-

tion is used, as 90º must be added, the

initial phases would be different.

• Root mean square: Root mean square

(rms) of a sinusoidal function is defined as the amplitude divided into 2 . For example (for a

voltage and for a intensity):

2

UU m

rms = 2

II mrms =

The root mean square of a sinusoidal function is very important in alternating current be-

cause this is the magnitude measured by the measuring devices (voltmeters, ammeters, mul-

timeters, etc…). Its physical meaning is that of a direct current dissipating the same power

that an alternating current of amplitude Um on a resistor. The demonstration of his equiva-

lence will be done when we deal with power on A.C., at the end of this unit. The current we

can found in the wall sockets at our homes is a current having a voltage Urms=220 V, and then

the amplitude of this voltage is Um=220*√2 =310 V.

• Phase lag is defined for two sinusoidal functions having the same angular speed. Phase lag is

the difference between phases of both functions at the same time; as the angular frequency is

the same for both functions, the difference between phases equals the difference between

initial phases.

Let’s consider, for example, a voltage )tcos(U)t(u um ϕω += and an intensity of current

)tcos(I)t(i im ϕω += . The phase lag between voltage and intensity is:

iuiuui )t()t( ϕϕϕωϕωϕ −=+−+=

Figure 10-2. Initial phase of several sinusoidal func-

tions (expressed as cosinus)

10-4

If we had defined phase lag between intensity and voltage ( iuϕ instead uiϕ ) the sign of phase

lag would be opposite. But we always will refer to a phase lag between voltage and intensity:

iu ϕϕϕ −=

10.3 Phasor diagram.

In order to simplify the analysis of A.C. circuits and make easier the graphical representa-

tion of sinusoidal functions, a special type of diagrams can be used, the phasor diagrams.

A phasor is a vector whose modulus (length) is proportional to the amplitude of sinusoidal

function we are representing. This vector is rotating around its origin in counterclockwise di-

rection, with an angular speed equal to the angular frequency of the wave (ω). The angle be-

tween this rotating vector and the positive direction of X axis at any moment is (ωt+φ). With

this graph, the horizontal projection of rotating vector corresponds to the function

( ) cos( )mu t U tω ϕ= + , and its vertical projection to the function ( ) sin( )mu t U tω ϕ= + .

But the position of such phasor is continuously changing (it is rotating) and to have a fixed

image of the vector, it is represented on time t=0, in such way that the angle with the X axis is

the initial phase of wave.

The phasor are useful because they can easily highlight the relationship between modulus

and initial phases of different waves. As an example, in the (Figure 10-4) can be seen two

couples of waves of voltage and intensity whit phases lag positive (voltage goes ahead intensi-

ty, or intensity goes behind voltage), and negative (voltage goes behind intensity, or intensity

goes ahead voltage) with their phasor diagrams.

If phase lag is zero, both functions go in phase.

Figure 10-4 a) Positive phase lag between voltage and intensity; voltage goes ahead intensity (or intensity goes behind voltage). On right, phasor diagram.

Figure 10-4 b) Negative phase lag between voltage and intensity; voltage goes behind intensity (or intensity goes ahead voltage). On right, phasor diagram

Figure 10-3 Rotating vector and phasor corresponding to function

( ) cos( )mu t U tω ϕ= + or

( ) sin( )mu t U tω ϕ= +

10-5

10.4 Behavior of basic dipoles facing an A.C.

At this point we’ll study the behavior of basic dipoles (resistor, inductor and capacitor) when

an alternating current is flowing along them. To do it, we’ll study two parameters:

• the rate between the amplitude of voltage on terminals of dipole and the intensity of cur-

rent

• the phase lag between voltage on terminals and intensity of current

1. Resistor

Let’s suppose a resistor (resistance R) flowed by a intensity of

current i(t) and having a difference of potential u(t) between its

terminals. In order to simplify the computations, we’ll take initial

phase zero for intensity (ϕi=0):

)tcos(I)t(i m ω=

At any time, on a resistor must be verified that )t(Ri)t(u = and then

)tcos(RI)t(u m ω=

The difference of potential on terminals of resistor is a new cosine function,

)tcos(U)t(u um ϕω += being mm RIU = and ϕu=0

Obviously, if we had taken a initial phase not zero for intensity, the initial phase of voltage

would have been the same of intensity.

On a resistor, then voltage and intensity go on phase

0

RI

U

iu

m

m

=−=

=

ϕϕϕ

2. Inductor

On an inductor, the main parameter is the self inductance

coefficient L. If we take initial phase zero for intensity (ϕi=0):

)tcos(I)t(i m ω=

At any time, on an inductor must be verified that

dt

)t(diL)t(u = and then

)2

tcos(IL)tsin(ILdt

))tcos(I(dL)t(u mm

m πωωωω

ω+=−==

The difference of potential on terminals of inductor is a new cosine function,

)tcos(U)t(u um ϕω += being mm ILU ω= and ϕu=π/2

10-6

On an inductor, then

2

LI

U

iu

m

m

πϕϕϕ

ω

=−=

=

Voltage goes 90º ahead intensity, or Intensity goes 90º behind voltage.

The factor Lω is known as Inductance or Inductive Reactance ( ωLXL = ); it has dimensions of

a resistance, and thus is measured in Ohms. Really, the inductance on inductor acts as the re-

sistance on a resistor, but the most important difference is that the inductance depends not only

on the self inductance coefficient, but also on the frequency. So, on a circuit, the rate between

amplitudes of voltage and intensity on inductor can change by only changing the frequency of

current.

3. Capacitor

In this case, for an easier computation, we’ll suppose the

voltage on terminals of capacitor having initial phase zero (ϕu=0):

)tcos(U)t(u m ω=

For a capacitor must be remembered that )t(u

)t(qC = and

dt

)t(dq)t(i =

Therefore

)2

tcos(CU)tsin(UCdt

))tcos(U(Cd

dt

)t(dCu

dt

)t(dq)t(i mm

m πωωωω

ω+=−====

The intensity of current on a capacitor is a new cosine function,

)tcos(I)t(i im ϕω += being ωCUI mm = and ϕi=π/2

On a capacitor, voltage goes 90º behind intensity, or Intensity goes 90º ahead voltage

2

C

1

I

U

iu

m

m

πϕϕϕ

ω

−=−=

=

The factor 1/Cω is known as Capacitance or Capacitive Reactance ( ωC1XC = ); it has di-

mensions of a resistance, and thus is measured in Ohms. Really, the capacitive reactance acts as a

resistor, but the most important difference is that the reactive capacitance not only depends on

the capacitance, but also on the frequency. So, in a circuit, the rate between amplitudes of volt-

age and intensity on capacitor can change by only changing the frequency of current.

10-7

Example 10-1

A resistor 5 Ω sized, an inductor 10 mH sized, and a capacitor 50 µF sized, are connected

in series. The voltage on terminals of inductor is V)4

t1000cos(100)t(uL

π+= . Compute

the intensity of current flowing along three dipoles, the voltage on terminals of resistor

and capacitor, and plot all these magnitudes on a graph and its phasor diagram.

Solution:

As we know voltage on inductor, it’s easy to get the intensity of current on inductor. The

angular speed of current is ω=1000 rad/s, and then:

A1010

100

X

UI1010001010LX

L

Lmm

3L ===Ω=⋅⋅== −ω

Phase lag on inductor is 90º, and then rad42424

ii

πππϕ

πϕ

πϕ −=−==−=

As all the dipoles are connected in series, the intensity of current on inductor is the same

for all the dipoles, being:

A)4

t1000cos(10)t(i)t(i)t(i)t(i CRL

π−====

From intensity, voltages on terminals of resistor and capacitor can be computed:

on resistor V50510RIU mRm =⋅==

and as phase lag on resistor is zero: V)4

t1000cos(50)t(uR

π−=

On capacitor: V2002010XIU2010001050

1

C

1X CmCm6C =⋅==Ω=

⋅⋅==

−ω

And phase lag rad4

3

422)

4( uu

πππϕ

ππϕϕ −=−−=−=−−=

V)4

3t1000cos(200)t(uC

π−=

Phasor diagram and graph with voltages and intensity are:

I

UL

UC UR

10-8

It can be noted that, obviously, intensity and voltage on resistor go in phase, phase lag

between voltages on inductor and capacitor is 180º (or -180º), voltage on inductor goes

ahead intensity, and voltage on capacitor goes behind intensity.

10.5 RLC series circuit. Impedance and phase lag.

A RLC series circuit is a circuit having

a resistor, an inductor and a capacitor

connected in series (RLC dipole). As we

have above seen, the intensity of current

is the same for all the basic dipoles on

circuit, and the voltage on each device can

be computed. In order to simplify compu-

tations, we have taken initial phase of

intensity zero (ϕi=0).

But the main problem is now to re-

late the intensity of current i(t) to the

voltage on terminals of RLC circuit, u(t). This voltage could be computed by only adding the volt-

ages on the devices on circuit:

)2

tcos(IC

1)

2tcos(IL)tcos(RI)t(u)t(u)t(u)t(u mmmCLR

πω

ω

πωωω −+++=++= (1)

The addition of sinusoidal functions is a new sinusoidal function, and u(t) is a sinusoidal func-

tion, but from this equation is difficult to assess its amplitude and its initial phase. Nevertheless,

the amplitude and phase of u(t) can be computed using two new features of a RLC circuit, its Im-

pedance and its phase lag. If we take null the initial phase of intensity (to simplify the computa-

tions), then the phasor diagram of this circuit will be that of next picture:

-250

-200

-150

-100

-50

0

50

100

150

200

250

-15

-10

-5

0

5

10

15

0 2 4 6 8 10

Vo

lta

ge

(V

)

Inte

nsi

ty (

A)

wt (rad)

i(t) UR(t) UL(t) UC(t)

R

u(t)

L C i(t)=Imcos(wt)

uR(t) uL(t) uC(t)

Figure 10-5. RLC series dipole

10-9

From the rectangle triangle of diagram on right, according Pythagoras’s theorem:

2 2 2 2 2 2 2 2 21 1( ) (( ) ) ( ) ( )m

m m m m m L C

m

UU RI L I U I R L R X X R X

C C Iω ω

ω ω= + − = + − = + − = +

The difference between the inductive reactance (XL) and the capacitive reactance (XC) is the

reactance (X), measured in Ohm: CL XXX −=

The rate between Um and Im is called the Impedance (Z) of RLC series circuit (it is, obviously

measured in Ohm and it plays the role of a “resistance” on an A.C.

2 2m

m

UZ R X

I= = +

The phase lag (ϕϕϕϕ) of circuit can be calculated from the rate of both catheti of before triangle:

1 1( ) ( )

( )m

L C

m

L I LX X XC Ctg

RI R R C

ω ωω ωϕ

− −−

= = = =

Obviously, the impedance and phase lag of a resistor equal R and 0º, those of an induc-

tor equal XL and 90º and those of a capacitor, XC and -90º.

The main magnitudes of an RLC series circuit can be easily summarized drawing the Imped-

ance triangle, a triangle being the horizontal cathetus the resistance of circuit, the vertical cathe-

tus the reactance of circuit, the hypotenuse the impedance of circuit, and the angle between Z

and R, the phase lag of circuit.

Depending on magnitudes of L and C, X can be positive (inductive circuit), negative (capac-

itive circuit), or zero (resonant circuit). R

ϕϕϕϕ

Figure 10-6 b) Impedance tri-

angle for a capacitive circuit

XZ

R

ϕϕϕϕ

Figure 10-6 a) Impedance tri-

angle for an inductive circuit

10-10

Example 10-2

On circuit of example 10-1, find the voltage on terminals of circuit.

Solution:

From the impedance of this circuit:

V8,11118,11*10ZIU18,11)2010(5Z mm22 ===Ω=−+=

Phase lag

º4,108454,63º4,6325

2010tg iu −=−−=+=−=−=

−= ϕϕϕϕϕ

And voltage on terminals of circuit: V)º4,108t1000cos(8,111)t(u −=

The full phasor diagram:

Review about phase lag and Impedance for some dipoles

Phase lag and impedance for basic dipoles (R, L and C) and for dipole RLC (RLC series

circuit) can be summarized on next table:

DDiippoollee PPhhaassee llaagg φφ==φφuu-- φφii IImmppeeddaannccee ZZ

RReessiissttoorr 00 RR

IInndduuccttoorr 9900ºº ωLXL =

CCaappaacciittoorr --9900ºº ωC1XC =

RRLLCC sseerriieess

−

=R

C

1L

arctg ωω

ϕ 2

2

m

m

C

1LR

I

UZ

−+==

ωω

TTaabbllee 1100--11 SSuummmmaarryy ooff pphhaassee llaagg aanndd iimmppeeddaannccee ffoorr bbaassiicc ddiippoolleess aanndd RRLLCC sseerriieess cciirr--

ccuuiitt..

I

UL

UC UR

U

10-11

10.6 Resonance on A.C.

A RLC series circuit is at resonant frequency (resonant circuit) when the reactance of circuit

X=XL-XC is zero. The reactance doesn’t only depend on magnitudes of L and C, but also on frequen-

cy of current, and then a circuit can be resonant at a frequency but not resonant at another dif-

ferent frequency. If a circuit is on resonance at an angular frequency ω0, it means that:

LC

1

2

1f

LC

1

C

1LXX 00

0

0CLπ

ωω

ω ====

At resonance, X=0, phase lag is also zero

(ϕ=0) and impedance equals the resistance of

circuit:

R0RZ 20 =+=

At resonance frequency, the impedance of

a circuit is the lowest, equal to the resistance of

circuit. Consequently, if impedance is the mini-

mum, the intensity of current flowing along the

circuit will be the maximum. For this reason,

when several currents with different frequencies

are flowing along a RLC series circuit, that cur-

rent whose frequency matches the resonant frequency reaches the maximum intensity.

On a resonant circuit, voltage on terminals of RLC series circuit and intensity go on phase,

because

0R

C

1L

tg 0

0

=

−

=ω

ω

ϕ

If we consider the circuit on picture, with

the generator working at resonant angular

frequency, ω0:

)tcos(RI)t(u 0mR ω=

)º90tcos(IL)t(u 0m0L += ωω

)º90tcos(C

I)t(u 0

0

mC −= ω

ω )t(u)t(u)t(u)t(u CLR ++=

In resonance, 00 C/1L ωω = and phase lag between uL(t) and uC(t) is 180º

uL(t) = - uC(t) and u(t) = uR(t)

R L C

uR(t) uL(t) uC(t)

i(t)=Imcosω0t

u(t)

f (Hz)

Z (

Oh

ms)

I (A

)

f r

IZ

resonancia

f0

Z0

resonance

Figure 10-7 Impedance and intensity on a RLC circuit as a function of frequency

10-12

Example 10-3

Represent the phasor diagram and graph the of intensity of current, voltage on resistor,

voltage on inductor, voltage on capacitor and voltage on terminals of a RLC series circuit

having a resistor 5 Ω sized, a 10 mH inductor and a 100 µF capacitor. The amplitude of

intensity is 10 A, initial phase of intensity can be taken zero, and as angular frequency must

be used that corresponding to the resonant angular frequency.

Solution:

The resonant angular frequency is:

s/rad100010*100*10*10

1

LC

1630 ===

−−ω

Magnitude of every voltage can be calculated in the same way than on exercises 10-1

and 10-2. After these calculations, with the resonant angular frequency, drawing of inten-

sity and voltages is:

And its phasor diagram:

Can be noted as uL and uC are in opposition of phase (phase lag 180º) and then their addi-

tion is always null (equal moduli). On the other hand, uR and i go on phase, being 90º the

phase lag between uL and i, and -90º between uC and i. Voltage on terminals of RLC circuit

equals uR.

-150

-100

-50

0

50

100

150

-15

-10

-5

0

5

10

15

0 2 4 6 8 10

Vo

lta

ge

(V

)

Inte

nsi

ty (

A)

wt (rad)

i(t) UR(t) UL(t) UC(t)

UL

I

U=UR

Uc

10-13

10.7 Quality factor of a RLC series circuit

When a RLC series circuit is at resonant frequency, capacitor and inductor are exchanging

their powers (none of them really consumes energy), being resistor the only device really losing

energy by Joule heating. Let’s imagine that we remove the generator from circuit while oscilla-

tions of exchange of energy are produced, and our circuit hasn’t any resistor. In this situation, the

energy stored on circuit would be oscillating forever between capacitor and inductor; but on a

circuit with resistor (as it occurs on real circuits), some energy is lost by cycle on resistor, and so

the oscillations are damping. To express the quantity of energy lost by oscillations and then to

give an idea about how quickly the damping of oscillations is produced, can be defined the quality

factor (or Q factor) of RLC series circuit as the rate at the resonant frequency between the maxi-

mum energy stored on inductor or capacitor (2

max

1

2stored mW I L= ) to the energy lost by cycle on re-

sistor 2π times. It can be easily demonstrated that the energy consumed by a resistor during a

cycle is 21

2R mW I RT= , being T the period of oscillation (

0

2T

ω

π= ). Therefore:

C

L

R

1

LC

1

R

L

R

L

RTI

LI2

W

W2Q 0

2m

2m

R

storedmax

0

=====

=

ωππ

ωω

Q is dimensionless, depending on R, L and C, and is high for low damped circuits. It is a very

important parameter to understand the behavior of a RLC series circuit as a filter.

10.8 Filters. Bandpass, highpass and lowpass filters.

The RLC series circuit is very useful because it can be employed as a filter of electric signals,

by filtering these signals according their frequency.

The idea of a filter involves a signal “entering” on filter and a second signal “exiting” from fil-

ter, and for this reason we’ll study the RLC series circuit as a device having two input terminals

(always the feeding of circuit) and two output terminals (these terminals will correspond to those

of resistor, inductor or capacitor, depending on the range of frequencies we are filtering). The

behaviour of such filter will be studied by comparing the amplitude of voltage on output related

to that on input.

Let’s start by considering the output of filter on terminals of resistor, the so called bandpass

filter.

• Bandpass filter

If we consider the terminals of resistor as output of fil-

ter, the amplitude of voltage on output for an angular fre-

quency ω is

22

mmmRouput

)C

1L(R

RU

Z

URRIUU

ωω −+

====

UR and Um are the amplitudes of voltages on resistor (output

of filter) and terminals of RLC circuit (input of filter); Im is the

amplitude of intensity along the RLC circuit. The rate be-

tween amplitudes of voltages on output and input is:

C

Ru(t)

L

uR(t)

Figure 10-8 RLC circuit as a bandpass

filter

10-14

22m

R

input

output

)C

1L(R

R

U

U

U

U

ωω −+

==

If we graph this equation (UR/Um against

ω or f), we get a curve showing a peak corre-

sponding to the resonant frequency; the

maximum at this frequency is 1U

U

0ffm

R =

=

The physical meaning of this curve is

that only those signals having the resonant

frequency (f0) will have the same amplitude

on output than on input of filter, and those

frequencies near to f0 will show on output a

amplitude similar to that on input (these frequencies “will pass” the filter).

On the other hand, frequencies very low or very high will show a very low amplitude on out-

put of filter (they “won’t pass” the filter).

It is usually considered as frequencies passing the filter (passing band or bandwidth) those

frequencies on range [f1, f2], being f1 and f2 the frequencies verifying 2

1

U

U

21 f,fm

R = .

It means that those frequencies whose amplitude on output is higher than 2

1 times the

amplitude on input are considered as passing the filter.

It can be demonstrated that the condition 2

1

U

U

m

R ≥ is equivalent to say that the power lost

on resistor must be at least a half of the maximum power lost on resistor (it occurs at the reso-

nant frequency): 2

1

RI

RI20

2m ≥ . I0 is the amplitude of intensity at the resonant frequency.

Obviously, a RLC circuit with a narrow passing band

will be more effective when we want to “select” a specific

frequency. An example is the tuning circuit of a radio,

when we want to listen to only one broadcasting station

(we want to select only the frequency corresponding to

that broadcasting station). To do it, there is a RLC circuit

inside the radio connected to the antenna, with a capaci-

tor having variable capacitance; by changing the capaci-

tance of capacitor, the resonant frequency of circuit can

be adjusted to that of broadcasting station, and in this

way we can select the station to be listened to.

The narrower the curve of UR/Um against frequency,

the better the radio quality. In fact, the shape of this curve is strongly related to the Q factor of

circuit; we can analytically calculate the frequencies f1 and f2 by only solving the equation

2

1

)C

1L(R

R

22

=

−+ω

ω

0

0,2

0,4

0,6

0,8

1

1,2

10 100 1000 10000 100000

f (Hz)

Us

/UUR/ U

m

f0 f

2 f

1

2

1

f

Figure 10-9 Amplitude of voltage on output related to amplitude on input for a bandpass filter

Figure 10-10 Variable capacitor of a radio tuning circuit

10-15

From this equation: 22222 )C

1L(R)

C

1L(RR2

ωω

ωω −=−+=

By introducing the variable 0

xω

ω= and taking in account that

RC

1

R

LQ

0

0

ω

ω== :

x)1x(Q1)x

1x(Q)

x

QRQRx(R)

C

1L(R 22222 ±=−±=−−=−=

ωω

By neglecting the negative solutions, we find out:

)4Q

1

Q

1(

2

1x

21 ++−= )4Q

1

Q

1(

2

1x

22 ++= Q

1xx 12 =−

L

R

Q

012 ==−

ωωω

L2

Rff 12

π=−

Therefore, higher quality factor

narrower passing band. It must also

be noted that passing band doesn’t

depend on capacitance of RLC circuit.

On figure can be seen the

curves of different RLC circuits acting

as bandpass filters. All of them have

the same resonant frequency (equiv-

alent to 1154,7 rad/s), but different

Q factors (1,15 1,73 and 3,46) and so

different passing bands (1000, 667

and 333 rad/s).

• Highpass filter

If we now consider the output of filter on termi-

nals of inductor, we have a highpass filter.

The amplitude of voltage on terminals of induc-

tor and so the amplitude on output related to the

amplitude on input for any frequency (or angular fre-

quency) can be calculated in the same way than for

the bandpass filter:

22m

L

input

ouput

22

mmmLouput

)C

1L(R

L

U

U

U

U

)C

1L(R

UL

Z

ULILUU

ωω

ω

ωω

ωωω

−+

==

−+

====

The drawing of this equation of UL/Um against w leads us to the characteristic curve of a

highpass filter. On figure 10-13 can be seen as a highpass filter is damping low frequencies

(UL/Um→0) and for high frequencies the amplitude on output is almost equal than that on input

(UL/Um→1). For frequencies around the resonant frequency, the voltage on output can even be

R

Lu(t)

C

uL(t)Input Output

Figure 10-12 RLC circuit as a highpass filter

Figure 10-11 Curves corresponding to three bandpass filters

with the same L=15 mH and C=50 µF but different resistors: 15,

10 and 5 Ω

10-16

higher than the voltage on input;

depending on Q factor of circuit,

this ratio can reach very high val-

ues, what is used to build genera-

tors of high voltages.

It must be noted that the

maximum of these curves is near

the resonant frequency, but they

don’t exactly match. On the other

hand, the ratio UL/Um at resonant

frequency is the Q factor:

QR

L

)C

1L(R

L

U

U 0

2

0

02

0

m

L

0

==

−+

=

=

ω

ωω

ω

ωω

and the Q factor can be calculated by only looking for the value of UL/Um at ω0.

In the same way than for the bandpass filter, the cutoff frequency f1 is arbitrarily taken as

that frequency verifying 2

1

U

U

1fm

L = . It Is, the bandwidth of a highpass filter is [f1, ∞] ; only high

frequencies will “pass” this filter.

• Lowpass filter

If we now consider the output of filter on termi-

nals of capacitor, we have a lowpass filter.

The amplitude of voltage on terminals of capaci-

tor and so the amplitude on output related to the

amplitude on input for any frequency (or angular fre-

quency) can be calculated in the same way than be-

fore:

22m

C

input

ouput

22

mmmCouput

)C

1L(RC

1

U

U

U

U

)C

1L(RC

U

Z

U

C

1I

C

1UU

ωωω

ωωω

ωω−+

==

−+

====

The drawing of this equation of UC/Um against w leads us to the characteristic curve of a

lowpass filter. On figure 10-15 can be seen as a lowpass filter is damping high frequencies

(UC/Um→0) and for low frequencies the amplitude on output is almost equal than that on input

(UL/Um→1). For frequencies around the resonant frequency, the voltage on output can even be

higher than the voltage on input; depending on Q factor of circuit, this ratio can reach very high

values, as it happened on the highpass filter.

It must be noted that the maximum of these curves is near the resonant frequency, but they

don’t exactly match. On the other hand, the ratio UC/Um at resonant frequency is the Q factor:

Figure 10-13 Curves corresponding to three highpass filters. Equal R L and C than on figure 10-11

Cu(t)

L

uC(t)

R

Input Output

Figure 10-14 RLC circuit as a lowpass filter

10-17

QRC

1

)C

1L(RC

1

U

U

02

0

02

0m

C

0

==

−+

=

=ω

ωωωωω

the Q factor can be calculated by only looking for the value of UC/Um at ω0.

And the cutoff frequency f1 is arbitrarily taken as that frequency verifying 2

1

U

U

1fm

C = . It is,

the bandwidth of a lowpass filter is [0, f1] ; only low frequencies will “pass” this filter.

10.9 Questions and problems

1. Consider a capacitive circuit. ¿How does the capacitive reactance change if the angular fre-

quency increases two times?

2. A capacitor C = 0,5μF is connected to an A.C. generator; the amplitude of voltage on terminals

of generator is Um = 300 V ¿Which is the amplitude of intensity of current along the capacitor, Im,

if angular frequency is

a) ω = 100 rad/s? b) ω = 1000 rad/s?

Sol: a) 0,015 A b) 0,15 A

3. An inductor L = 45 mH is connected to an A.C. generator; the amplitude of voltage on terminals

of generator is Um = 300 V and the inductive reactance of inductor is XL = 1300 Ω. ¿Which is

a) the angular frequency ω and the frequency f of generator?

b) ¿The amplitude of intensity, Im?

c) Find that angular frequency that equals the inductive reactance of inductor to the ca-

pacitive reactance of problem 2.

Sol:

10-18

a) ω = 2,9.104

rad/s, f= 181514,67 Hz b) 0,23 A c) 21081,85 rad/s

4. A circuit is made up by two basic dipoles in series. The terminals of this circuit are connected to

an A.C. generator giving a voltage u(t) = 150 cos(500 t+ 10º) V, and flowing along the circuit an

intensity of current i(t) = 13,42cos(500t - 53,4º) A. Determine the two basic dipoles and their

magnitudes.

Sol: R = 5 Ω L = 0,02 H

5. Along a circuit made up by two basic dipoles in series and a generator giving a voltage u(t) =

200sen(2000t + 50º) V flows an intensity i(t) = 4cos(2000t + 13,2º) A. Determine such dipoles and

their magnitudes.

Sol: R = 29,7 Ω C = 12,4 µF

6. On a RL series circuit, R = 5 Ω and L = 0,06 H, drop of potential on terminals of inductor is uL(t)

= 15cos200t V. Compute:

a) the intensity of current b) phase lag and impedance of circuit

c) voltage on terminals of circuit

Sol:

a) i(t) = 1,25 cos(200t - 90º) A; b) ϕ = 67,4º Z = 13 Ω

c) u(t) = 16,3cos(200t - 22,6º) V

7. Along a RLC series circuit, R = 2 Ω, L = 1,6 mH and C = 20 µF, flows an intensity of current i(t) =

3cos(5000t - 60º) A.

Compute and draw the phasor diaagram of functions corresponding to the drop of potential on

each device and the drop of potential on terminals of RLC circuit.

Sol:

uR(t) = 6cos(5000t - 60º) V, uL(t) = 24cos(5000t + 30º) V,

uC(t) = 30cos(5000t - 150º) V, u(t) = 6(2)1/2

cos(5000t - 105º) V

8. A resistor 5 Ω sized and a capacitor are connected in series. Voltage on resistor is uR(t) = 25

cos(2000t + 30º) V. If voltage on terminals of RC dipole goes 60º behind current, ¿which is the

capacitance C of capacitor?

Sol: C = 57,7 µF

9. The voltage applied on terminals of a RLC series circuit goes 30º ahead the current. Amplitude

of voltage on inductor is two times the amplitude of voltage on capacitor, and uL(t) = 10cos1000t

V. If R = 20 Ω find values of L and C.

Sol: L = 23,1 mH, C = 86,6 µF

10. Along a RL series circuit, having L = 0,05 H, flows an intensity of current i(t)=2(2)1/2

cos500t A.

Voltage measured with a voltmeter on terminals of resistor is VR = 50 V. Determine:

a) Magnitude of R

b) u(t) on terminals of generator

10-19

c) if a capacitor is added in series with R and L, compute the capacitance of this capacitor

in order to get 30º of phase lag between voltage on generator and intensity of current.

d) the new i(t) after the capacitor has been added.

Sol:

a) R = 25 Ω b) v(t) = 100 cos(500t + 45º) V,

c) C = 189 µF, d) I1(t) = 2,45(2)1/2

cos(500t + 15º) A

11. A RLC series circuit made up by L=2 H, C=2 μF and R=20 Ohm is connected to an adjustable

frequency generator, giving an amplitude Um=100 V.

a. Find the resonant frequency f0.

b. Find the amplitude of intensity at resonant frequency.

c. Find the amplitudes of voltage on resistor, inductor and capacitor at resonant frequency.

d. Find the quality factor of circuit.

e. Find the bandwith of circuit.

12. If the voltage of a RLC circuit goes ahead regarding the current, is the frequency of current

higher or lower than the resonant frequency?

GLOSSARY

Frequency is the number of cycles by second of sinusoidal function.

Initial phase is the phase of sinusoidal function on time t = 0.

Phase lag ϕ is the difference of phases between the initial phases of two sinus-

oidal functions, usually voltaje and intensity iu ϕϕϕ −=

Root mean square (rms): square root of average value of square of sinusoidal

function along a cycle:

2

UU m

rms = 2

II mrms =

Impedance: Rate between amplitudes of drop of potential and intensity in a

RLC dipole

m

m

I

UZ =

Inductive reactance: Rate between amplitudes of difference of potential and

intensity on an inductor.

ωLXL =

Capacitive reactance: Rate between amplitudes of difference of potential

and intensity on a capacitor.

ωC

1XC =

10-20

Resonant frequency: is that frequency of a RLC series circuit producing the

minimum impedance. It depends on L and C.

LC2

1f0

π=

Quality factor: of a RLC series circuit is the rate at the resonant frequency be-

tween the maximum energy stored on inductor or capacitor and the lost energy

by cycle on resistor:

C

L

R

1

R

LQ 0 ==

ω

Bandwith of a bandpass filter: range of frequencies [f1, f2] verifying that volt-

age on output of filter related to that on input at resonant frequency is high-

er than 21

2

1

U

U

21 f,fm

R ≥

Cutoff frequency: of a highpass/lowpass filter is the lower/higher frequency

verifying that voltage on output of filter related to that on input at resonant

frequency equals 21