ct1 unit 2

7/30/2019 CT1 Unit 2

1/29

Unit 2: AC Electricity 2-1

AC ELECTRICITY

2

AIMS

The aims of this unit are to describe and quantify the behaviour of

resistors, inductors and capacitors in AC circuits.

INTRODUCTION

In Unit 1 we studied direct current or DC electricity. In DC circuits, the

voltage remains constant with respect to time and the current flows in

one direction only through the circuit.

In this unit we will be studying another form of electricity, called

alternating current or AC. In AC circuits, the value of the voltage is

constantly changing with time, as is the direction of current flow through

the circuit. In particular, we will be looking at the behaviour of simple

AC circuits containing resistances, inductances and capacitances. For

the purpose of this course, this unit will complete our study of electricity.

Why use AC? Why build circuits in which the magnitude and direction

of the voltage is constantly changing, and in which the current is

oscillating back and forth, rather than flowing steadily in one direction?

DC circuits seem so straightforward in comparison: all that are needed

in analysing them are Ohms and Kirchoffs laws. So why not use DC in

all electric and electronic applications?

The answer is that AC is the only form in which electrical power can be

economically transmitted over long distances. The following example

demonstrates this point very clearly. Suppose we wish to supply a city

with ten million watts of electrical power. Our generating station is

located some distance away, and we wish to minimise the power

dissipation along the transmission lines.

Remember that electrical power (in watts) equals the line voltage (in

volts) multiplied by the current (in amps). To supply our city with power,

we can use any combination of voltage and current, so long as theproduct of the two gives us the figure of ten million watts. Shall we send

the power out from our generating station in the form of high-voltage and

low-current? Or as low-voltage and high-current?

If we set the voltage at 100V, then the current must be 100,000A to

satisfy the electricity consumers in the city. Alternatively, if we increase

the voltage to 100,000V, then 100A of current will suffice.

However, even the best conducting transmission lines will present some

resistance to the flow of current, and accordingly, some power will be

dissipated. Assuming line resistance is 20, the power lost in transmission

is given by the equation, P = I2 R. If we generate 100,000A at 100V, then

the power loss equals 200,000 million watts. This is some 20 thousand

times the power required by the city!

7/30/2019 CT1 Unit 2

2/29

2-2 Communications Technology 1

On the other hand, if the power is sent out along the lines at 100,000V

and only 100A, then the lost power is just 200,000 watts or 2% of the load

required by the city. The reason for these hugely varying results is that

power dissipated in the transmission of electricity increases not just with

the current, but with the square of the current.

It is for this reason that electrical power is distributed at high-voltagesand low currents. But could we not transmit the high-voltage, low-

current power in DC form? Yes, we could; but then we would be unable

to step-down the line voltage with transformers before passing it on to

the power consumers. Only AC can be manipulated in this way. Today,

most power stations produce electricity at 100,000-400,000V (100-

400kV) AC, which transformer stations near the point of usage step-

down to the mains line voltage of 240V AC.

Radio waves are of their nature also constantly changing in voltage and

current. Their transmission, detection, processing, amplification and

translation into audible sound are all other important areas in which an

understanding of AC is essential. In summary, AC is a vital element inour electric and electronic world.

SINUSOIDAL

WAVEFORMS

OBJECTIVES

After studying this sub-unit, you should be able to:

1. Define or explain the meaning of the following terms:

. alternating current and voltage

. amplitude

. angular frequency

. cycle

. frequency

. hertz

. instantaneous current and voltage

. leading and lagging phase angles

. peak value

. periodic time

. phase angle and phase difference

. sinusoidal waveform

2. Write the mathematical equation for the instantaneous value of avoltage or current sinusoidal waveform, given the appropriate data.

We learnt in Unit 1 that DC voltage and current remains constant with

respect to time, as shown in Figure 2.1. As indicated above, very often

in electric and electronic devices voltages and currents are used which

are not constant with respect to time. The variation of these voltages or

currents with respect to time are known as AC (alternating current )

wavefoms. By far the most important type of AC waveform is the

sinusoid as shown in Figure 2.2. Note that although the abbreviation AC

literally stands for alternating current, it is used generally to describe

both current and voltage waveforms.

7/30/2019 CT1 Unit 2

3/29


A sinusoidal waveform is continuously changing in direction from

positive to negative to positive and so on, ie the magnitude and polarity

(direction) both change with time. It consists of a basic pattern, repeated

every T seconds. We refer to a single and complete basic pattern as a

cycle of the waveform. The duration of the cycle is called the period,

denoted by the capital letter T. The sinusoid is therefore said to be

periodic.

Voltage

Current

V

I

time t

time t

Figure 2.1

Time variation of a DC

voltage or current

Figure 2.2

Sinusoidal Waveforms

The frequency, f, is defined as the number of complete cycles it goes

through every second. Frequency was formerly expressed in cycles per

second, but the unit hertz (Hz) is now universally used. Thus, when the

frequency of the mains electricity supply is said to be 50Hz, this means

that 50 complete cycles of the sinusoidal voltage waveform take place

every second. The relationship between frequency, f, in hertz and period,

T, in seconds, is given by the expression:

f = 1/T

Mathematically, we can represent the sinusoid as a function of time

using the equation

v t V ft V ft ( ) sin sin ,= =2 2 =

where V is the peak value of the voltage waveform andfis its frequency

in Hz.

.

voltageor

currentT

t

Sinusoidal Wave

0

+

-

7/30/2019 CT1 Unit 2

4/29


This is simply the sine function of an angle measured in radians and

where the angle varies with time in accordance with the frequency of the

voltage , ie = 2ft.

If this voltage is connected across a resistor R the current flowing in the

resistor will follow the same pattern of periodic variation as the emf, and

its value at any instant is expressed as follows:

i t I ft ( ) sin= 2 , where

IV

R= is the peak value of the current flowing in

the circuit.

It is important to note that this equation for instantaneous current applies

only in what is termed a pure resistive circuit. A pure resistive circuit is

one in which there is no capacitance or inductance, only resistance.

Another way of expressing how fast the sinusoidal waveform is changing

is in relation to its angular frequency in radians per second or Hz. Since

one complete cycle corresponds to an angle of 2 radians (360o), a

frequency of f Hz (cycles per second) corresponds to an angular

frequency of 2f radians per second. Angular frequency is denoted by

the symbol . Thus:

= 2f or f = /2

As f = 1/T, where T is the time for one period, we can write:

= 2/T or T = 2/

We now have two equivalent expressions for both the instantaneous

voltage and instantaneous current of an AC sinusoidal waveform:

v t V ft V t ( ) sin sin= =2

i t I ft I t ( ) sin sin= =2

where V and I are the peak voltage and peak current respectively, and

is the angular frequency in radians per second andfis the waveform

frequency in cycles per second (hertz).

Voltage and current in DC circuits are represented usually by the capital

letters, V and I. AC voltage and current are represented by the same

letters, but in lower case form, v and i. When instantaneous values of ACvoltage and current are discussed, the symbols v(t) and i(t) are most

commonly used. The peak value (amplitude) of an AC voltage or current

waveform is shown by the symbols

.

V or I

7/30/2019 CT1 Unit 2

5/29


An emf, e, is given by the expression e(t) = 20 sin (2000t) Volts

Determine:

(a) The amplitude of the waveform.

(b) The angular frequency of the waveform.

(c) The frequency of the waveform in Hz.

(d) The time for one complete period of the wave.

0.5 1 1.5

e (t)

+20V

-20V

T = 1ms

t (ms)

Figure 2.3

Graph of the emf waveform

The waveform is illustrated in Figure 2.3.

Comparing the equation e(t) = 20 sin (2000t) to the general equation:

t


we can deduce that:

(a) The amplitude of the waveform is 20V

(b) The angular frequency, = 2000 rad/s

(c) The frequency, f = /2

= 2000/2

= 1000Hz

(d) The periodic time, T = 1/f

= 1/1000 s

= 1ms

EXAMPLE

SOLUTION

SAQ 1In a domestic light bulb connected to the mains electricity

supply, the instantaneous current is zero twice in each cycle of

the current. Why doesnt the light go out during these times of

zero current?

7/30/2019 CT1 Unit 2

6/29


SAQ 2A sinusoidal current, i(t), has a maximum value of 10A and a

frequency of 100Hz. Calculate:

(a) The angular frequency in radians per second.

(b) The time for one complete cycle of the current.

(b) Write down an expression for the instantaneous current,

i(t), given that i(t) = 0 when t = 0 and sketch the waveform.

SAQ 3A sinusoidal voltage v(t) is described by the equation

v(t) = 100 sin (100t) Volts.

Calculate:

(a) The instantaneous value of the voltage after 6ms.

(b) The first time after t = 0 that the instantaneous voltage is

50V.

SUMMARY

1. A voltage or current which varies with some particular pattern is

called a waveform. Waveforms which vary in magnitude and which

change from positive to negative and vice versa are known asalternating or AC waveforms.

2. A periodic waveform is composed of the repetition of a single basic

waveform pattern.

3. The following parameters may be defined for any periodic waveform:

amplitude - the maximum value of the waveform

cycle - a complete basic wave pattern

period - the time T for one cycle

frequency - the number of cycles in one second

The unit of frequency is the hertz (Hz), where f = 1/T and T is theperiodic time.

4. The instantaneous current and voltage of an AC sinusoidal

waveform are given by the expressions:


i t I ft I t ( ) sin sin= =2

5. Two sinusoidal waveforms are said to be in phase when they have

their maximum and minimum values at the same instant of time;otherwise, they are described as out of phase.

7/30/2019 CT1 Unit 2

7/29


ANSWERS TO SAQSSAQ 1

One of the reasons that the frequency of the mains electricity

was chosen at 50 Hz throughout Europe was that this frequency

was high enough to stop electric lights from flickering. In

other words, the current does not stay at zero long enough toallow the light to diminish.

Other electric devices for which the frequency of the mains

supply is important are those driven by synchronous motors,

such as record players and clocks. 50Hz is also convenient for

them. In the United States, the mains frequency is slightly

higher at 60 Hz.

SAQ 2

Peak value or amplitude of current, I =10Afrequency, f = 100Hz

(a) angular frequency, = 2f

= 2 100

= 628 rads/s

(b) periodic time, T = 1/f

= 1/100

= 0.01s

= 10ms

(c) instantaneous current, i(t) = I sin ti(t) = 10 sin (200t)

SAQ 3

(a) At t = 6ms = 6 10-3 s and measuring angles in radians,

v(t) = 100 sin (100 6 10-3)

= 100 sin (1.88)

= 100 0.951

= 95.1V

(b) Let t1

= time at which v(t) is 50V

50 = 100 sin (100t1)

1

2100 1= sin( )t

sin

=1 11

2100t

0.52 = 100t1

giving t1

= 0.52/100

= 1.67 ms.

7/30/2019 CT1 Unit 2

8/29


POWER DISSIPATION

AND RMS VALUES

OBJECTIVES



. average power

. effective value of AC current or voltage

. pure resistance

. rms voltage and current

2. Calculate rms and peak values of sinusoidal AC waveforms, given

the appropriate data.

AVERAGE POWERDISSIPATION IN AN AC

CIRCUIT

If a DC current, I, flows in a pure resistance, R, and is constant with

respect to time, as shown in Figure 2.4 then the power dissipation, P, isgiven by the expression P = I2R

CurrentI

I2R)

Power

(

t

t

(a) Current variation through a resistance R in a DC circuit

(b) Power dissipation in a resistance R in a DC circuit

Figure 2.4

Power dissipation in a DC

circuit

However, consider a sinusoidal AC current, i(t), flowing through a pure

resistance, R. The resulting instantaneous power dissipation, P(t), is

given by the expression:

P(t) = i2(t)R

In the case of an AC current, the instantaneous power, P(t), is clearly not

constant with respect to time. Figure 2.5 shows how both the instantaneous

current, i(t), and the square of the instantaneous current i2(t), continually

vary with respect to time.

Figure 2.5

Power dissipation in an AC

circuit

Current

I

I

I

2

21/2

i (t)2

i2

i (t)

Time, t

^

^

^

2

3

7/30/2019 CT1 Unit 2

9/29


Note that squaring the instantaneous current, i(t), has the effect of

removing the negative sign and making the result permanently positive;

that is, the graph of the square of the instantaneous current, i2(t), never

passes beneath the horizontal axis. Thus, the instantaneous power

dissipation, P(t), is always positive, regardless of whether the

instantaneous current, i(t), isflowing in the positive or negative direction.

In this situation, it is more convenient to deal with average power

dissipation, Pav , rather than with the constantly changing instantaneous

power, P(t). The average power dissipation is defined as the average

value of the instantaneous power dissipation over one complete period.

Pav = average value of P(t) over one period, T = 2/

To find Pav we perform the following, four-step operation:

1. We take the various positive and negative values of the instantaneous

current, i(t), over one complete cycle.

2. We square all these values, making them all positive.

3. We then calculate the average of the squared values, i2 .

4. Finally, we multiply the average of the squared values over one

complete cycle by the value of the resistance, R, thus finding the

average power dissipated in that resistance.

Therefore Pav = i2 R where i

2 is the average of the squares of the

instantaneous current values over one complete period.

For a sinusoidal AC current, such as that in Figure 2.5, we can draw astraight, horizontal line which is proportional to the constant value of the

average power, given by the expression

Pav = i2 R

It can be shown that: the average value of the square of the instantaneous

current is equal to half the square of the peak value of that current:

i I2 21

2=

Thus, the average power dissipation, Pav, can be written as:

P iI

Rav = R2

2

2=

The average power dissipation of a pure resistance in an AC circuit is

therefore equal to half the square of the peak current multiplied by the

resistance.

7/30/2019 CT1 Unit 2

10/29


ROOT-MEAN SQUARE

(RMS) VALUES OF AC

CURRENTS AND

VOLTAGES

We now have an expression for the average power dissipation for a

sinusoidal AC current flowing in a pure resistance. Is this expression

related to the average current? Reference to Figure 2.5 reveals that the

answer to this question clearly is no. The area enclosed by the i(t) graph

above the t-axis is equal to the area under the t-axis enclosed by the

negative part of the i(t) graph. The average value of i(t) is therefore zero.

Instead of average values for AC currents (or voltages), we use so-calledeffective values, where:

An effective value of an AC current or voltage is that value which, if in

DCform, would produce the same average power dissipation.

The equation for average power dissipation in a purely resistive AC

circuit contained the expression for the average value of the square of the

instantaneous current, i 2 . If we find the square root of this, we then have

the effective current:

effective current =i I I

2 2

2 0 707= =

/ .

Note how we obtained this effective value: we found the square root of

the average of the square. Another word for average is mean. The

effective value of an AC current is therefore the square root of the mean

of the square; or, as it is commonly known, the root-mean square(rms)

value.

We can follow the same reasoning and derive a similar expression for the

rms value of an AC voltage. These expressions for the rms values of AC

current and voltage are important ones and you should memorise them:

i I

v V==

0 707

0 707

.

.

RMS values are almost universally used to specify AC voltages and

currents. For example, the voltage of the mains supply is quoted as 240V.

By this is meant that the rms value of the voltage is 240V. Since the AC

mains voltage is sinusoidal, is peak value is:

240 2 339 = V

SAQ 4 Calculate the rms values of a sinusoidal voltage with a peakvalue 10V.

SAQ 5Calculate the peak values of a sinusoidal voltage waveform

with an rms value of 20V.

7/30/2019 CT1 Unit 2

11/29


SUMMARY

1. When an AC current, i(t), is applied across apure resistance, R, the

power dissipation at any instant, P(t), continually varies as the

instantaneous current varies. Mathematically:

P(t) = i2(t)R

2. Theaverage power dissipation of an AC current in a pure resistanceis equal to half the square of the maximum value of that current,

multiplied by the resistance:

P I RAV =1

2

2

3. The effective value orroot-mean square (rms) value of a sinusoidal

AC current or voltage is that value, which if in DC form, would

produce the same average power dissipation.

4. The rms values of a sinusoidal AC current and voltageare related to

its peak amplitude values by the expressions:

i I

v V

=

=

0 707

0 707

.

.

7/30/2019 CT1 Unit 2

12/29



For a sinusoidal voltage, the rms value, v , is 0.707 times the

peak value. That is:

v =/v 2

v = 0.707 V= 0.707 10

= 707V

SAQ 5

For a sinusoidal waveform:

.V v v= =2 1 14

= 1.414 20

= 28.3V

7/30/2019 CT1 Unit 2

13/29


COMPONENT

BEHAVIOUR IN AC

CIRCUITS

OBJECTIVES



. capacitive reactance

. impedance

. inductive reactance

. reactance

. phasor diagram

2. Calculate the rms values of current, voltage and average power in

simple resistive AC circuits.

3. Describe and calculate the response of a pure resistance to an AC

voltage.

4. Describe and calculate the response of a pure capacitance to an AC

voltage.

5. Describe and calculate the response of a pure inductance to an AC

voltage.

RESISTANCE IN AC

CIRCUITS

When an AC voltage, v, is applied across a resistance R, as shown in

Figure 2.6, an alternating current, i, flows.

Figure 2.6AC current and voltage

relationship in a pure

resistance

If the applied voltage is sinusoidal, then the voltage at any instant, v(t),

is given by the expression:

v t V t ( ) sin=

voltageor

current V

I

v

i t

(a) graphical relationship of voltage and current

(b) phasor relationship of voltage and current

^

^

i v

7/30/2019 CT1 Unit 2

14/29


From Ohms law, the instantaneous current will be:

i tv t

R

V

Rt I t( )

( ) sin sin= =

=

where

IV

R=

Therefore the current and voltage are in phase as shown in Figure 2.6(a).

Generally phase relationshipscan be conveniently indicated on whats

called a phasor diagram In a phasor diagram, each sinusoidal waveform

is represented by a single line called a phasor. The length of the phasor

is proportional to the amplitude of the wave; the greater the amplitude,

the longer the line. All lines are drawn from a single point, which may

be imagined to be the centre of a clock face. The phase difference in

radians between each waveform is represented by a corresponding angle

between the lines. Leading phase angles are drawn in an anti-clockwise

sense from the reference phasor, while lagging phase angles are drawn

in a clockwise direction. In this case the voltage and current are in phase

and therefore the phase angle between the phasors is zero as shown in

Figure 2.6(b).

For one half-cycle, the current is flowing in one direction through the

resistor and for the next half-cycle, it is flowing in the opposite direction

and so on. Furthermore, at two instants in time during each cycle, the

current is zero.

Multiplying both sides of the equation V = IR by 0.707 converts the

peak values of voltage and current to their rms values:

which is the same basic form of Ohms law as used in DC circuits. Thus,

if rms values of voltage and current are used, a resistor behaves the same

for AC as it does for DC. The techniques already encountered in our

study of DC circuits - Ohms law, Kirchoffs laws - are equally

applicable to AC resistor networks.

From the definition of the root-mean square (rms) value of AC waveforms,

it follows that the average power, Pav , dissipated in a resistance, R, is

given by the expression:

Pav = v i

= i2R

= v 2/R

where v is the rms value of the potential difference across the resistor

and i is the rms value of the current through it. So: power calculations

in AC resistive circuits are exactly the same as in DC circuits, when the

rms values of voltages and currents are used. For this reason we will drop

the bar over the symbols for voltage v and current i , and proceed on the

basis that when we use these lower case symbols in ac circuits, we are

As we shall see, AC circuits which contain capacitors or inductorsbehave quite differently assuming that they are rms values.

0 707 0 707. . V IR

v i R

==

7/30/2019 CT1 Unit 2

15/29


In conclusion: in AC resistive circuits, the current and voltage are in

phase. If rms values of current and voltage are used, Ohms and

Kirchoffs laws can be applied, and the calculations of power dissipation

are the same as those in DC circuits.

EXAMPLE For the circuit shown in Figure 2.7, calculate:

(a) The rms current supplied by the generator.

(b) The rms voltage across the 2.2k resistor.

(c) The average power dissipated in the 6.8k resistor.

18Vrms1kHz

V

R2

=2.2k

R1

=6.8k

Figure 2.7

AC circuit with two

resistances

SOLUTION (a) The total circuit resistance, RT = 6.8 + 2.2 = 9k

The rms current, i = v/R

= 18/(9 103

)= 2mA

(b) Let v2 = rms voltage across R2 = 2.2k

v2 = iR2= 2 10- 3 2.2 103

= 4.4V

(c) Let P = average power dissipation in R1 = 6.8k

P = i2Ri= (2 10-3)2 6.8 103

= 27.2 10-3

= 27.2mW

SAQ 6In the circuit shown in Figure 2.8, calculate:

(a) The rms current supplied by the generator.

(b) The average power dissipated in the 5.6 k resistor.

(c) The frequency of the AC current in Hz through the 2.7 k

resistor.

7/30/2019 CT1 Unit 2

16/29


Figure 2.8

Circuit for SAQ 6

3.3k

v(t) = 30 sin (400t) volts

2.7k

5.6k

The effect of placing a capacitor in a DC circuit is to stop the flow of

electric current. Electrons cannot travel across the gap between the

capacitor plates and so the continuous conducting path required for a

current to flow is interrupted. For example, if a capacitor is connected in

series with an electric light bulb and a DC power supply, the bulb will

not illuminate as the circuit is open at the capacitor.

If we remove the DC power supply, however, and replace it with an AC

one, then the electric bulb does light up! If a variable capacitor is used,

it can also be seen that the brightness of the bulb is directly proportional

to the value of the capacitance and to the waveform frequency of the

applied voltage. In this section, we shall examine in some detail the

behaviour of a capacitance in an AC circuit.

In our study of capacitance in Unit 1, we learnt that a capacitor of

capacitance C, charged to a DC voltage V, stores an amount of charge,

Q, given by the expression:

Q = CV

If an alternating voltage v is applied across a pure capacitance C, as

shown in Figure 2.9, the capacitor will be continually charged and

discharged.

CAPACITANCE IN AC

CIRCUITS

Figure 2.9

AC voltage applied across a

capacitor C

X

Y

CV sin t

i

^

vc

7/30/2019 CT1 Unit 2

17/29


During the positive half-cycle of the voltage waveform, plate X of the

capacitor becomes positively charged and plate Y negatively charged.

During the negative half-cycle, X receives a negative charge and Y a

positive one. There is therefore an alternating flow of charge or alternating

current, i, through the capacitor Thus, unlike the case for a dc voltage

where the capacitor eventually charges up to the value of the dc supply

voltage and no more current flows, a capacitor continually conducts anac current. Thus we may say that a capacitor passes ac and blocks dc.

It may be shown that the alternating current is also sinusoidal but leads

the supply voltage by a phase angle of/2 radians (90o). The expression

for the current i(t) maybe shown to be

i t I t I t I CV ( ) sin( / ) cos , = + = = 2 where

Thus the graph of the instantaneous voltage across the capacitor is a sine

function and the graph of the instaneous current through it is a cosine

function.

Figure 2.10 shows the plots of v and i as functions of time, illustrating

the current/voltage relationship in a capacitor.

voltage

or

current

iCvC

tFigure 2.10

AC current and voltagerelationship in a pure

capacitance

The equations for instantaneous current and voltage are:

v t V t i(t) CV t ( ) sin sin( / )= = + and 2

These equations, together with their corresponding graphs, indicate very

clearly that the instantaneous voltage and current in a capacitor are out

of phase by a quarter of a cycle - /2 radians or 90o - with the current

leading the voltage. Thus, the phasor diagram is as shown in Figure 2.11.

The peak value of the current , I, is related to the peak value of the

voltage, V , by the expression.

I CV

V

I C

=

=

1

7/30/2019 CT1 Unit 2

18/29


Figure 2.11

Phasor diagram showing therelationship between the

instantaneous voltage and

current in a capacitor

90

VC

iC

Converting to rms values gives:

i Cv

v

i C fC

=

= =

1 1

2

where f is the waveform frequency in hertz.

Note that this expression is in the form of Ohms law, V/I = R. The

quantity 1/C or 1/2fC is the factor resisting the flow of current

through the capacitor and is called capacitive reactance. The term

resistance is not used to denote opposition in this case, for the following

reason: power is dissipated in a resistance, but not in a capacitor.

However, the units of capacitive reactance are ohms.

The symbol for capacitive reactance is XC, where:

X V I v ic c c c c= = / /= 1/C

= 1/2fC ohms

Note that XC is inversely proportional to both the capacitance, C, and the

waveform frequency, f. Hence, if either (or both) f or C increase, the

capacitive reactance decreases and vice versa. In other words, the greater

the waveform frequency or the capacitance, the smaller the reactance

and the larger will be the current flowing across the capacitor. The

relationship between capacitive reactance and waveform frequency is

shown in Figure 2.12.

Figure 2.12

Variation of capacitive

reactance with waveform

frequency

capacitivereactance

X c

X =c

frequency f(Hz)

1

2fC

7/30/2019 CT1 Unit 2

19/29


EXAMPLE A 1F capacitor has an rms current of 2mA flowing through it at afrequency of 1000Hz. Calculate the rms voltage across the capacitor.

SOLUTION Capacitive reactance, XC = 1/2fC= 1/(2 1000 10-6)

= 159

rms voltage across capacitor, v= iXc= 2 10-3 159

= 0.32V

= 320mV

SAQ 7Why do capacitive reactances become small in high-frequency

circuits, such as those in FM radios or TV sets?

SAQ 8A sinusoidal source of emf of 10V rms and frequency 2kHz is

applied across a 0.1F capacitor. Calculate the rms value of

the current flowing through the capacitor.

In conclusion we can state the following points concerning the behaviour

of a capacitance in an AC circuit:

If a sinusoidal waveform is applied across a capacitor, the current will

leadthe voltage by 90o. Capacitive reactance is the opposition which

every capacitor presents to the flow of current. Its units are ohms.

INDUCTANCE IN AC

CIRCUITS

A piece of wire wound in the form of a coil possesses an electrical

property known as inductance. The property arises from the observable

phenomenon that if a current flowing through the coil changes for some

reason then an emfe is induced in the coil which tries to oppose the

current change. The magnitude of the induced emf is proportional to the

rate of change of current. The constant of proportionality is known as the

inductance of the coil L and is measured in a unit called the henry.

Mathematically we can summarise this with the equation

e Ldi

dt=

Where di/dt denotes the charge of current with respect to time. Considera sinusoidal voltage, v, applied across a lossless (zero resistance)

inductor of inductance L, as shown in Figure 2.13.

Figure 2.13

Instantaneous voltage and

current relationship in a pure

inductance

voltage

or

current

vL

vL

iLiL

(t)(t)

7/30/2019 CT1 Unit 2

20/29


This results in a sinusoidal current flowing through the inductor, i , which

is taken as the reference waveform with equation

i t I t ( ) sin=

It may be shown that the voltage across the inductor is given by

Figure 2.13 plots v and i as functions of time. Again, the equations for

v and i indicate that the instantaneous voltage and current in an inductor

are out of phase by an angle of 90o (/2 radians), with the voltage leading

the current. The phasor diagram will be as shown in Figure 2.14.

90

vL

iL

Figure 2.14Phasor diagram showing the

relationship between the

instantaneous voltage and

current in an inductor

Since

=

/ /

V LI

V I v i L

fL

=

= =

2

where f is the waveform frequency in hertz.

The quantity L = 2fL is the factor resisting the flow of current and is

called the inductive reactance, XL , measured in ohms.

XL = v/i

= L

= 2fL

XL is directly proportional to frequency; the higher the frequency, the

greater the inductive reactance. See Figure 2.15.

inductivereactance

XLXL = 2 fL

frequency (Hz)

Figure 2.15Variation of inductive

reactance with waveform

frequency

v t V t V t LI ( ) cos sin( / ), = = + 2 where V =

7/30/2019 CT1 Unit 2

21/29


Like capacitive reactance, XC, inductive reactance, XL, cannot be added

directly to resistance in an electric circuit. No power is dissipated in a

pure inductance; like a pure capacitance, it presents no resistance to the

flow of current. Energy is repeatedly stored in the magnetic field

surrounding the inductor and released back to the supply. A device that

behaves in this way is called a reactor. Lossless capacitors and inductors

are pure reactors. Practical capacitors and inductors are never quitelossless and contain some resistance, which dissipates some of the

supplied power.

SAQ 9Manufacturers of resistors, those circuit components in circuits

which offer specified resistances to the flow of current,

always show the resistance value of their component by

means of colour-coded bands on the outer surface of each

resistor. In this way, the user can readily distinguish between

- say - a 30k resistor and a 2 one. Why dont capacitor and

inductor manufacturers similarly display the reactance oftheir components?

SAQ 10When an AC voltage is applied across a pure inductance of

0.1H, a sinusoidal current, i = 300 sin (100t) mA, flows.

Calculate:

(a) the rms current

(b) the rms value of the applied voltage

(c) the peak value of the supply voltage

Write a mathematical expression for the instantaneous value

of the supply voltage.

PHASE DIFFERENCE

BETWEEN SINUSOIDAL

WAVEFORMS

In circuits containing resistors and capacitors and/or inductors the phase

difference between the voltages and currents may vary in a general way

and do not necessarily have a phase difference of just 90, or one quarter

of a cycle, (lag or lead), as is the case for a pure inductance or

capacitance. Consider the two sinusoidal waveforms, v1and v2, sketched

in Figure 2.16(a).

7/30/2019 CT1 Unit 2

22/29


Figure 2.16

Illustration of phase difference

(a) in phase

(b) out of phase

2

2

t

=

3 =

2 3

=

t

2 t

3 = t3

td

v2

v1

v2v1

We can say that v1 and v2 are in phase because v1 has its maximum and

minimum values at exactly the same instants of time as v2. For the

waveform in Figure 2.16 (b), however, this is not the case. The waveforms

are displaced relative to each other and are said to be out of phase. In

other words, there is a phase difference between v1 and v2.

The amount of phase difference is expressed in terms of the phase angle

, measured in radians on the t-axis. If we define v1 as the reference

waveform, then v2

lags behind v1

(since v2

reaches it maximum value

after v1) by a phase angle . Thus:

A phase lag on the-axis corresponds to a time lag td on the t-axis. Both

voltages are sinusoidal waveforms with v1 = 0 when t = 0 and v2 = 0 when

t = td. That is:

v2 (td ) = 0

td - = 0

and td =

thus td = /

We can therefore conclude that: the time delay between two out-of-phase

sinusoidalwaveforms is equal to the phase angle between them divided

by their angular frequency.

v t V t

v t V t

1 1

2 2

( ) sin

( ) sin( )

=

=

v t V t t d2 2( ) sin( )=

7/30/2019 CT1 Unit 2

23/29


EXAMPLE A voltage v1 in an electric circuit is sinusoidal, and has a peak value of4V and a frequency of 200Hz. Another voltage, v2, has the same

frequency as v1 and a peak value of 2V. Also it reaches its peak 1ms

before v1 does.

(a) Determine the phase angle between v1 and v2 .

(b) Write mathematical expressions for v1 and v2 using v1 as the

reference waveform.

SOLUTION (a) Time lead of v2 over v1 = 1ms

angular frequency of v2 and v1, = 2f

= 2 200

= 400

phase lead of v2

over v1, = td= 400 10-3

= 0.4 radians

(b) If v1 is the reference voltage, then:

v t V t 1 1( ) sin=

where V V1 4=

and = 400 rads/s

thus v1(t) = 4 sin (400t)V

Since v2 leads v1 by a phase angle:

v t V t 2 2( ) sin( )= +

where V V2 2=

. = 400 rads/s

= 0.4 rads

v2(t) = 2 sin (400t + 0.4)V

= 2 sin (400t + 0.4)V

SAQ 11A current in an AC circuit, i1, is sinusoidal and has a peak

value of 100mA and a frequency of 500Hz. Another current in

the same circuit, i2 , has the same frequency but is twice as

large as i1 and lags it by a phase angle of/4 radians (45).

(a) Write the mathematical expressions for i1 and i2.

(b) Calculate the time interval between a peak in the waveform

of i1 and a peak in the waveform of i2.

(c) Sketch roughly i1 and i2 on the same axis.

7/30/2019 CT1 Unit 2

24/29


SUMMARY

1. In a purely resistive AC circuit, the current and voltage are in phase

with one another. The rms values of voltage and current are related

to resistance by Ohms law:

v = iR i = v/R R = v/i

2. The average power dissipation in a purely resistive circuit is givenby the expressions:

P = vi or P = i2R or P = v2/R

where v and i are rms values.

3. In a purely capacitive AC circuit, the current leads the voltage by

90o or /2 radians.

4. Capacitive reactance, XC , measured in ohms, is the opposition

that a pure capacitance, C, presents to the flow of AC current. It is

inversely proportional to the angular or waveform frequency of theapplied voltage.

XC = 1/C or XC = 1/2fC

5. The voltage and current in a purely capacitive circuit are related by

the expression:

v = iXC

where v and i are rms. values.

6. In a purely inductive AC circuit, the current lags the voltage by 90o

or /2 radians.

7. Inductive reactance, XL, measured in ohms, is the opposition that

a pure inductance, L, presents to the flow of AC current. It is directly

proportional to the angular or waveform frequency of the applied

voltage.

XL = L or XL = 2fL

8. The voltage and current in a purely inductive circuit are related by

the expression:

v = iXL

where v and i are rms values.

9. Capacitive and inductive reactances cannot be added directly to

resistance, as a component which offers reactance does not dissipate

power.

10. The phase difference between two out-of-phase sinusoidal

waveforms is expressed in terms of the phase angle between them.

The phase angle is equal to the time difference, td, between the two

waveforms multiplied by their angular frequency:

= t d

7/30/2019 CT1 Unit 2

25/29


11. When comparing two out-of-phase waveforms, one is described as

the reference waveform and the other waveform as lagging or

leading the reference one.

7/30/2019 CT1 Unit 2

26/29



(a) The equivalent resistance, Rp, of 5.6k and 2.7k in

parallel is:

Rp = (5.6 2.7)/(5.6 + 2.7)= 1.82k

Total circuit resistance, RT = 3.3 + 1.82

= 5.12k

RMS voltage of generator, C = 30 0.707

= 21.2V

RMS current of generator, i = v/R

= 21.2/(5.12 103)

= 4.14 10-3

= 4.14mA

(b) Let v1 be the rms voltage across R1, the 3.3k resistor

and let v2 be the rms voltage across Rp, the parallel

combination of the 5.6k and the 2.7k resistor.

v1 = iR1= 4.14 10-3 3.3 103

= 13.7V

v2 = v - v1= 21.2 - 13.7

= 7.5

The average power dissipation in the 5.6k resistor is

given by:

= (7.5)2 /(5.6 103)

= 0.01W

= 10mW

(c) The frequency of the AC current through the 2.7k

resistor is the same as the frequency of the supply

voltage.

angular frequency, = 400twaveform frequency, f = /2

= 400/2

= 200Hz

7/30/2019 CT1 Unit 2

27/29


SAQ 7

Because capacitive reactance decreases as the frequency

across the capacitor increases; the higher the frequency, the

lower the reactance.

SAQ 8

Capacitive reactance, XC = 1/2fC

= 1/(2 2000 0.1 10 -6)

= 796

rms current through capacitor, i = v/Xc= 10/796

= 0.0126A

= 12.6mA

SAQ 9

The resistive value of a resistor remains constant, regardless

of the frequency of the AC current passing through it. However,

the reactances of capacitors and inductors vary according to

the AC frequency applied across them. As a result,

manufacturers of capacitors and inductors cannot possibly

predict their reactances in every circumstance; that depends

on how the components are used.

7/30/2019 CT1 Unit 2

28/29


SAQ 10

(a) The rms current = 0.707 I

= 0.707 300

= 212mA

(b) Frequency of supply, f = /2

= 100 /2

= 50Hz

inductive reactance, XL = 2fL

= 2 50 0.1

= 31.4

rms value of supply voltage, v= iXL= 212 10-3 31.4

= 6.7V

(c) Peak value of supply voltage,

V v=2

= 6.7 1.414

= 6.7V

The voltage across a pure inductance leads the current

through it by /2 radians or by 90o.

v(t) = 9.5 sin (100t + /2) volts

7/30/2019 CT1 Unit 2

29/29

SAQ 11

(a) i t I t I ft 1 1 1 2( ) sin sin= =

= 100 sin (2 500 t)= 100 sin (1000t) mA

If i2 is twice as large as i1 , then:

I I2 12== 2 100

= 200mA

i2 lags i1 by a phase angle of = /4 radians (45o).

Therefore i2 is of the form:

i2 = 200 sin (1000t - /4) mA

(b) Time delay between waveforms, td =/

= (/4)/1000

= 2.5 10-4

= 0.25ms

(c) Your sketch of i 1 and i 2 should look like that shown in

Figure 2.17.

1 2 3

200

100

0

100

0.25ms

t (ms)

current(mA)

200

-

-

i2

ii

Figure 2.17

Waveforms in SAQ 11

ct1 unit 2

Documents