2
AC ELECTRICITY
AIMS
INTRODUCTION
The aims of this unit are to describe and quantify the behaviour of
resistors, inductors and capacitors in AC circuits.
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 Ohm’s and Kirchoff’s 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 the
product 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 20Ω, 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!
Unit 2: AC Electricity
2-1
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-voltages
and low currents. But could we not transmit the high-voltage, lowcurrent 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 (100400kV) AC, which transformer stations near the point of usage stepdown 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 in
our electric and electronic world.
SINUSOIDAL
WAVEFORMS
After studying this sub-unit, you should be able to:
OBJECTIVES
1.
Define or explain the meaning of the following terms:
.
.
.
.
.
.
.
.
.
.
.
.
2.
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
Write the mathematical equation for the instantaneous value of a
voltage 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.
2-2
Communications Technology 1
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
V
time t
Current
I
Figure 2.1
Time variation of a DC
voltage or current
time t
voltage
or
current
+
T
0
t
Figure 2.2
Sinusoidal Waveforms
Sinusoidal Wave
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 ) = Vˆ sin 2πft = Vˆ sin θ ,
θ = 2πft
where V̂ is the peak value of the voltage waveform and f is its frequency
in Hz.
.
Unit 2: AC Electricity
2-3
This is simply the sine function of an angle θ measured in radians and
where the angle varies with time in accordance with the frequency of the
voltage ω, ie θ = 2πft.
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:
Vˆ
i(t ) = Iˆ sin 2πft , where Iˆ = is the peak value of the current flowing in
R
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
o
one complete cycle corresponds to an angle of 2π radians (360 ), a
frequency of f Hz (cycles per second) corresponds to an angular
frequency of 2πf radians per second. Angular frequency is denoted by
the symbol ω. Thus:
ω = 2πf or f = ω/2π
As f = 1/T, where T is the time for one period, we can write:
ω = 2π/T or T = 2π/ω
We now have two equivalent expressions for both the instantaneous
voltage and instantaneous current of an AC sinusoidal waveform:
v(t ) = Vˆ sin 2πft = Vˆ sin ωt
i(t ) = Iˆ sin 2πft = Iˆ sin ωt
where V̂ and Î are the peak voltage and peak current respectively, and
ω is the angular frequency in radians per second and f is 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 AC
voltage 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
.
V̂ or Î
2-4
Communications Technology 1
EXAMPLE
An emf, e, is given by the expression e(t) = 20 sin (2000πt) 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.
e (t)
+20V
0.5
Figure 2.3
Graph of the emf waveform
SOLUTION
1
1.5
t (ms)
-20V
T = 1ms
The waveform is illustrated in Figure 2.3.
Comparing the equation e(t) = 20 sin (2000πt) to the general equation:
t
v(t ) = Vˆ sin 2πft = Vˆ sin ωt
we can deduce that:
(a) The amplitude of the waveform is 20V
(b) The angular frequency, ω = 2000π rad/s
(c) The frequency, f = ω/2π
= 2000π/2π
= 1000Hz
(d) The periodic time, T = 1/f
= 1/1000 s
= 1ms
SAQ 1
Unit 2: AC Electricity
In a domestic light bulb connected to the mains electricity
supply, the instantaneous current is zero twice in each cycle of
the current. Why doesn’t the light go out during these times of
zero current?
2-5
SAQ 2
A 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 3
A sinusoidal voltage v(t) is described by the equation
v(t) = 100 sin (100πt) Volts.
Calculate:
(a) The instantaneous value of the voltage after 6ms.
(b) The first time after t = 0 that the instantaneous voltage is
50V.
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 as
alternating 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:
SUMMARY
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 the
periodic time.
4.
The instantaneous current and voltage of an AC sinusoidal
waveform are given by the expressions:
v(t ) = Vˆ sin 2πft = Vˆ sin ωt
i(t ) = Iˆ sin 2πft = Iˆ sin ωt
5.
2-6
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.
Communications Technology 1
SAQ 1
ANSWERS TO SAQS
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 to
allow 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 =10A
frequency, f = 100Hz
(a) angular frequency, ω = 2πf
= 2π × 100
= 628 rads/s
(b) periodic time, T = 1/f
= 1/100
= 0.01s
= 10ms
(c) instantaneous current,
i(t) = Î sin ωt
i(t) = 10 sin (200πt)
SAQ 3
(a) At t = 6ms = 6 × 10-3 s and measuring angles in radians,
v(t)
= 100 sin (100π × 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 (100πt1)
1
= sin(100πt1 )
2
1
sin −1   = 100πt1
 2
0.52 = 100πt1
giving
Unit 2: AC Electricity
t1 = 0.52/100π
= 1.67 ms.
2-7
POWER DISSIPATION
AND RMS VALUES
After studying this sub-unit, you should be able to:
OBJECTIVES
1.
Define or explain the meaning of the following terms:
.
.
.
.
2.
AVERAGE POWER
DISSIPATION IN AN AC
CIRCUIT
average power
effective value of AC current or voltage
pure resistance
rms voltage and current
Calculate rms and peak values of sinusoidal AC waveforms, given
the appropriate data.
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, is
given by the expression P = I2R
Current
I
t
(a) Current variation through a resistance R in a DC circuit
Power
(I2R)
Figure 2.4
Power dissipation in a DC
circuit
t
(b) Power dissipation in a resistance R 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.
Current
^2
I
i 2 (t)
^2
1/2 I
Figure 2.5
Power dissipation in an AC
circuit
2-8
i2
i (t)
^
I
π
ω
2π
ω
3π
ω
Time, t
Communications Technology 1
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), is flowing 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 = 2π/ω
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, i 2 .
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 = i 2 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 a
straight, horizontal line which is proportional to the constant value of the
average power, given by the expression
Pav = i 2 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:
i2 =
1 ˆ2
I
2
Thus, the average power dissipation, Pav, can be written as:
Pav = i 2 R =
Iˆ 2
R
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.
Unit 2: AC Electricity
2-9
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-called
effective values, where:
An effective value of an AC current or voltage is that value which, if in
DC form, 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 2 = Iˆ 2 / 2 = 0.707 Iˆ
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 = 0.707 Iˆ
v = 0.707Vˆ
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 = 339V
SAQ 4
Calculate the rms values of a sinusoidal voltage with a peak
value 10V.
SAQ 5
Calculate the peak values of a sinusoidal voltage waveform
with an rms value of 20V.
2-10
Communications Technology 1
1.
SUMMARY
When an AC current, i(t), is applied across a pure resistance, R, the
power dissipation at any instant, P(t), continually varies as the
instantaneous current varies. Mathematically:
P(t) = i2(t)R
2.
The average power dissipation of an AC current in a pure resistance
is equal to half the square of the maximum value of that current,
multiplied by the resistance:
PAV =
1 ˆ2
I R
2
3.
The effective value or root-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 = 0.707 Iˆ
v = 0.707Vˆ
Unit 2: AC Electricity
2-11
SAQ 4
ANSWERS TO SAQS
For a sinusoidal voltage, the rms value,
peak value. That is:
v =
v =
=
=
v , is 0.707 times the
vˆ / 2
0.707 V̂
0.707 × 10
707V
SAQ 5
For a sinusoidal waveform:
Vˆ = 2 v = 1.14v
= 1.414 × 20
= 28.3V
2-12
Communications Technology 1
COMPONENT
BEHAVIOUR IN AC
CIRCUITS
After studying this sub-unit, you should be able to:
OBJECTIVES
1.
Define or explain the meaning of the following terms:
.
.
.
.
.
RESISTANCE IN AC
CIRCUITS
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.
When an AC voltage, v, is applied across a resistance R, as shown in
Figure 2.6, an alternating current, i, flows.
voltage
or
^
current V
^I
v
t
i
(a) graphical relationship of voltage and current
Figure 2.6
AC current and voltage
relationship in a pure
resistance
i
v
(b) phasor relationship of voltage and current
If the applied voltage is sinusoidal, then the voltage at any instant, v(t),
is given by the expression:
v(t ) = Vˆ sin ωt
Unit 2: AC Electricity
2-13
From Ohm’s law, the instantaneous current will be:
i (t ) =
v(t )  Vˆ 
=   sin ωt = Iˆ sin ωt
R
 R
Vˆ
where Iˆ =
R
Therefore the current and voltage are in phase as shown in Figure 2.6(a).
Generally phase relationshipscan be conveniently indicated on what’s
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 V̂ = ÎR by 0.707 converts the
peak values of voltage and current to their rms values:
ˆ
0.707Vˆ = 0.707IR
v = iR
which is the same basic form of Ohm’s 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 - Ohm’s law, Kirchoff’s 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 inductors
behave quite differently assuming that they are rms values.
2-14
Communications Technology 1
In conclusion: in AC resistive circuits, the current and voltage are in
phase. If rms values of current and voltage are used, Ohm’s and
Kirchoff’s 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:
V
Figure 2.7
AC circuit with two
resistances
SOLUTION
R1 = 6.8kΩ
18V
rms
1kHz
R2 = 2.2 kΩ
(a) The rms current supplied by the generator.
(b) The rms voltage across the 2.2kΩ resistor.
(c) The average power dissipated in the 6.8kΩ resistor.
(a) The total circuit resistance, RT = 6.8 + 2.2 = 9kΩ
The rms current, i = v/R
= 18/(9 × 103)
= 2mA
(b) Let v2 = rms voltage across R2 = 2.2kΩ
v2 = iR2
= 2 × 10- 3 × 2.2 × 103
= 4.4V
(c) Let P = average power dissipation in R1 = 6.8kΩ
P = i 2R i
= (2 × 10-3)2 × 6.8 × 103
= 27.2 × 10-3
= 27.2mW
SAQ 6
In 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 kΩ resistor.
(c) The frequency of the AC current in Hz through the 2.7 kΩ
resistor.
Unit 2: AC Electricity
2-15
2.7kΩ
3.3kΩ
5.6kΩ
Figure 2.8
Circuit for SAQ 6
CAPACITANCE IN AC
CIRCUITS
v(t) = 30 sin (400πt) volts
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.
i
Figure 2.9
AC voltage applied across a
capacitor C
2-16
^ sin ω t
V
X
C
vc
Y
Communications Technology 1
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 an
ac 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 π/2 radians (90o). The expression
for the current i(t) maybe shown to be
i(t ) = Iˆ sin(ωt + π / 2) = Iˆ cos ωt, where Iˆ = ωCVˆ
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.
iC
voltage
or
current
vC
Figure 2.10
AC current and voltage
relationship in a pure
capacitance
t
The equations for instantaneous current and voltage are:
v(t ) = Vˆ sin ωt and i(t) = ωCVˆ sin(ωt + π / 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 - π/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 , Î , is related to the peak value of the
voltage, V̂ , by the expression.
Iˆ = ωCVˆ
1
Vˆ
=
ˆI ωC
Unit 2: AC Electricity
2-17
iC
Figure 2.11
Phasor diagram showing the
relationship between the
instantaneous voltage and
current in a capacitor
90°
VC
Converting to rms values gives:
i = ωCv
v
1
1
=
=
i ωC 2πfC
where f is the waveform frequency in hertz.
Note that this expression is in the form of Ohm’s law, V/I = R. The
quantity 1/ωC or 1/2πfC 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:
Xc = Vˆc / Iˆc = vc / ic
= 1/ωC
= 1/2πfC 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.
capacitive
reactance
Xc
Figure 2.12
Variation of capacitive
reactance with waveform
frequency
2-18
Xc=
1
2 πfC
frequency f(Hz)
Communications Technology 1
EXAMPLE
A 1µF capacitor has an rms current of 2mA flowing through it at a
frequency of 1000Hz. Calculate the rms voltage across the capacitor.
SOLUTION
Capacitive reactance, XC = 1/2πfC
= 1/(2π × 1000 × 10-6)
= 159Ω
rms voltage across capacitor, v= iXc
= 2 × 10-3 × 159
= 0.32V
= 320mV
SAQ 7
Why do capacitive reactances become small in high-frequency
circuits, such as those in FM radios or TV sets?
SAQ 8
A sinusoidal source of emf of 10V rms and frequency 2kHz is
applied across a 0.1µF 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
lead the 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 emf e 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 = −L
di
dt
Where di/dt denotes the charge of current with respect to time. Consider
a 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
Unit 2: AC Electricity
vL (t)
iL
voltage
or
current
iL
(t)
vL
2-19
This results in a sinusoidal current flowing through the inductor, i , which
is taken as the reference waveform with equation
i(t ) = Iˆ sin ωt
It may be shown that the voltage across the inductor is given by
ˆ = ωLIˆ
v (t ) = Vˆ cos ωt = Vˆ sin(ωt + π / 2), where V
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 (π/2 radians), with the voltage leading
the current. The phasor diagram will be as shown in Figure 2.14.
vL
Figure 2.14
Phasor diagram showing the
relationship between the
instantaneous voltage and
current in an inductor
90°
iL
Since Vˆ = ωLIˆ
Vˆ / Iˆ = v / i = ωL
= 2πfL
where f is the waveform frequency in hertz.
The quantity ωL = 2πfL is the factor resisting the flow of current and is
called the inductive reactance, XL , measured in ohms.
XL = v/i
= ωL
= 2πfL
XL is directly proportional to frequency; the higher the frequency, the
greater the inductive reactance. See Figure 2.15.
inductive
reactance
Figure 2.15
Variation of inductive
reactance with waveform
frequency
2-20
XL
XL = 2 π fL
frequency (Hz)
Communications Technology 1
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 quite
lossless and contain some resistance, which dissipates some of the
supplied power.
SAQ 9
Manufacturers 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 30kΩ resistor and a 2Ω one. Why don’t capacitor and
inductor manufacturers similarly display the reactance of
their components?
SAQ 10
When an AC voltage is applied across a pure inductance of
0.1H, a sinusoidal current, i = 300 sin (100πt) 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
Unit 2: AC Electricity
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, v1 and v2, sketched
in Figure 2.16(a).
2-21
v2
v1
3π
3π
ω
2π
π
π
ω
2π
ω
θ = ωt
t =θ
ω
(a) in phase
v2
v1
π
π
ω
φ
Figure 2.16
Illustration of phase difference
td
2π
3π
2π
ω
3π
ω
θ = ωt
t= θ
ω
(b) out of phase
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
φ, measured in radians on the ωt-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 φ . Thus:
v1 (t ) = Vˆ1 sin ωt
v 2 (t ) = Vˆ2 sin(ωt − φ )
A phase lag φ on the θ-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:
v 2 (t ) = Vˆ2 sin(ωt − ωtd )
v2 (td ) = 0
and
thus
ωtd - φ = 0
ωtd = φ
td = φ/ω
We can therefore conclude that: the time delay between two out-of-phase
sinusoidal waveforms is equal to the phase angle between them divided
by their angular frequency.
2-22
Communications Technology 1
EXAMPLE
A voltage v1 in an electric circuit is sinusoidal, and has a peak value of
4V 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, ω = 2πf
= 2π × 200
= 400π
phase lead of v2 over v1, φ
= ωtd
= 400π × 10-3
= 0.4π radians
(b) If v1 is the reference voltage, then:
v1 (t ) = Vˆ1 sin ωt
where V̂1 = 4V
and ω = 400π rads/s
thus v1(t) = 4 sin (400πt)V
Since v2 leads v1 by a phase angle:
v2 (t ) = Vˆ2 sin(ωt + φ )
where V̂2 = 2V
.
ω = 400π rads/s
φ = 0.4π rads
v2(t) = 2 sin (400πt + 0.4π)V
= 2 sin π(400t + 0.4)V
SAQ 11
A 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 π/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.
Unit 2: AC Electricity
2-23
1.
SUMMARY
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 Ohm’s law:
v = iR i = v/R R = v/i
2.
The average power dissipation in a purely resistive circuit is given
by 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 π/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 the
applied voltage.
XC = 1/ωC or XC = 1/2πfC
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 π/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 = ωL or XL = 2πfL
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 φ between them.
The phase angle is equal to the time difference, td, between the two
waveforms multiplied by their angular frequency:
φ = ωt d
2-24
Communications Technology 1
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.
Unit 2: AC Electricity
2-25
SAQ 6
ANSWERS TO SAQS
(a) The equivalent resistance, Rp, of 5.6kΩ and 2.7kΩ in
parallel is:
Rp = (5.6× 2.7)/(5.6 + 2.7)
= 1.82kΩ
Total circuit resistance, RT = 3.3 + 1.82
= 5.12kΩ
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 3.3kΩ resistor
and let v2 be the rms voltage across Rp, the parallel
combination of the 5.6k Ω and the 2.7kΩ resistor.
v1 = iR1
= 4.14 × 10-3 × 3.3 × 103
= 13.7V
v 2 = v - v1
= 21.2 - 13.7
= 7.5
The average power dissipation in the 5.6kΩ resistor is
given by:
= (7.5)2 /(5.6 × 103)
= 0.01W
= 10mW
(c) The frequency of the AC current through the 2.7kΩ
resistor is the same as the frequency of the supply
voltage.
angular frequency, ω = 400πt
waveform frequency, f = ω/2π
= 400π/2π
= 200Hz
2-26
Communications Technology 1
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 = 1/2πfC
= 1/(2π × 2000 × 0.1× 10 -6)
= 796Ω
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.
Unit 2: AC Electricity
2-27
SAQ 10
(a) The rms current = 0.707 Î
= 0.707 × 300
= 212mA
(b) Frequency of supply, f = ω/2π
= 100π /2π
= 50Hz
inductive reactance, XL = 2πfL
= 2π × 50 × 0.1
= 31.4Ω
rms value of supply voltage, v = iXL
= 212 × 10-3 × 31.4
= 6.7V
(c) Peak value of supply voltage,
V̂ = v 2
= 6.7 × 1.414
= 6.7V
The voltage across a pure inductance leads the current
through it by π/2 radians or by 90o.
v(t) = 9.5 sin (100πt + π/2) volts
2-28
Communications Technology 1
SAQ 11
(a) i1 (t ) = Iˆ1 sin ωt = Iˆ1 sin 2πft
=
=
100 sin (2π × 500 × t)
100 sin (1000πt) mA
If i2 is twice as large as i1 , then:
Iˆ2 = 2 Iˆ1
= 2 × 100
= 200mA
i2 lags i1 by a phase angle of φ = π/4 radians (45o).
Therefore i2 is of the form:
i2 = 200 sin (1000πt - π/4) mA
(b) Time delay between waveforms, td = φ/ω
= (π/4)/1000π
= 2.5 × 10-4
= 0.25ms
(c) Your sketch of i 1 and i 2 should look like that shown in
Figure 2.17.
current
(mA)
i2
200
100
0
1
- 100
Unit 2: AC Electricity
3 t (ms)
ii
- 200
Figure 2.17
Waveforms in SAQ 11
2
0.25ms
2-29