Alternating voltages and currents

Alternating Voltages and Currents

Chapter 15

 Introduction
 Voltage and Current
 Reactance of Inductors and Capacitors
 Phasor Diagrams
 Impedance
 Complex Notation

Storey: Electrical & Electronic Systems © Pearson Education Limited 2004

OHT 15.1

Introduction

15.1

 From our earlier discussions we know that
v = Vp sin (ω t + φ )
where

Vp is the peak voltage
ω is the angular frequency

φ is the phase angle
 Since ω = 2πf it follows that the period T is given by
T =

1 2π
=
f
ω


OHT 15.2

 If φ is in radians, then a time delay t is given by φ /ω
as shown below


OHT 15.3

Voltage and Current

15.2

 Consider the voltages across a resistor, an inductor
and a capacitor, with a current of
i = IP sin(ωt )

 Resistors
– from Ohm’s law we know
v R = iR

– therefore if i = Ipsin(ωt)

v R = IP R sin(ωt )

OHT 15.4

Voltage and Current
 Inductors - in an inductor

15.2

di
vL = L
dt

– therefore if i = Ipsin(ωt)
d(I sin(ωt ))
vL = L P
= ωLIP cos(ωt )
dt
1
v C = ∫ idt
C

 Capacitors - in a capacitor
Ip
1
v if = I ∫ IP sin
– therefore C i= Cpsin(ωt)(ωt ) = − ωC cos(ωt )

OHT 15.5


OHT 15.6

Reactance of Inductors and Capacitors

15.3

 Let us ignore, for the moment the phase angle and
consider the magnitudes of the voltages and currents
 Let us compare the peak voltage and peak current
 Resistance
Peak value of voltage Peak value of (IP Rsin(ωt )) IP R
=
=
=R
Peak value of current
Peak value of (IP sin(ωt ))
IP


OHT 15.7

 Inductance
Peak value of voltage Peak value of (ωLIP cos(ωt )) ωLIP
=
=
= ωL
Peak value of (IP sin(ωt ))
IP

 Capacitance
Peak value of voltage
=

Peak value of ( −

Ip

cos(ωt ))

ωC
Peak value of (I p sin(ωt ))


Ip

1
= ωC =
Ip
ωC
OHT 15.8

 The ratio of voltage to current is a measure of how
the component opposes the flow of electricity
 In a resistor this is termed its resistance
 In inductors and capacitors it is termed its reactance
 Reactance is given the symbol X
 Therefore
Re ac tan ce of an inductor, X L = ωL
Reactance
1
Re ac tan ce of a capacitor, X C =
Reactance
ωC

OHT 15.9

 Since reactance represents the ratio of voltage to
current it has units of ohms
 The reactance of a component can be used in much
the same way as resistance:
– for an inductor
V = I XL

– for a capacitor
V = I XC

OHT 15.10

 Example – see Example 15.3 from course text
A sinusoidal voltage of 5 V peak and 100 Hz is applied across
an inductor of 25 mH. What will be the peak current?
At this frequency, the reactance of the inductor is given by

X L = ωL = 2πfL = 2 × π × 100 × 25 × 10 − 3 = 15.7 Ω
Therefore

VL
5
IL =
=
= 318 mA peak
X L 15.7

OHT 15.11

Phasor Diagrams

15.4

 Sinusoidal signals are characterised by their
magnitude, their frequency and their phase
 In many circuits the frequency is fixed (perhaps at
the frequency of the AC supply) and we are
interested in only magnitude and phase
 In such cases we often use phasor diagrams which
represent magnitude and phase within a single
diagram


OHT 15.12

 Examples of
phasor diagrams
(a) here L represents
the magnitude and φ
the phase of a
sinusoidal signal
(b) shows the voltages
across a resistor, an
inductor and a
capacitor for the same
sinusoidal current

OHT 15.13

 Phasor diagrams can be used to represent the
addition of signals. This gives both the magnitude
and phase of the resultant signal


OHT 15.14

 Phasor diagrams can also be used to show the
subtraction of signals


OHT 15.15

 Phasor analysis of an RL circuit

 See Example 15.5 in the text for a numerical example

OHT 15.16

 Phasor analysis of an RC circuit

 See Example 15.6 in the text for a numerical example

OHT 15.17

 Phasor analysis of an RLC circuit


OHT 15.18

 Phasor analysis
of parallel
circuits
in such circuits the
voltage across each
of the components is
the same and it is the
currents that are of
interest


OHT 15.19

Impedance

15.5

 In circuits containing only resistive elements the
current is related to the applied voltage by the
resistance of the arrangement
 In circuits containing reactive, as well as resistive
elements, the current is related to the applied voltage
by the impedance, Z of the arrangement
– this reflects not only the magnitude of the current but
also its phase
– impedance can be used in reactive circuits in a similar
manner to the way resistance is used in resistive circuits

OHT 15.20

 Consider the following circuit and its phasor diagram


OHT 15.21

 From the phasor diagram it is clear that that the
magnitude of the voltage across the arrangement V
is
2
2
V = VR + VL

= (IR )2 + (IX L )2
= I R 2 + XL2
= IZ

Z = R 2 + XL2

where
 Z is the magnitude of the impedance, so Z =|Z|

OHT 15.22

 From the phasor diagram the phase angle of the
impedance is given by
φ = tan −1

VL
IX
X
= tan-1 L = tan-1 L
VR
IR
R

 This circuit contains an inductor but a similar analysis
can be done for circuits containing capacitors
 In general
2

Z= R +X

2

and

φ = tan

−1


X
R
OHT 15.23

 A graphical representation of impedance


OHT 15.24

Complex Notation

15.6

 Phasor diagrams are similar to Argand Diagrams
used in complex mathematics
 We can also represent impedance using complex
notation where
 Resistors:

ZR

=

R

 Inductors:

ZL

=

jXL

=

 Capacitors:

ZC

=

-jXC

=

jω L

1
1
−j
=
ωC j ωC


OHT 15.25

 Graphical representation of complex impedance


OHT 15.26

 Series and parallel
combinations of
impedances
– impedances combine
in the same way as
resistors


OHT 15.27

 Manipulating complex impedances
– complex impedances can be added, subtracted,
multiplied and divided in the same way as other
complex quantities
– they can also be expressed in a range of forms such
as the rectangular, polar and exponential forms
– if you are unfamiliar with the manipulation of complex
quantities (or would like a little revision on this topic)
see Appendix D of the course text which gives a
tutorial on this subject

OHT 15.28

 Example – see Example 15.7 in the course text
Determine the complex
impedance of this circuit
at a frequency of 50 Hz.
At 50Hz, the angular frequency ω = 2πf = 2×π ×50 = 314 rad/s
Therefore
Z = ZC + Z R + Z L = R + j( X L − X C ) = R + j(ωL −
= 200 + j(314 × 400 × 10 −3 −
= 200 + j62 ohms

1
314 × 50 × 10

−6

1
)
ωC

)


OHT 15.29

 Using complex impedance
 Example – see Section 15.6.4
in course text
Determine the current in this circuit.
Since v = 100 sin 250t , then ω = 250
Therefore

Z = R − j XC
=R−j

1
ωC

= 100 − j

1

250 × 10 − 4
= 100 − j40

OHT 15.30

 Example (continued)
The current is given by v/Z and this is easier to compute in
polar form
Z = 100 − j40
Z = 100 2 + 40 2 = 107.7
∠Z = tan − 1

− 40
= −21.8
100

Z = 107.7∠ − 21.8 

Therefore
i=

v
100∠0
=
= 0.93∠21.8
Z 107.7∠ − 21.8


OHT 15.31

 A further example
A more complex task is to
find the output voltage of this
circuit.
The analysis of this circuit,
and a numerical example
based on it, are given in
Section 15.6.4 and
Example 15.8 of the
course text

OHT 15.32

Key Points
 A sinusoidal voltage waveform can be described by the
equation v = Vp sin (ω t + φ )
 The voltage across a resistor is in phase with the current,
the voltage across an inductor leads the current by 90°, and
the voltage across a capacitor lags the current by 90°
 The reactance of an inductor XL = ωL
 The reactance of a capacitor XC = 1/ωC
 The relationship between current and voltage in circuits
containing reactance can be described by its impedance
 The use of impedance is simplified by the use of complex
notation

OHT 15.33

Alternating voltages and currents

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