Download Chapter 15

Alternating Voltages and Currents
Chapter 15
 Introduction
 Voltage and Current
 Reactance of Inductors and Capacitors
 Phasor Diagrams
 Impedance
 Complex Notation
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Introduction
15.1
 From our earlier discussions we know that
v  Vp sin ( t   )
where
Vp is the peak voltage
 is the angular frequency
 is the phase angle
 Since  = 2f it follows that the period T is given by
T
1 2

f 
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 If  is in radians, then a time delay t is given by  /
as shown below
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Voltage and Current
15.2
 Consider the voltages across a resistor, an inductor
and a capacitor, with a current of
i  IP sin(t )
 Resistors
– from Ohm’s law we know
v R  iR
– therefore if i = Ipsin(t)
v R  IP R sin(t )
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Voltage and Current
 Inductors - in an inductor
15.2
di
vL  L
dt
– therefore if i = Ipsin(t)
d(IP sin(t ))
 LIP cos(t )
dt
1
 Capacitors - in a capacitor vC   idt
C
– therefore if i = Ipsin(t)
vL  L
Ip
1
vC   IP sin(t )  
cos(t )
C
C
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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(t )) IP R


R
Peak value of current
Peak value of (IP sin(t ))
IP
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 Inductance
Peak value of voltage Peak value of (LIP cos(t )) LIP


 L
Peak value of current
Peak value of (IP sin(t ))
IP
 Capacitance
Peak value of voltage

Peak value of current
Peak value of ( 
Ip
C
cos(t ))
Peak value of (I p sin(t ))
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Ip
1
 C 
Ip
C
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 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  L
Reactance
1
Reactance
Re
ac tan ce of a capacitor, XC 
C
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 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
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 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  L  2fL  2    100  25  10  3  15.7 
Therefore
VL
5
IL 

 318 mA peak
X L 15.7
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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
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 Examples of
phasor diagrams
(a) here L represents
the magnitude and 
the phase of a
sinusoidal signal
(b) shows the voltages
across a resistor, an
inductor and a
capacitor for the same
sinusoidal current
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 Phasor diagrams can be used to represent the
addition of signals. This gives both the magnitude
and phase of the resultant signal
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 Phasor diagrams can also be used to show the
subtraction of signals
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 Phasor analysis of an RL circuit
 See Example 15.5 in the text for a numerical example
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 Phasor analysis of an RC circuit
 See Example 15.6 in the text for a numerical example
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 Phasor analysis of an RLC circuit
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 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
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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
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 Consider the following circuit and its phasor diagram
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 From the phasor diagram it is clear that that the
magnitude of the voltage across the arrangement V is
V  VR 2  VL 2
 (IR )2  (IX L )2
 I R 2  XL2
 IZ
where Z  R 2  X L 2
 Z is the magnitude of the impedance, so Z =|Z|
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 From the phasor diagram the phase angle of the
impedance is given by
  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
Z R X
2
2
and
  tan
1
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X
R
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 A graphical representation of impedance
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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
=
jL
1
1
j

C jC
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 Graphical representation of complex impedance
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 Series and parallel
combinations of
impedances
– impedances combine
in the same way as
resistors
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 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
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 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  = 2f = 2 50 = 314 rad/s
Therefore
Z  Z C  Z R  Z L  R  j( X L  XC )  R  j(L 
 200  j(314  400  10 3 
1
314  50  10
6
1
)
C
)
 200  j62 ohms
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 Using complex impedance
 Example – see Section 15.6.4
in course text
Determine the current in this circuit.
Since v = 100 sin 250t , then  = 250
Therefore
Z  R  j XC
Rj
1
C
 100  j
1
250  10  4
 100  j40
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 Example (continued)
The current is given by v/Z and this is easier to compute in
polar form
Z  100  j40
Z  1002  402  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
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 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
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Key Points
 A sinusoidal voltage waveform can be described by the
equation v  Vp sin ( t   )
 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 = L
 The reactance of a capacitor XC = 1/C
 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
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