Chapter 6
Alternating Voltages and Currents
6.1 Introduction
6.2 Voltage and Current
6.3 Reactance of Inductors and Capacitors
6.4 Phasor Diagrams
6.5 Impedance
6.6 Complex Notation
6.7 Problems
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
Chapter 6
6.1
6.1 Introduction (1/1)
 We know that alternating Voltage is given by
v  Vp sin ( t   )
Vp = peak voltage= Maximum voltage,  = angular
frequency in rd/s,  = phase angle in rd
1 2
 Since  = 2f  T  
f

 If  is in radians, then a time delay t is given by  /:
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.2
6.2 Voltage and Current (1/2)
 Consider the voltages across a resistor, an inductor and a
capacitor, with a current of i  IP sin(t )
 Resistors: Ohm’s law  v R  iR  v R  IP R sin(t )
di
 Inductors: v L  L
dt
𝒍𝒆𝒂𝒅𝒔
d( I P sin(t ))
ฑ
vL  L
 LI P cos(t ) = 𝝎𝑳𝑰𝑷 𝒔𝒊𝒏 𝝎𝒕+𝟗𝟎
dt
1
 idt
C
𝒍𝒂𝒈𝒔
𝑰𝑷
Ip
1
ฑ
vC   IP sin(t )  
cos(t ) =
𝒔𝒊𝒏 𝝎𝒕−𝟗𝟎
C
C
𝝎𝑪
 Capacitors: vC 
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.3
6.2 Voltage and Current (2/2)




R: I and V in phase
Capacitors: I leads V by 90 (or V lags I by 90); 𝜃𝑖 −𝜃𝑣 = 90°
Inductors: V leads I by 90 (or I lags V by 90); 𝜃𝑣 −𝜃𝑖 = 90°
CIVIL: in C I leads V and V leads I in L
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.4
6.3 Reactance of Inductors and Capacitors (1/4)
 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 𝑷𝒆𝒂𝒌
 Resistance
𝑷𝒆𝒂𝒌
Peak value of voltage Peak value of (IP Rsin(t )) IP R


R
Peak value of current
Peak value of (IP sin(t ))
IP
 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 ))
Ip
1

C


Ip
C
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.5
6.3 Reactance of Inductors and Capacitors (2/4)
 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
Re
ac tan ce of a capacitor, XC 
Reactance
1
C
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.6
6.3 Reactance of Inductors and Capacitors (3/4)
 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 a resistor
V I R
V  I ωL
– for an inductor V  I X L
or
– for a capacitor V  I XC
I
or V 
ωC
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.7
6.3 Reactance of Inductors and Capacitors (4/4)
 Example 6.3: 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:
X L  L  2fL  2    100  25  10 3  15.7 
IL 
VL
5

 318 mA peak
X L 15.7
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.8
6.4 Phasor Diagrams (1/8)
 Sinusoidal signals are characterized 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
 phasor diagrams are similar to vectors
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.9
6.4 Phasor Diagrams (2/8)
𝒍𝒆𝒂𝒅𝒔
 Examples of
phasor diagrams
(a) here L represents the
magnitude and  the
phase of a sinusoidal
signal
ฑ
𝒗𝑳 = 𝝎𝑳𝑰𝑷 𝒔𝒊𝒏 𝝎𝒕+𝟗𝟎
𝒍𝒂𝒈𝒔
𝑰𝑷
ฑ
𝒗𝑪 =
𝒔𝒊𝒏 𝝎𝒕−𝟗𝟎
𝝎𝑪
V Leads I in L
V Lags I in C
(b) shows the voltages
across a resistor, an
inductor and a capacitor
for the same sinusoidal
current (Reference)
Assuming  of the current =0
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.10
6.4 Phasor Diagrams (3/8)
 Phasor diagrams can be used to represent the additionor
subtraction of signals. This gives both the magnitude and
phase of the resultant signal (same frequency)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.11
6.4 Phasor Diagrams (4/8)
 Phasor analysis of an RL circuit
 See Example 6.5 in the text for a numerical example
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.12
6.4 Phasor Diagrams (5/8)
 Example 6.5
 Answers
– V= 63.6 V
– = 38.2
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.13
6.4 Phasor Diagrams (6/8)
 Phasor analysis of an RC circuit
 See Example 6.6 in the text for a numerical example
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.14
6.4 Phasor Diagrams (7/8)
 Phasor analysis of an RLC circuit
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.15
6.4 Phasor Diagrams (8/8)
 Phasor analysis
of parallel
circuits
in such circuits the
voltage across each
of the components is
the same (reference)
and it is the currents
that are of interest
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.16
6.5 Impedance (1/4)
 In circuits containing only resistive elements the
current is related to the applied voltage by the
resistance of the arrangement (Req)
 In circuits containing reactive, as well as resistive
elements, the current is related to the applied voltage
by the impedance, Z of the arrangement
– 𝐙 = 𝑽Τ𝑰 : 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
– Impedance changes with frequency.
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.17
6.5 Impedance (2/4)
 Consider the following circuit and its phasor diagram
 From the phasor diagram it is clear that that the
magnitude of the voltage across the arrangement V is
V  V R 2 V L 2  (IR ) 2  (IX L ) 2  I R 2  X L 2  IZ ; Z  R 2  X L 2
 Z is the magnitude of the impedance, so Z =|Z|
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.18
6.5 Impedance (3/4)
 From the phasor diagram the phase angle of the
impedance is given by
θv  θ i  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

Z  R2  X 2
  tan 1
X
R
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.19
6.5 Impedance (4/4)
 A graphical representation of the impedance
equivalent of
R in series with L
R in series with C
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.20
6.6 Complex Notation (1/7)
 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
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.21
6.6 Complex Notation (2/7)
 Graphical representation of complex impedance vs. phasor
Complex
Amplitude
and Phase
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.22
6.6 Complex Notation (3/7)
 Series and parallel
combinations of
impedances
– impedances combine
in the same way as
resistors
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.23
6.6 Complex Notation (4/7)
 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
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.24
6.6 Complex Notation (5/7)
 Example 6.7: Determine the complex impedance of this
circuit at a frequency of 50 Hz
At 50 Hz, 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
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.25
6.6 Complex Notation (6/7)
 Using complex impedance
 Example in Section 6.6.4
Determine the current in this circuit
Since v = 100 sin 250t , then  = 250
Z  R  jX
C
Rj
1
ωC
 100  j
1
 100  j 40
4
250 10
The current is given by v/Z; Using polar form:
Z  100  j 40  Z  100 2  40 2  107.7 & Z  tan  1
i
 40
 21.8  Z  107.7  21.8
100
v
1000

 0.9321.8
Z 107.7  21.8
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.26
6.6 Complex Notation (7/7)
 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 6.6.4
and Example 6.8 of the
course text
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.27
6.7 Problems (1/5)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.28
6.7 Problems (2/5)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.29
6.7 Problems (3/5)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.30
6.7 Problems (4/5)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.31
6.7 Problems (5/5)
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.32
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
Neil Storey, Electronics: A Systems Approach, 4th Edition © Pearson Education Limited 2009
6.33