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ELECTRICAL TECHNOLOGY
EET 103/4
 Define and explain sine wave, frequency,
amplitude, phase angle, complex number
 Define, analyze and calculate impedance,
inductance, phase shifting
 Explain and calculate active power,
reactive power, power factor
 Define, explain, and analyze Ohm’s law,
KCL, KVL, Source Transformation,
Thevenin theorem.
1
BASIC ELEMENTS
AND PHASORS
(CHAPTER 14)
2
14.2 The Derivative
 To understand the response of the basic R, L,
and C elements to a sinusoidal signal, you
need to examine the concept of the derivative.
 The derivative dx/dt is defined as the rate of
change of x with respect to time. If x fails to
change at a particular instant, dx = 0, and the
derivative is zero.
 For the sinusoidal waveform, dx/dt is zero only
at the positive and negative peaks (wt = p/2
and 3p/2) since x fails to change at these
instants of time.
3
14.2 The Derivative
 The derivative dx/dt is actually the slope of
the graph at any instant of time.
 The greatest change in x will occur at the
instants wt = 0, p, and 2p.
 For various values of wt between these
maxima and minima, the derivative will exist
and will have values from the minimum to the
maximum inclusive.
 The derivative of a sine wave is a cosine
wave; it has the same period and frequency
as the original sinusoidal waveform.
4
14.2 The Derivative
5
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
 Resistor
 For power-line frequencies and frequencies up
to a few hundred kilohertz, resistance is, for all
practical purposes, unaffected by the frequency
of the applied sinusoidal voltage or current.
 For a purely resistive element, the voltage
across and the current through the element are
in phase, with their peak values related by Ohm’s
law.
6
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
7
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
 Inductor
 The magnitude of the voltage across the
element is determined by the opposition of the
element to the flow of charge, or current i.
 The inductive voltage, therefore, is directly
related to the frequency (or, more specifically,
the angular velocity of the sinusoidal ac current
through the coil) and the inductance of the coil.
For an inductor, vL leads iL by 90°, or iL lags vL
by 90°.
8
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
iL  I m sin wt
diL
d
vL  L
 L I m sin wt   wLI m cos wt
dt
dt
Or;



vL  wLI m sin wt  90  Vm sin wt  90

Where; Vm  wLI m
Hence;
Vm
 wL  X L  Inductive reactance
Im
9
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
 Inductive reactance is the opposition to the flow of
current, which results in the continual interchange of
energy between the source and the magnetic field of the
inductor.
10
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
 Capacitor
 Since capacitance is a measure of the rate at
which a capacitor will store charge on its
plates, for a particular change in voltage
across the capacitor, the greater the value of
capacitance, the greater will be the resulting
capacitive current.
11
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
The fundamental equation relating the voltage
across a capacitor to the current of a capacitor
[i = C(dv/dt)] indicates that for a particular
capacitance, the greater the rate of change of
voltage across the capacitor, the greater the
capacitive current.
12
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
vC  Vm sin wt
dvC
d
iC  C
 C Vm sin wt   wCI m cos wt
dt
dt
Or;



iC  wCI m sin wt  90  I m sin wt  90

Where; I m  wCVm
Hence;
Vm
1

 X C  Capative reactance
I m wC
13
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
14
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
 Capacitive reactance is the opposition to the
flow of charge, which results in the continual
interchange of energy between the source
and the electric field of the capacitor.
 If the source current leads the applied
voltage, the network is predominantly
capacitive, and if the applied voltage leads the
source current, it is predominantly inductive.
15
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(a)
The voltage across a 10- resistor is given by the
expression;
v  100 sin 377t
Find the expression for the current i through the
resistor and sketch the curves for v and i.
16
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(a) – solution
v  Vm sin wt  100 sin 377t
Vm  100 V
and
w  377 rad/s  2pf
Vm 100
Im 

 10 A
R
10
Hence;
im  I m sin wt  10 sin 377t A
17
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(a) – solution (cont’d)
The waveform;
18
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(a) – solution (cont’d)
The alternative waveform;
v, i
100 V
v
In phase
10 A
0
i
8.35
t (ms)
16.7
19
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(b)
The voltage across a 10- resistor is given by the
expression;

v  25 sin 377t  60

Find the expression for the current i through the
resistor and sketch the curves for v and i.
20
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(b) – solution

v  Vm sin wt     25 sin 377t  60
Vm  25 V;
w  377 rad/s  2pf

and
  60 
Vm 25
Im 

 2.5 A
R 10
Hence;


im  I m sin wt     2.5 sin 377t  60 A
21
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(b) – solution (cont’d)
The waveform;
22
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.1(b) – solution (cont’d)
The alternative waveform;
v, i
v
25 V
i
In phase
8.33
12.5
2.5 A
-4.17
2.78
0
4.17
16.67
t (ms)
23
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(a)
The current a 0.1-H coil is given by the expression;
i  10 sin 377t
Find the expression for the voltage v across the coil
and sketch the curves for v and i.
24
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(a) – solution
The voltage across a coil leads the current throuh it
by 90.
Hence, if the current is;
iL  I m sin wt


the voltage will be; vL  Vm sin wt  90

where;
Vm  I m X L
and;
X L  wL  2pfL
25
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(a) – solution (cont’d)
From the expression;
i  10 sin 377t
we obtain;
and;
Also;
I m  10 A;
w  377 rad/s;
377
f 
 60 Hz;
2p
1
1
T 
 16.67 ms;
f 60
X L  wL  377  0.1  37.7 
26
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(a) – solution (cont’d)
Vm  I m X L  10  37.7  377 V
Hence the expression for the voltage is;



v  Vm sin wt  90  377 sin 377t  90

27
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(a) – solution (cont’d)
The waveforms;
16.67
t (ms)
-4.17
4.17
8.33
12.5
28
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(b)
The current a 0.1-H coil is given by the expression;

i  7 sin 377t  70

Find the expression for the voltage v across the coil
and sketch the curves for v and i.
29
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(b) – solution
The voltage across a coil leads the current throuh it
by 90.
Hence, if the current is;
the voltage will be;
iL  I m sin wt   

vL  Vm sin wt    90

where;
Vm  I m X L
and;
X L  wL  2pfL
30
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(b) – solution (cont’d)
From the expression;

i  7 sin 377t  70
we obtain;
I m  7 A;

w  377 rad/s;
1
1
T 
 16.67 ms;
f 60
Also;
377
f 
 60 Hz;
2p
and;   70 rad/s;
X L  wL  377  0.1  37.7 
31
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(b) – solution (cont’d)
Vm  I m X L  7  37.7  263.9 V
Hence the expression for the voltage is;

v  Vm sin wt    90


 263.9 sin 377t  70  90



 263.9 sin 377t  20 V
32
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.3(b) – solution (cont’d)
The waveforms;
16.67
4.17
0.93
8.33
12.5
t (ms)
3.24
v leads i by 90  4.17 ms
33
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.5
The voltage across a 1-F capacitor is given by the
expression;
v  30 sin 400t V
Find the expression for the current i through the
capacitor and sketch the curves for v and i.
34
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.5 – solution
The current through a capacitor leads the voltage
across it by 90.
Hence, if the voltage is;
the current will be;
where;
Vm
Im 
XC
vC  Vm sin wt

iC  I m sin wt  90
and;

1
1
XC 

wC 2pfC
35
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.5 – solution (cont’d)
From the expression;
v  30 sin 400t
we obtain; Vm  30 V; w  400 rad/s;
and;
Also;
400
f 
 63.7 Hz;
2p
1
1
T 
 15.7 ms;
f 63.7
1
1
XC 

 2500 
6
wC 400 110
36
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.5 – solution (cont’d)
Vm
30
Im 

 12 mA
X C 2500
Hence the expression for the current is;

i  I m sin wt  90



 12 sin 400t  90 mA
37
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.5 – solution (cont’d)
The waveforms;
t (ms)
7.85
3.93
3.93
11.78
15.7
i leads v by 3.93 ms  90
38
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.6
The current through a 100-F capacitor is given by
the expression;


i  40 sin 500t  60 V
Find the expression for the voltage v across the
capacitor.
39
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.6 – solution
The voltage across a capacitor lags the current
through it by 90.
Hence, if the current is;
the voltage will be;
iC  I m sin wt   

vC  Vm sin wt    90
where;
Vm  I m X C
and;

1
1
XC 

wC 2pfC
40
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.6 – solution (cont’d)
From the expression;

i  40 sin 500t  60
we obtain; I m  40 A;

w  500 rad/s;
1
1
T 
 12.57 ms;
f 79.6
Also;
500
f 
 79.6 Hz;
2p
and;
  60 
1
1
XC 

 20 
6
wC 500 100 10
41
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.6 – solution (cont’d)
Vm  I m X C  40  20  800 V
Hence the expression for the current is;

v  Vm sin wt    90



 800 sin 500t  60  90 V


 800 sin 500t  30 V
42
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(a)
Determine the type of element in the box (C, L or
R) and calculate its value if;


v  100 sin wt  40 V
and


i  20 sin wt  40 A
43
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(a) – solution


v  100 sin wt  40 V

and


i  20 sin wt  40 A
The voltage and current are in phase – the element
is a resistor (R).
Vm 100
R

5
Im
20
44
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(b)
Determine the type of element in the box (C, L or
R) and calculate its value if;


v  1000 sin 377t  10 V

and


i  5 sin 377t  80 A
45
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(b) – solution


v  1000 sin 377t  10 V

and


i  5 sin 377t  80 A
The voltage leads the current by 90 – the element
is an inductor (L).
Vm 1000
XL 

 200 
Im
5
XL
200
L

 0.53 H
w 377
46
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(c)
Determine the type of element in the box (C, L or
R) and calculate its value if;


v  500 sin 157t  30 V

and


i  1sin 157t  120 A
47
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(c) – solution


v  500 sin 157t  30 V

and


i  1sin 157t  120 A
The voltage lags the current by 90 – the element
is a capacitor (C).
Vm 500
XC 

 500 
Im
1
1
1
C

 12.74 F
wX C 157  500
48
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(d)
Determine the type of element in the box (C, L or
R) and calculate its value if;


v  50 cos wt  20 V

and


i  5 sin wt  110 A
49
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(d) – solution




v  50 cos wt  20 V  50 sin wt  20  90 V


 50 sin wt  110 V


i  5 sin wt  110 A
The voltage and current are in phase – the element
is a resistor (R).
50
14.3 Response of Basic R, L and C Elements
to a Sinusoidal Voltage or Current
Example 14.7(d) – solution (cont’d)
Vm 50
R

 10 
Im
5
51
14.4 Frequency Response of Basic Elements
Resistance
 In the real world, each resistive element has stray
capacitance levels and lead inductance that are
sensitive to the applied frequency.
Usually the capacitive and inductive levels involved
are so small that their real effect is not noticed until
the frequency is in the megahertz range.
 Frequency does have impact on the resistance of an
element, but for our purpose the resistance level of a
resistor is independent of frequency (below 15 MHz).
52
14.4 Frequency Response of Basic Elements
As the applied frequency increases, the resistance of
a resistor remains constant
53
14.4 Frequency Response of Basic Elements
As the applied frequency increases, the reactance of
an inductor increases linearly.
X L  wL  2pfL
X L  kf
k  2pL
54
14.4 Frequency Response of Basic Elements
As the applied frequency increases, the reactance of
a capacitor decreases nonlinearly.
1
1
XC 

wC 2pfC
k
XC 
f
1
k
2pC
55
14.5 Average Power and Power Factor
 For any load in a sinusoidal ac network, the
voltage across the load and the current through
the load will vary in a sinusoidal nature.
 The average power (real power) is the power
delivered to and dissipated by the load.
 It corresponds to the power calculations performed for dc
networks.
 The angle (v - i) is the phase angle between v and i.
 The magnitude of average power delivered is
independent of whether v leads i or i leads v, since cos(a) = cos a.
56
14.5 Average Power and Power Factor
Power versus time for a purely
resistive load.
57
14.5 Average Power and Power Factor
For a purely resistive load;
Vm I m
Pav 

2
2Vrms 2 I rms
2
Pav  Vrms I rms
2
Pav  I rms
R
2
Vrms
Pav 
R
58
14.5 Average Power and Power Factor
Instantaneous power;
p  vi
 Vm sin wt  v I m sin wt  i 
 Vm I m sin wt  v sin wt  i 
i = I m sin(w t +  i )
v = V m sin(w t +  v )
cos v   i   cos2wt   v   i 




sin wt   v sin wt   i 
2
59
14.5 Average Power and Power Factor
Hence;
Vm I m
cos v  i   cos2wt   v  i 
p
2
Vm I m
Vm I m

cos v   i  
cos2wt   v   i 
2
2
Fixed value
Time-varying
60
14.5 Average Power and Power Factor
61
14.5 Average Power and Power Factor
The average value of the term;
Vm I m
cos2wt   v   i 
2
is zero
Therefore the average power delivered to the load is
contributed by the first term only i.e.;
Pav 
Vm I m
cos v   i 
2
 v   i is the angle difference between v and i
62
14.5 Average Power and Power Factor
Defining v  i   , the average power delivered to
the load becomes;
Vm I m
Pav 
cos 
2
watt, W 
Pav  Vrms I rms cos 
watt, W
or;
63
14.5 Average Power and Power Factor
For purely resistive load
 0
and
cos   1
Pav  Vrms I rms cos  Vrms I rms
cos is known
as the power
factor (Fp)
For purely inductive or purely capacitive load
  90
and
cos  0
Pav  Vrms I rms cos  0
64
14.5 Average Power and Power Factor
Power factor can be leading or lagging
A power factor is said to be leading if the current leads
the corresponding voltage
When the current lags the voltage, the corresponding
power factor is known as a lagging power factor
In terms of the average power and terminal voltage and
current;
P
Fp  cos  
Vrms I rms
65
14.5 Average Power and Power Factor
Example 4.10
Find the average power dissipated in a network
whose input current and voltage are;


i  5 sin wt  40 A


v  10 sin wt  40 V
66
14.5 Average Power and Power Factor
Example 4.10 – solution
Since the current and voltage are in phase;
 0
and
cos   1
Vm I m 10  5
P

 25 W
2
2
Or;
Vm 10
R

 2
Im
5
and
I rms
Im
5


2
2
67
14.5 Average Power and Power Factor
Example 4.10 – solution (cont’d)
2
PI
Or;
Vrms
2
rms
 5 
R
  2  25 W
 2
Vm 10


2
2
2
V
 10  1
P

   25 W
R  2 2
2
rms
68
14.5 Average Power and Power Factor
Example 4.11(a)
Determine the average power delivered to a
network whose input voltage and current are;


v  100 sin wt  40 V


i  20 sin wt  70 A
69
14.5 Average Power and Power Factor
Example 4.11(a) – solution
Given;


v  100 sin wt  40 V


i  20 sin wt  70 A
Hence;
  70  40  30
cos   cos 30  0.866
Vm I m
100  20
P
cos  
 0.866  866 W
2
2
70
14.5 Average Power and Power Factor
Example 4.11(b)
Determine the average power delivered to a
network whose input voltage and current are;


v  150 sin wt  70 V


i  3 sin wt  50 A
71
14.5 Average Power and Power Factor
Example 4.11(b) – solution
Given;


v  150 sin wt  70 V


i  3 sin wt  50 A
Hence;
  70  50  20
cos   cos 20  0.9397
Vm I m
150  3
P
cos  
 0.9397  211.4 W
2
2
72
14.5 Average Power and Power Factor
Example 4.12(a)
Determine the power factor and indicate whether
it is leading or lagging
73
14.5 Average Power and Power Factor
Example 4.12(a) – solution


i  2 sin wt  40 A


v  50 sin wt  20 V
Therefore;
  40   20   60
Fp  cos 60  0.5
(i leads v)
(leading)
74
14.5 Average Power and Power Factor
Example 4.12(b)
Determine the power factor and indicate whether
it is leading or lagging
75
14.5 Average Power and Power Factor
Example 4.12(b) – solution


i  5 sin wt  30 A


v  120 sin wt  80 V
Therefore;
  30  80  50
Fp  cos 50  0.64
(i lags v)
(lagging)
76
14.5 Average Power and Power Factor
Example 4.12(c)
Determine the power factor and indicate whether
it is leading or lagging
77
14.5 Average Power and Power Factor
Example 4.12(c) – solution
P  100 W
Veff I eff  20  5  100 W
P
100
Fp 

1
Veff I eff 100
(neither leading nor lagging)
78