Lecture 1.2.ppt

Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Lecture 02
AC Theory
THOMAS L. FLOYD
DAVID M. BUCHLA

Floyd/Buchla
Chapter 8
Summary
The sinusoidal waveform (sine wave) is the fundamental
alternating current (ac) and alternating voltage waveform.
Sine waves
Electrical sine waves are
named from the
mathematical function
with the same shape.

Floyd/Buchla
Chapter 8
Summary
Sine waves are characterized by the amplitude and period.
The amplitude is the maximum value of a voltage or current;
the period is the time interval for one complete cycle.
Sine waves
0 V
10 V
-10 V
15 V
-15 V
-20 V
t ( s)

0 25 37.5 50.0
20 V
The amplitude (A)
of this sine wave
is 20 V
The period is 50.0 s
A
T

Floyd/Buchla
Chapter 8
Summary
The period of a sine wave can be measured between
any two corresponding points on the waveform.
Sine waves
T T T T
T T
By contrast, the amplitude of a sine wave is only
measured from the center to the maximum point.
A

Floyd/Buchla
Chapter 8
3.0 Hz
Summary
Summary
Frequency
Frequency ( f ) is the number of cycles that a sine wave
completes in one second.
Frequency is measured in hertz (Hz).
If 3 cycles of a wave occur in one second, the frequency
is 1.0 s

Floyd/Buchla
Chapter 8
Summary
The period and frequency are reciprocals of each other.
Summary
Period and frequency
T
f
1
 and f
T
1

Thus, if you know one, you can easily find the other.
If the period is 50 s, the frequency is 0.02 MHz = 20 kHz.
(The 1/x key on your calculator is handy for converting between f and T.)

Floyd/Buchla
Chapter 8
Summary
Sinusoidal voltages are produced by ac generators and
electronic oscillators.
Summary
Sinusoidal voltage sources
Generation of a sine wave
N S
Motion of conductor Conductor
B
C
D
A
A
B
C
D
A
B
B
C
D
A
C
B
C
D
A
D
When a conductor rotates in a constant magnetic
field, a sinusoidal wave is generated.
When the conductor is moving parallel with
the lines of flux, no voltage is induced.
When the loop is moving perpendicular to the
lines of flux, the maximum voltage is induced.
B
C
D
A

Floyd/Buchla
Chapter 8
0 V
10 V
-10 V
15 V
-15 V
-20 V
t ( s)

0 25 37.5 50.0
20 V
The voltage of a sine wave can also be specified as
either the peak-to-peak or the rms value. The peak-to-
peak is twice the peak value. The rms value is 0.707
times the peak value.
Sine wave voltage and current values
The peak-to-peak
voltage is 40 V.
The rms voltage
is 14.1 V.
VPP
Vrms
Summary
Summary

Floyd/Buchla
Chapter 8
0 V
10 V
-10 V
15 V
-15 V
-20 V
t ( s)

0 25 37.5 50.0
20 V
For some purposes, the average value (actually the half-
wave average) is used to specify the voltage or current.
By definition, the average value is as 0.637 times the
peak value.
Sine wave voltage and current values
The average value for
the sinusoidal voltage
is 12.7 V.
Vavg
Summary
Summary

Floyd/Buchla
Chapter 8
Angular measurements can be made in degrees (o) or
radians. The radian (rad) is the angle that is formed when
the arc is equal to the radius of a circle. There are 360o or
2p radians in one complete revolution.
Angular measurement
R
R
1
.0
-1
.0
0
.8
-0
.8
0
.6
-0
.6
0
.4
-0
.4
0
.2
-0
.2
0 0 2
p
p
p
2
p
4
p
4
3 p
2
3
p
4
5 p
4
7
Summary
Summary

Floyd/Buchla
Chapter 8
There are 2p radians in one complete revolution and 360o
in one revolution. To find the number of radians, given
the number of degrees: 2 rad
rad degrees
360
p
 

180
deg rad
rad
p

 
To find the number of degrees, given the number of
radians:
Angular measurement
Summary
Summary
This can be simplified to:
rad
rad degrees
180
p
 


Floyd/Buchla
Chapter 8
How many radians are in 45o?
rad
rad degrees
180
rad
= 45 0.785 rad
180
p
p
 

  

180
deg rad
rad
180
1.2 rad =
rad
69
p
p

 

  
Angular measurement
Summary
Summary
How many degrees are in 1.2 radians?

Floyd/Buchla
Chapter 8
Instantaneous values of a wave are shown as v or i. The
equation for the instantaneous voltage (v) of a sine
wave is
Sine wave equation
where
If the peak voltage is 25 V, the instantaneous
voltage at 50 degrees is

sin
p
V
v 
Vp =
 =
Peak voltage
Angle in rad or degrees
19.2 V
Summary
Summary

Floyd/Buchla
Chapter 8
Sine wave equation
v = = 19.2 V
Vp sin
Vp
90
50
0
= 50
Vp
Vp
= 25 V
A plot of the example in the previous slide (peak at
25 V) is shown. The instantaneous voltage at 50o is
19.2 V as previously calculated.
Summary
Summary

Floyd/Buchla
Chapter 8
0
0 90
90
180
180 360
The sine wave can be represented as the projection of a
vector rotating at a constant rate. This rotating vector is
called a phasor. Phasors are useful for showing the
phase relationships in ac circuits.
Phasors
Summary
Summary

Floyd/Buchla
Chapter 8
Phase shift
where
f = Phase shift
The phase of a sine wave is an angular measurement
that specifies the position of a sine wave relative to a
reference. To show that a sine wave is shifted to the
left or right of this reference, a term is added to the
equation given previously.
 
f
 
 sin
P
V
v
Summary
Summary

Floyd/Buchla
Chapter 8
Phase shift
Voltage
(V)
270 360
0 90 180
40
45 135 225 315
0
Angle ()
30
20
10
-20
-30
- 40
405
P
eak voltage
Reference
Notice that a lagging sine
wave is below the axis at 0o
Example of a wave that lags the
reference
v = 30 V sin ( - 45o)
…and the equation
has a negative phase
shift
Summary
Summary

Floyd/Buchla
Chapter 8
Phase shift
Voltage
(V)
270 360
0 90 180
40
45 135 225 315
0
Angle ()
30
20
10
-20
-30
-40
P
eak v
oltage
Reference
-45
-10
Notice that a leading sine
wave is above the axis at 0o
Example of a wave that leads the
reference
v = 30 V sin ( + 45o)
…and the equation
has a positive phase
shift
Summary
Summary

Floyd/Buchla
Chapter 8
An important application of phase-shifted sine waves is in
electrical power systems. Electrical utilities generate ac with
three phases that are separated by 120° as illustrated.
Phase shift
Summary
Summary
120o
Normally, 3-phase power is delivered to the user with three hot lines plus
neutral. The voltage of each phase, with respect to neutral is 120 V.
120o 120o

Floyd/Buchla
Chapter 8
The power formulas are:
Power in resistive AC circuits
rms rms
2
2
rms
rms
P V I
V
P
R
P I R



ac or dc
source
Bulb
Summary
Summary
The power relationships developed for dc circuits apply
to ac circuits except you must use rms values in ac
circuits when calculating power.
0 V
0 V
For example, the dc and the ac sources
produce the same power to the bulb:
120 Vdc
170 Vp
= 120 Vrms

Floyd/Buchla
Chapter 8
Assume a sine wave with a peak value of 40 V is
applied to a 100 W resistive load. What power is
dissipated?
Power in resistive AC circuits
2 2
28.3 V
100
rms
V
P
R
  
W
Voltage
(V)
40
0
30
20
10
-10
-20
-30
- 40
Vrms = 0.707 x Vp = 0.707 x 40 V = 28.3 V
8 W
Summary
Summary

Floyd/Buchla
Chapter 8
Frequently dc and ac voltages are together in a waveform.
They can be added algebraically, to produce a composite
waveform of an ac voltage “riding” on a dc level.
Superimposed dc and ac voltages
Summary
Summary

Floyd/Buchla
Chapter 8
Pulse definitions
Amplitude
Pulse
width
Baseline
Amplitude
Pulse
width
Baseline
(a) P
ositive-going pulse (b) Negative-going pulse
Leading (rising) edge
Trailing (falling) edge
Leading (falling) edge
Trailing (rising) edge
Ideal pulses
Summary
Summary

Floyd/Buchla
Chapter 8
Pulse definitions
Non-ideal pulses
A
0.9 A
0.1A
tr t
t
f
W
t t
0.5 A
A
(a) (b)
Rise and fall times Pulse width
Notice that rise and fall times are measured between
the 10% and 90% levels whereas pulse width is
measured at the 50% level.
Summary
Summary

Floyd/Buchla
Chapter 8
Harmonics
All repetitive non-sinusoidal waveforms are composed
of a fundamental frequency (repetition rate of the
waveform) and harmonic frequencies.
Odd harmonics are frequencies that are odd multiples
of the fundamental frequency.
Even harmonics are frequencies that are even multiples
of the fundamental frequency.
Summary
Summary

Floyd/Buchla
Chapter 8
Harmonics
A square wave is composed only of the fundamental
frequency and odd harmonics (of the proper amplitude).
Summary
Summary

Floyd/Buchla
Chapter 8
Quiz
1. In North America, the frequency of ac utility voltage is
60 Hz. The period is
a. 8.3 ms
b. 16.7 ms
c. 60 ms
d. 60 s

Floyd/Buchla
Chapter 8
Quiz
2. The amplitude of a sine wave is measured
a. at the maximum point
b. between the minimum and maximum points
c. at the midpoint
d. anywhere on the wave

Floyd/Buchla
Chapter 8
Quiz
3. An example of an equation for a waveform that lags the
reference is
a. v = -40 V sin ()
b. v = 100 V sin ( + 35o)
c. v = 5.0 V sin ( - 27o)
d. v = 27 V

Floyd/Buchla
Chapter 8
4. In the equation v = Vp sin  , the letter v stands for the
a. peak value
b. average value
c. rms value
d. instantaneous value
Quiz

Floyd/Buchla
Chapter 8
7. The number of radians in 90o are
a. p/2
b. p
c. 2p/3
d. 2p
Quiz

Floyd/Buchla
Chapter 8
8. For the waveform shown, the same power would be
delivered to a load with a dc voltage of
a. 21.2 V
b. 37.8 V
c. 42.4 V
d. 60.0 V
Quiz
0 V
30 V
-30 V
45 V
-45 V
-60 V
t ( s)

0 25 37.5 50.0
60 V

Floyd/Buchla
Chapter 8
9. A square wave consists of
a. the fundamental and odd harmonics
b. the fundamental and even harmonics
c. the fundamental and all harmonics
d. only the fundamental
Quiz

Floyd/Buchla
Chapter 8
Quiz
Answers:
1. b
2. a
3. c
4. d
5. b
6. b
7. a
8. c
9. a
10. b

Floyd/Buchla
chapter 9
Capacitors
THOMAS L. FLOYD
DAVID M. BUCHLA

Floyd/Buchla
Chapter 8
Capacitance is the ratio of charge to voltage
Q
C
V

Rearranging, the amount of charge on a
capacitor is determined by the size of the
capacitor (C) and the voltage (V).
Q CV

If a 22 F capacitor is connected to
a 10 V source, the charge is 220 C
Capacitance

Floyd/Buchla
Chapter 8
A capacitor stores energy in the form of an electric field
that is established by the opposite charges on the two
plates. The energy of a charged capacitor is given by the
equation
Capacitance
2
2
1
CV
W 
where
W = the energy in joules
C = the capacitance in farads
V = the voltage in volts

Floyd/Buchla
Chapter 8
Summary
Capacitive reactance
Capacitive reactance is the opposition to ac by a
capacitor. The equation for capacitive reactance is
1
2π
C
X
fC

The reactance of a 0.047 F capacitor when a
frequency of 15 kHz is applied is 226 W

Floyd/Buchla
Chapter 8
Summary
Power in a capacitor
Voltage and current are always 90o out of phase.
For this reason, no true power is dissipated by a capacitor,
because stored energy is returned to the circuit.
The rate at which a capacitor stores or returns
energy is called reactive power. The unit for reactive
power is the VAR (volt-ampere reactive).
Energy is stored by the capacitor during a portion of the ac
cycle and returned to the source during another portion of
the cycle.

Floyd/Buchla
Chapter 8
Summary
Power supply filtering
There are many applications for capacitors. One is in
filters, such as the power supply filter shown here.
Rectifier
60 Hz ac
C Load
resistance
The filter smoothes the
pulsating dc from the
rectifier.

Floyd/Buchla
Chapter 8
Summary
Coupling capacitors
Coupling capacitors are used to pass an ac signal
from one stage to another while blocking dc.
The capacitor isolates dc
between the amplifier stages,
preventing dc in one stage from
affecting the other stage.
0V 0V
3V
Amplifier
stage 1
Amplifier
stage 2
R2
R1
C
+V
Input Output

Floyd/Buchla
Chapter 8
Summary
Bypass capacitors
Another application is to bypass an ac signal to ground
but retain a dc value. This is widely done to affect gain
in amplifiers.
The bypass capacitor places
point A at ac ground, keeping
only a dc value at point A.
R2
R1
C
0V
dc plus
ac
Point in circuit where
only dc is required
0V
dc only
A

Floyd/Buchla
Chapter 8

Floyd/Buchla
Chapter 8
Summary
Sinusoidal response of RC circuits
When both resistance and capacitance are in a series
circuit, the phase angle between the applied voltage and
total current is between 0 and 90, depending on the
values of resistance and reactance.
R
VR
C
VR leads VS VC lags VS
I leads VS
I
VS
VC

Floyd/Buchla
Chapter 8
Summary
Impedance of series RC circuits
In a series RC circuit, the total impedance is the phasor
sum of R and XC.
R is plotted along the positive x-axis.
R
XC is plotted along the negative y-axis.
XC
1
tan C
X
R
 -  
  
 
Z
It is convenient to reposition the
phasors into the impedance triangle.
R
XC
Z
Z is the diagonal
 

Floyd/Buchla
Chapter 8
Summary
Impedance of series RC circuits
R = 1.2 kW
XC =
960 W
Sketch the impedance triangle and show the
values for R = 1.2 kW and XC = 960 W.
   
2 2
1.2 k + 0.96 k
1.33 k
Z  W W
 W
1 0.96 k
tan
1.2 k
39
 - W

W
 
Z = 1.33 kW
39o


Floyd/Buchla
Chapter 8
Summary
Analysis of series RC circuits
Ohm’s law is applied to series RC circuits using Z,
V, and I.
V V
V IZ I Z
Z I
  
Because I is the same everywhere in a series circuit,
you can obtain the voltages across different
components by multiplying the impedance of that
component by the current as shown in the following
example.

Floyd/Buchla
Chapter 8
Summary
Analysis of series RC circuits
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasor diagram. The impedance
triangle from the previous example is shown for reference.
VR = 12 V
VC =
9.6 V
The voltage phasor diagram can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
x 10 mA
=
R = 1.2 kW
XC =
960 W
Z = 1.33 kW
39o
VS = 13.3 V
39o
 

Floyd/Buchla
Chapter 8
Summary
Variation of phase angle with frequency
Phasor diagrams that have reactance phasors can only
be drawn for a single frequency because X is a
function of frequency.
As frequency changes,
the impedance triangle
for an RC circuit changes
as illustrated here
because XC decreases
with increasing f. This
determines the frequency
response of RC circuits.

Z3
XC1
XC2
XC3
Z2
Z1
1
2
3
1
2
f
f
f
3
Increasing f


R

Floyd/Buchla
Chapter 8

(phase lag)
f
f
(phase lag)
V
Summary
Applications
R
VR
Vout
C Vout
Vin
Vin
Vout
Vin
For a given frequency, a series RC circuit can be used to
produce a phase lag by a specific amount between an
input voltage and an output by taking the output across
the capacitor. This circuit is also a basic low-pass filter, a
circuit that passes low frequencies and rejects all others.

Floyd/Buchla
Chapter 8


(phase lead)
(phase lead)
V
Summary
Applications
R
VC
Vout
C
Vout
Vin
Vin
Vout
Vin
Reversing the components in the previous circuit produces
a circuit that is a basic lead network. This circuit is also a
basic high-pass filter, a circuit that passes high frequencies
and rejects all others. This filter passes high frequencies
down to a frequency called the cutoff frequency.

Floyd/Buchla
Chapter 8
Summary
Applications
An application showing how the phase-shift network is
useful is the phase-shift oscillator, which uses a
combination of RC networks to produce the required 180o
phase shift for the oscillator.
Amplifier
Rf
R R R
C C C
Phase-shift network

Floyd/Buchla
Chapter 8
Summary
Sinusoidal response of parallel RC circuits
In a parallel RC circuit, the admittance phasor is the sum
of the conductance and capacitive susceptance phasors.
The magnitude can be expressed as 2 2
+ C
Y G B

VS G BC
Y
G
BC

From the diagram, the phase angle is 1
tan C
B
G
 -  
  
 

Floyd/Buchla
Chapter 8
Summary
VS G BC
Y
G
BC

G is plotted along the positive x-axis.
BC is plotted along the positive y-axis.
1
tan C
B
G
 -  
  
 
Y is the diagonal
Some important points to notice are:

Floyd/Buchla
Chapter 8
Summary
VS C
0.01 F
Y =
1.18 mS
G = 1.0 mS
BC = 0.628 mS
Draw the admittance phasor diagram for the circuit.
f = 10 kHz
R
1.0 kW
1 1
1.0 mS
1.0 k
G
R
  
W
  
2 10 kHz 0.01 F 0.628 mS
C
B p 
 
The magnitude of the conductance and susceptance are:
   
2 2
2 2
+ 1.0 mS + 0.628 mS 1.18 mS
C
Y G B
  

Floyd/Buchla
Chapter 8
Summary
Analysis of parallel RC circuits
Ohm’s law is applied to parallel RC circuits using
Y, V, and I.
I I
V I VY Y
Y V
  
Because V is the same across all components in a
parallel circuit, you can obtain the current in a given
component by simply multiplying the admittance of
the component by the voltage as illustrated in the
following example.

Floyd/Buchla
Chapter 8
Summary
Analysis of parallel RC circuits
If the voltage in the previous example is 10 V, sketch the
current phasor diagram. The admittance diagram from the
previous example is shown for reference.
The current phasor diagram can be found from
Ohm’s law. Multiply each admittance phasor by 10 V.
x 10 V
=
Y =
1.18 mS
G = 1.0 mS
BC = 0.628 mS
IR = 10 mA
IC = 6.28 mA
IS =
11.8 mA

Floyd/Buchla
Chapter 8
Summary
Phase angle of parallel RC circuits
Notice that the formula for capacitive susceptance is
the reciprocal of capacitive reactance. Thus BC and IC
are directly proportional to f: 2
C
B fC
p

IR
IC
IS

As frequency increases, BC
and IC must also increase,
so the angle between IR and
IS must increase.

Floyd/Buchla
Chapter 8
Summary
Equivalent series and parallel RC circuits
For every parallel RC circuit there is an equivalent
series RC circuit at a given frequency.
The equivalent resistance and capacitive
reactance are shown on the impedance triangle:
Z
Req = Z cos 
XC(eq) = Z sin 


Floyd/Buchla
Chapter 8
Summary
Series-Parallel RC circuits
Series-parallel RC circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
series and parallel circuits.
For example, the
components in the
green box are in
series:
The components in
the yellow box are
in parallel: 2 2
2 2 2
2 2
C
C
R X
Z
R X


R1 C1
R2 C2
Z1
Z2
The total impedance can be found by
converting the parallel components to an
equivalent series combination, then
adding the result to R1 and XC1 to get the
total reactance.
2 2
1 1 1
C
Z R X
 

Floyd/Buchla
Chapter 8
Summary
Measuring Phase Angle
An oscilloscope is commonly used to measure phase angle in
reactive circuits. The easiest way to measure phase angle is to
set up the two signals to have the same apparent amplitude and
measure the period. An example of a Multisim simulation is
shown, but the technique is the same in lab.
Set up the oscilloscope so that
two waves appear to have the
same amplitude as shown.
Determine the period. For the
wave shown, the period is
20 μs
8.0 div 160 μs
div
T
 
 
 
 

Floyd/Buchla
Chapter 8
Summary
Measuring Phase Angle
Next, spread the waves out using the SEC/DIV control in order
to make an accurate measurement of the time difference between
the waves. In the case illustrated, the time difference is
5 μs
4.9 div 24.5 μs
div
t
 
  
 
 
The phase shift is calculated from
24.5 μs
360 360
160 μs
t
T

 

 
    
 
 
   
55o

Floyd/Buchla
Chapter 8
Summary
The power triangle
Recall that in a series RC circuit, you could multiply
the impedance phasors by the current to obtain the
voltage phasors. The earlier example is shown for
review:
VR = 12 V
VC =
9.6 V
x 10 mA
=
R = 1.2 kW
XC =
960 W
Z = 1.33 kW
39o
VS = 13.3 V
39o
 

Floyd/Buchla
Chapter 8
Summary
The power triangle
Multiplying the voltage phasors by Irms gives the power triangle
(equivalent to multiplying the impedance phasors by I2). Apparent
power is the product of the magnitude of the current and magnitude of
the voltage and is plotted along the hypotenuse of the power triangle.
The rms current in the earlier example was 10 mA.
Show the power triangle.
Ptrue = 120 mW
Pr = 96
mVAR
x 10 mA
= 39o
Pa = 133 mVA
VR = 12 V
VC =
9.6 V
VS = 13.3 V
39o
 

Floyd/Buchla
Chapter 8
Summary
Power factor
The power factor is the relationship between the
apparent power in volt-amperes and true power in
watts. Volt-amperes multiplied by the power factor
equals true power.
Power factor is defined mathematically as
PF = cos 
The power factor can vary from 0 for a purely reactive
circuit to 1 for a purely resistive circuit.

Floyd/Buchla
Chapter 8
Summary
Apparent power
Apparent power consists of two components; a true
power component, that does the work, and a
reactive power component, that is simply power
shuttled back and forth between source and load.
Ptrue (W)
Pr (VAR)
Some components such
as transformers, motors,
and generators are rated
in VA rather than watts.
Pa (VA)


Floyd/Buchla
Chapter 8
10V dc
Vout
Vin
100 W
1 F

10V dc
0
10V dc
0
Summary
Frequency Response of RC Circuits
When a signal is applied to an RC circuit, and the output is taken
across the capacitor as shown, the circuit acts as a low-pass filter.
As the frequency increases, the output amplitude decreases.
Plotting the response:
Vout (V)
9.98
8.46
1.57
0.79
0.1 1 10 20 100
f (kHz)
9
8
7
6
5
4
3
2
1
1
ƒ = 1 kHz
8.46V rms
10V rms W
100
F

1.57V rms
10V rms
1
ƒ = 10 kHz
W
100
F

0.79V rms
10V rms
1
ƒ = 20 kHz
W
100
F


Floyd/Buchla
Chapter 8
Vin
10V dc
0
Vout
0V dc
10V dc 100 W
1 F

Summary
Frequency Response of RC Circuits
Reversing the components, and taking the output across the resistor as
shown, the circuit acts as a high-pass filter.
As the frequency increases, the output amplitude also increases.
Plotting the response:
ƒ = 100 Hz
0.63V rms
10V rms
100 W
1 F

ƒ = 1 kHz
5.32V rms
10V rms
100 W
1 F

ƒ = 10 kHz
9.87V rms
10V rms
100 W
1 F

Vout (V)
f (kHz)
9.87
5.32
0.63
0
0.01 0.1 1
10
9
8
7
6
5
4
3
2
1
10

Floyd/Buchla
Chapter 8
Quiz
3. A series RC circuit is driven with a sine wave. If you
measure 7.07 V across the capacitor and 7.07 V across the
resistor, the voltage across both components is
a. 0 V
b. 5 V
c. 10 V
d. 14.1 V

Floyd/Buchla
Chapter 8
Quiz
Answers:
1. a
2. b
3. c
4. c
5. d
6. a
7. d
8. d
9. a
10. b

Floyd/Buchla
Chapter 8
When a length of wire is formed into a coil., it
becomes a basic inductor. When there is current in
the inductor, a three-dimensional magnetic field is
created.
Summary
The Basic Inductor
A change in current
causes the magnetic
field to change. This in
turn induces a voltage
across the inductor that
opposes the original
change in current.
N
S

Floyd/Buchla
Chapter 8
Summary
Factors affecting inductance
Four factors affect the amount of inductance for a
coil. The equation for the inductance of a coil is
2
N A
L
l


where
L = inductance in henries
N = number of turns of wire
 = permeability in H/m (same as Wb/At-m)
l = coil length on meters

Floyd/Buchla
Chapter 8
Summary
What is the inductance of a 2 cm long, 150
turn coil wrapped on an low carbon steel core that
is 0.5 cm diameter? The permeability of low
carbon steel is 2.5 x10-4 H/m (Wb/At-m).
 
2
2 5 2
π π 0.0025 m 7.85 10 m
A r -
   
2
N A
L
l


22 mH
    
2 4 5 2
150 t 2.5 10 Wb/At-m 7.85 10 m
0.02 m
- -
 



Floyd/Buchla
Chapter 8
Summary
Inductive reactance
Inductive reactance is the opposition to
ac by an inductor. The equation for
inductive reactance is
The reactance of a 33 H inductor when a
frequency of 550 kHz is applied is 114 W
2π
L
X fL


Floyd/Buchla
Chapter 8
Summary
Inductive phase shift
When a sine wave
is applied to an
inductor, there is a
phase shift between
voltage and current
such that voltage
always leads the
current by 90o.
VL 0
0
90
I

Floyd/Buchla
Chapter 8
True Power: Ideally, inductors do not dissipate power.
However, a small amount of power is dissipated in
winding resistance given by the equation:
Ptrue = (Irms)2RW
Reactive Power: Reactive power is a measure of the rate
at which the inductor stores and returns energy. One form
of the reactive power equation is:
Pr=VrmsIrms
The unit for reactive power is the VAR.
Power in an inductor

Lecture 1.2.ppt

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