Transcript Chapter Title

DC/AC Fundamentals: A Systems
Approach
Thomas L. Floyd
David M. Buchla
RL Circuits
Chapter 12
Ch.12 Summary
Sinusoidal Response of RL Circuits
When both resistance and inductance 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.
VS
VR
VS
VL
VR lags VS
VL leads VS
R
L
VS
VS
I
I lags VS
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
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Ch.12 Summary
Series RL Circuit Impedance
In a series RL circuit, the total impedance is the
phasor sum of R and XL.
R is plotted along the positive x-axis.
XL is plotted along the positive y-axis.
Z is the diagonal.
It is convenient
to reposition the
phasors into an
impedance
triangle.
XL
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
Z
q  tan
1
XL
R
XL
Z
q
q
R
R
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Ch.12 Summary
Series RL Circuit Impedance
Sketch the impedance triangle and show the values
for R = 1.2 kW and XL = 960 W.
Z  R 2  X L2  (1.2 kW)2  (0.96 kW)2  1.33 kΩ
X 
q  tan 1 L 
 R 
1  0.96 kΩ 
 tan 

 1.2 kΩ 
 39
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
Z = 1.33 kW
XL = 0.96 kW
q = 39o
R = 1.2 kW
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Ch.12 Summary
Series RL Circuit Analysis
Ohm’s law is applied to series RL circuits using
quantities of Z, V, and I.
V  IZ
V
I
Z
V
Z
I
Because I is the same everywhere in a series circuit,
you can multiply the impedance phasor values by
the circuit current to obtain the voltage phasor
values.
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Series RL Circuits Analysis
Assume the current in the previous example is 10 mA. Sketch
the voltage phasors. (The impedance triangle from the previous
example is shown for reference.)
The voltage phasor diagram can be found using Ohm’s law.
Multiply each impedance phasor by 10 mA (as shown below):
Z = 1.33 kW
XL = 0.96 kW
x 10 mA
=
VS = 13.3 V
VL = 9.6 V
q = 39o
q = 39o
R = 1.2 kW
VR = 12 V
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Phase Angle Vs. 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 RL
circuit changes (as illustrated
here) because XL is directly
proportional to f. This
determines the frequency
response of RL circuits.
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
R
R
q1
q3
q2
q2
Increasing f
q1
Z
Z33
q3
X
XC3
ff 3
3
X
XC2
f 22
X
XC1
ff1
Z
Z22
Z
Z11
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Ch.12 Summary
Phase Shift
A series RL circuit can be used to produce a specific phase lead
between an input voltage and an output by taking the output
across the inductor. This circuit is a basic high-pass filter, a
circuit that passes high frequencies and rejects all others. This
filter passes frequencies that are above a specific frequency,
called the cutoff frequency.
V
Vin
R
Vout
(phase lead)
Vin
L
Vin
Vout
Vout
VR
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
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Ch.12 Summary
Phase Shift
Reversing the components in the previous circuit produces
a circuit that is a basic lag network. This circuit is a lowpass filter, a circuit that passes low frequencies and
rejects all others. This filter passes low frequencies up to a
frequency called the cutoff frequency.
L
VL
Vin
Vin
Vout
Vin
R
Vout
(phase lag)
f
Vout
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
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Ch.12 Summary
AC Response of Parallel RL Circuits
For parallel circuits, it is useful to review conductance,
susceptance and admittance, introduced in Chapter 10.
Conductance is the reciprocal of
resistance.
1
G
R
1
XL
Inductive susceptance is the
reciprocal of inductive reactance.
BL 
Admittance is the reciprocal of
impedance.
1
Y 
Z
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
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Ch.12 Summary
AC Response of Parallel RL Circuits
In a parallel RL circuit, the admittance phasor is the sum of the
conductance and capacitive susceptance phasors:
Y  G 2  BL2
From the
diagram, the
phase angle
is:
1  BL 
q
VS
q  tan  
G
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
G
G
BL
BL
Y
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Ch.12 Summary
AC Response of Parallel RL Circuits
Draw the admittance phasor diagram for the circuit below.
The magnitudes of conductance, susceptance, and admittance
are:
1
1
1
1
B


 .629 mS
G 
 1 mS
C
X L 2(10 kHz)(25.3 mH)
R 1 kΩ
Y  G 2  BL2  (1 mS) 2  (0.629 mS) 2  1.18 mS
q
VS
f =10 kHz
R
1 kW
L
25.3 mH
G=
1.0 mS
Y=
1.18 mS
BL =
0.629 mS
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Analysis of Parallel RL Circuits
Ohm’s law is applied to parallel RL circuits using
quantities of Y, V, and I.
I
V
Y
I  VY
I
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.
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Analysis of Parallel RL 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.
G = 1.0 mS
BL =
0.629 mS
Y=
1.18 mS
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
IR = 10 mA
x 10 V
=
IL =
6.29 mA
IS =
11.8 mA
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Ch.12 Summary
Phase Angle of Parallel RL Circuits
Notice that the formula for inductive susceptance is the
reciprocal of inductive reactance. Thus BL and IL are
inversely proportional to f:
1
BL 
2fL
As frequency increases, BL
and IL decrease, so the angle
between IR and IS must
decrease as well.
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
q
IL
IR
IS
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Ch.12 Summary
Series-Parallel RL Circuits
Series-parallel RL circuits are combinations of both series and
parallel elements. These circuits can be solved by methods from
Z1
series and parallel circuits.
Z2
For example, the
components in the
green box are in series:
RR1
L1 C
RR22
L 2C2
Z1  R12  X L2
The components in
the yellow box are in
parallel:
R2 X L 2
Z2 
R22  X L22
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
The total impedance can be found by
converting the parallel components to
an equivalent series combination,
then adding the result to R1 and XL1
to get the total reactance.
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Ch.12 Summary
The Power Triangle
As shown earlier, you can multiply the impedance phasors
for a series RL circuit by the current to obtain the voltage
phasors. The earlier example is shown below for review:
Z = 1.33 kW
XL = 0.96 kW
x 10 mA
=
VS = 13.3 V
VL = 9.6 V
q = 39o
q = 39o
R = 1.2 kW
VR = 12 V
Multiplying each value in the left-hand triangle gives you the
corresponding value in the right-hand triangle.
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
The Power Triangle (Cont’d)
Multiplying the voltage phasors by Irms (10 mA) gives the
power triangle values (because P = V  I ). 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.
x 10 mA =
VS = 13.3 V
VL = 9.6 V
q = 39o
VR = 12 V
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
Pa = 133 mVA
q  39o
Pr =
96 mVAR
Ptrue = 120 mW
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Ch.12 Summary
Power Factor
The power factor was introduced in Chapter 10 and applies
to RL circuits (as well as RC circuits). Recall that it is the
relationship between the apparent power (in VA) and true
power (in W). True power equals the product of Voltamperes and power factor.
Power factor can be determined using:
PF  cos q
Power factor can vary from 0 (for a purely reactive circuit) to
1 (for a purely resistive circuit).
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Apparent Power
Apparent power consists of two components; the true power
component, which does the work, and a reactive power
component, that is simply power shuttled back and forth
between source and load.
Components such as
transformers, motors, and
generators are rated in VA
rather than watts.
Pa (VA)
Pr ( VAR)
q
Ptrue (W)
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Frequency Response of RL Circuits
The response of a series RL circuit is similar to that of a series RC
circuit. In the case of the low-pass response shown here, the output
is taken across the resistor.
Vout
Vin
10
10V
V rms
rms
10 V dc
0
10 V rms
10 V dc
10
mH
10mH
mH
10
ƒ ==
110kHz
ƒƒ
kHz
= 20
kHz
8.46
1.57 V
V rms
rms V rms
10 V dc 0.79
100
W
100W
W
100
100
W
0
V out (V)
Plotting the response:
9.98
9.98
8.46
8.46
9
8
7
6
5
4
3
1.57
1.57
0.79
0.79
2
1
0.1
0.1
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
11
10
10 20
100
f (kHz)
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Ch.12 Summary
Frequency Response of RL Circuits
Reversing the position of the R and L components produces a high-pass
response. In this case, the output is taken across the inductor.
Vin
Vout
10VV rms
rms
10
10 V dc
0
10 V rms
10 V dc
100
100 W
W
100
W
ƒ
=
100
Hz
1
kHz
ƒ = 10 kHz
10
10
mH
10mH
mH
10
mH
9.87 V rms
5.32 V rms
0.63 V rms
0 V dc
Vout (V)
9.87
9.87
Plotting the response:
5.32
5.32
10
9
8
7
6
5
4
3
0.63
0.63
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
2
1
0
0.01
0.1
0.1
11
10
10
f (kHz)
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Ch.12 Summary
Key Terms
Inductive The ability of an inductor to permit
susceptance (BL) current; the reciprocal of inductive
reactance, measured in siemens (S).
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
1. If the frequency is increased in a series RL circuit,
the phase angle will
a. increase
b. decrease
c. remain unchanged
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
2. If you multiply each of the impedance phasors in a
series RL circuit by the current, the result is the
a. voltage phasors
b. power phasors
c. admittance phasors
d. none of the above
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
3. For the circuit shown, the output voltage
a. is in phase with the input voltage
b. leads the input voltage
c. lags the input voltage
d. leads the resistor voltage
Vin
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
Vout
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Ch.12 Summary
Quiz
4. Which of the equations below can be used to
calculate the phase angle in a series RL circuit?
a. q  tan 1 X L R 


b. q  tan 1VL V 
R

c. both of the above are correct
d. none of the above is correct
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
5. In a series RL circuit, if the inductive reactance is
equal to the resistance, the source current will lag
the source voltage by
a. 0o
b. 30o
c. 45o
d. 90o
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
6. Susceptance is the reciprocal of
a. resistance
b. reactance
c. admittance
d. impedance
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
7. In a parallel RL circuit, the magnitude of the
admittance can be expressed as
a. Y 
1
1
1

G BL
b. Y  G 2  BL2
c. Y  G  B L
d. Y  G 2  BL2
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
8. If you increase the frequency in a parallel RL
circuit,
a. the total admittance will increase
b. the total current will increase
c. both a and b
d. none of the above
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
9. The unit used for measuring true power is the
a. volt-ampere
b. watt
c. volt-ampere-reactive (VAR)
d. kilowatt-hour
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Quiz
10. A power factor of zero implies that the
a. circuit is entirely reactive
b. reactive and true power are equal
c. circuit is entirely resistive
d. maximum power is delivered to the load
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved
Ch.12 Summary
Answers
1. a
6. b
2. a
7. d
3. c
8. d
4. a
9. b
5. c
10. a
DC/AC Fundamentals: A Systems Approach
Thomas L. Floyd
© 2013 by Pearson Higher Education, Inc
Upper Saddle River, New Jersey 07458 • All Rights Reserved