AC Circuits Analysis 2 RL (Lecture notes)

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chapter 12
RL Circuits
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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.
VL
VR
V R lags V S
VL lead s VS
R
L
VS
I
I lags V S
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Impedance of series RL circuits
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.
 XL 

 R 
  tan 1 
Z
Z is the diagonal
Z
XL
XL

R

R
It is convenient to reposition the
phasors into the impedance triangle.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Impedance of series RL circuits
Sketch the impedance triangle and show the
values for R = 1.2 kW and XL = 960 W.
Z
1.2 kW 
2
+  0.96 kW 
 1.33 kW
  tan 1
0.96 kW
1.2 kW
 39
Electronics Fundamentals 8th edition
Floyd/Buchla
2
Z = 1.33 kW

39o
XL =
960 W
R = 1.2 kW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Analysis of series RL circuits
Ohm’s law is applied to series RL circuits using
quantities of Z, V, and I.
V  IZ
I
V
Z
Z
V
I
Because I is the same everywhere in a series circuit,
you can obtain the voltage phasors by simply
multiplying the impedance phasors by the current.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Analysis of series RL circuits
Assume the current in the previous example is 10 mArms.
Sketch the voltage phasors. The impedance triangle from
the previous example is shown for reference.
The voltage phasors can be found from Ohm’s
law. Multiply each impedance phasor by 10 mA.
Z = 1.33 kW

39o
R = 1.2 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XL =
960 W
VS = 13.3 V

39o
VL =
9.6 V
VR = 12 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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.
Increasing f
As frequency changes,
X
Z
the impedance triangle
for an RL circuit changes
Z
X
as illustrated here
because XL increases with
Z
X
increasing f. This



determines the frequency
R
response of RL circuits.
3
2
1
1
Electronics Fundamentals 8th edition
Floyd/Buchla
2
L3
L2
L1
3
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Phase shift
For a given frequency, a series RL circuit can be used to
produce a phase lead by a specific amount between an
input voltage and an output by taking the output across
the inductor. This circuit is also a basic high-pass filter, a
circuit that passes high frequencies and rejects all others.
R
Vout
Vin
Vout
f
Vin
L
Vout
(phase lead)
Vin
f

Electronics Fundamentals 8th edition
Floyd/Buchla
VR
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Phase shift
Reversing the components in the previous circuit
produces a circuit that is a basic lag network. This circuit
is also a basic low-pass filter, a circuit that passes low
frequencies and rejects all others.
L
Vin
VL
R
Vin
Vin
Vout
f (phase lag)
Vout
f
Vout
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Sinusoidal response of parallel RL circuits
For parallel circuits, it is useful to review conductance,
susceptance and admittance, introduced in Chapter 10.
G
Conductance is the reciprocal of resistance.
Inductive susceptance is the reciprocal
of inductive reactance.
BL 
1
R
1
XL
1
Admittance is the reciprocal of impedance. Y 
Z
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Sinusoidal response of parallel RL circuits
In a parallel RL circuit, the admittance phasor is the sum
of the conductance and inductive susceptance phasors.
The magnitude of the susceptance is Y  G 2 + BL 2
The magnitude of the phase angle is   tan 1 
G
VS
G
BL
BL
Electronics Fundamentals 8th edition
Floyd/Buchla
BL 

G
 
Y
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Sinusoidal response of parallel RL circuits
Some important points to notice are:
G is plotted along the positive x-axis.
BL is plotted along the negative y-axis.
 BL 

G
  tan 1 
Y is the diagonal
G
VS
G
BL
BL
Electronics Fundamentals 8th edition
Floyd/Buchla
Y
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Sinusoidal response of parallel RL circuits
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
1
1
1
B

 0.629 mS
G 
 1.0 mS
L
2 10 kHz  25.3 mH 
R 1.0 kW
Y  G 2 + BL 2 
1.0 mS +  0.629 mS  1.18 mS
2
2
G = 1.0 mS
VS
f = 10 kHz
Electronics Fundamentals 8th edition
Floyd/Buchla
R
1.0 kW
L
25.3 mH
BL =
0.629 mS
Y=
1.18 mS
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Analysis of parallel RL circuits
Ohm’s law is applied to parallel RL circuits using
quantities of Y, V, and I.
I
Y
V
I
V
Y
I  VY
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.
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Analysis of parallel RL circuits
Assume the voltage in the previous example is 10 V.
Sketch the current phasors. The admittance diagram
from the previous example is shown for reference.
The current phasors 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
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 V
=
IL =
6.29 mA
IR = 10 mA
IS =
11.8 mA
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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: BL  1
2 fL
As frequency increases, BL
and IL decrease, so the angle
between IR and IS must
decrease as well.

IL
Electronics Fundamentals 8th edition
Floyd/Buchla
IR
IS
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Series-Parallel RL circuits
Series-parallel RL circuits are combinations of both series and
parallel elements. The solution of these circuits is similar to
resistive combinational circuits but you need to combine reactive
elements using phasors.
The components in the
yellow box are in series and
those in the green box are
also in series.
Z1  R12  X L21
and
Z 2  R22  X L22
Electronics Fundamentals 8th edition
Floyd/Buchla
R2
R1
Z1
L1
Z2
L2
The two boxes are in parallel. You can
find the branch currents by applying
Ohm’s law to the source voltage and
the branch impedance.
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
The power triangle
Recall that in a series RC or RL circuit, you could
multiply the impedance phasors by the current to
obtain the voltage phasors. The earlier example from
this chapter is shown for review:
Z = 1.33 kW
39o
R = 1.2 kW
Electronics Fundamentals 8th edition
Floyd/Buchla
x 10 mA
=
XL =
960 W
VS = 13.3 V
39o
VL =
9.6 V
VR = 12 V
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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.
x 10 mA =
VS = 13.3 V
39o
VR = 12 V
Electronics Fundamentals 8th edition
Floyd/Buchla
VL =
9.6 V
Pa = 133 mVA
Pr =
96 mVAR
39o
Ptrue = 120 mW
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Power factor
The power factor was discussed in Chapter 15 and
applies to RL circuits as well as RC circuits. Recall
that it 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 as
PF = cos 
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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.
Power factor corrections
for an inductive load
(motors, generators, etc.)
are done by adding a
parallel capacitor, which
has a canceling effect.
Electronics Fundamentals 8th edition
Floyd/Buchla
(W)
PPatrue
(VA)
Pr (VAR)
Pt (W)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Frequency Response of RL Circuits
Series RL circuits have a frequency response similar to series RC
circuits. 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
10 V rms
0
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
Vout (V)
9.98
Plotting the response:
8.46
1.57
0.79
9
8
7
6
5
4
3
2
1
0.1
Electronics Fundamentals 8th edition
Floyd/Buchla
1
10 20
100
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Frequency Response of RL Circuits
Reversing the position of the R and L components, produces the
high-pass response. The output is taken across the inductor.
Vin
Vout
10VV rms
rms
10
10 V dc
10 V rms
0
10 V dc
100
100 W
W
100
W
ƒ
=
100
Hz
1
kHz
ƒ = 10 kHz
9.87 V rms
5.32 V rms
0.63 V rms
10
10
mH
10mH
mH
10
mH
0 V dc
Vout (V)
9.87
Plotting the response:
5.32
0.63
Electronics Fundamentals 8th edition
Floyd/Buchla
10
9
8
7
6
5
4
3
2
1
0
0.01
0.1
1
10
f (kHz)
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Quiz
1. If the frequency is increased in a series RL circuit, the
phase angle will
a. increase
b. decrease
c. be unchanged
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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
Vin
Vout
d. none of the above
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Quiz
4. In a series RL circuit, the phase angle can be found from
the equation
 XL 

 R 
a.
  tan 1 
b.
 VL 
  tan  
 VR 
1
c. both of the above are correct
d. none of the above is correct
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Quiz
6. Susceptance is the reciprocal of
a. resistance
b. reactance
c. admittance
d. impedance
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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  BL 2
c. Y = G + BL
d. Y  G 2 + BL 2
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Quiz
9. The unit used for measuring true power is the
a. volt-ampere
b. watt
c. volt-ampere-reactive (VAR)
d. kilowatt-hour
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
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
Electronics Fundamentals 8th edition
Floyd/Buchla
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
Chapter 12
1
Quiz
Answers:
Electronics Fundamentals 8th edition
Floyd/Buchla
1. a
6. b
2. a
7. d
3. c
8. d
4. a
9. b
5. c
10. a
© 2010 Pearson Education, Upper Saddle
River, NJ 07458. All Rights Reserved.
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