Transcript NTUST-EE-2013S

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• Course overview and information
09/16/2010
© 2010 NTUST
Sinusoidal Response of RL
•
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
Impedance of Series RL
Impedance of Series RL
• 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
It is convenient to reposition the
phasors into the impedance triangle.

R
Examples
Impedance of Series RL
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
 39
0.96 kW
1.2 kW
2
Z = 1.33 kW

39o
R = 1.2 kW
XL =
960 W
Analysis of Series RL
• 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 obtain the voltage phasors by simply multiplying the
impedance phasors by the current.
Examples
Analysis of Series RL
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
x 10 mA
=
XL =
960 W
VS = 13.3 V

39o
VR = 12 V
VL =
9.6 V
Phase Relationships
Variation of Phase Angle
•
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, the
X
Z
impedance triangle for an RL
circuit changes as illustrated
Z
X
here because XL increases with
increasing f. This determines
Z
X
the frequency response of RL



circuits.
3
2
1
1
2
3
R
L3
L2
L1
Variation of Impedance and Phase Angle
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)
f

VR
Vin
RL Lead Circuit
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
Vout
f
RL Lag Circuit
Examples
Sinusoidal Response of Parallel RL
• For parallel circuits, it is useful to review conductance,
susceptance and admittance, introduced in Chapter 10.
•
Conductance is the reciprocal of resistance.
•
Inductive susceptance is the reciprocal of
inductive reactance.
•
Admittance is the reciprocal of impedance.
G
BL 
1
XL
Y
1
Z
1
R
Admittance
Sinusoidal Response of Parallel RL
• 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
 BL 
  tan  
G 
1
G
VS
G
BL
BL
Y
Sinusoidal Response of Parallel RL
•
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
Y
Sinusoidal Response of Parallel RL
Draw the admittance phasor diagram for the circuit.
The magnitude of the conductance and susceptance are:
G
1
1
1
 0.629 mS

 1.0 mS BL 
2

10
kHz
25.3
mH



R 1.0 kW
Y  G 2 + BL 2 
1.0 mS
2
+  0.629 mS  1.18 mS
2
G = 1.0 mS
VS
f = 10 kHz
R
1.0 kW
L
25.3 mH
BL =
0.629 mS
Y=
1.18 mS
Analysis of Parallel RL Circuit
• Ohm’s law is applied to parallel RL circuits using
quantities of Y, V, and I.
Y
I
V
V
I
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.
Analysis of Parallel RL Circuit
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
x 10 V
=
IL =
6.29 mA
IR = 10 mA
IS =
11.8 mA
Phase Angle of Parallel RL
• Notice that the formula for inductive susceptance is the
reciprocal of inductive reactance. Thus BL and IL are
1
inversely proportional to f:
BL 
2 fL
• As frequency increases, BL and IL
decrease, so the angle between IR
and IS must decrease as well.

IL
IR
IS
Examples
Phase Relationships
Examples
Series-Parallel RL
•
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
R2
R1
box are in series and those in the
Z1
Z2
green box are also in series.
L
L
1
Z1  R12  X L21
and
Z 2  R22  X L22
2
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.
The Power
• 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
x 10 mA
=
XL =
960 W
VS = 13.3 V
39o
VR = 12 V
VL =
9.6 V
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
VL =
9.6 V
Pa = 133 mVA
Pr =
96 mVAR
39o
Ptrue = 120 mW
The Power Triangle
Examples
Power
• 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 voltamperes and true power in watts. Volt-amperes
multiplied by the power factor equals true power.
Power factor is defined as
PF = cos 
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.
Ptrue (W)
Pr (VAR)
Pa (VA)
Selected Key Terms
Inductive
susceptance (BL)
The ability of an inductor to permit
current; the reciprocal of inductive
reactance. The unit is the siemens (S).
Quiz
1. If the frequency is increased in a series RL circuit, the
phase angle will
a. increase
b. decrease
c. be unchanged
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
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 V
in
d. none of the above
Vout
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
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
Quiz
6. Susceptance is the reciprocal of
a. resistance
b. reactance
c. admittance
d. impedance
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  G2  BL 2
c. Y = G + BL
d. Y  G2 + BL 2
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
Quiz
9. The unit used for measuring true power is the
a. volt-ampere
b. watt
c. volt-ampere-reactive (VAR)
d. kilowatt-hour
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
Quiz
Answers:
1. a
6. b
2. a
7. d
3. c
8. d
4. a
9. b
5. c
10. a