Transcript Introduction and Digital Images

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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
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
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,
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
2
3
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.
G
Conductance is the reciprocal of resistance.
Inductive susceptance is the reciprocal of
inductive reactance.
BL 
1
XL
Admittance is the reciprocal of impedance.
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   tan 1 
G
VS
G
BL 

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 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
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
R2
R1
yellow box are in series and
Z1
Z2
those in the green box are
L1
L2
also in series.
Z1  R12  X L21
and
Z 2  R22  X L22
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 volt-amperes and true power in watts.
Volt-amperes multiplied by the power factor equals
true power.
Power factor is defined as
PF = cos 
Apparent
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  G 2  BL 2
c. Y = G + BL
2
2
d. Y  G + BL
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