Download Inductance in an AC Circuit

SERIES AC CIRCUITS
Graphs of sinusoidal functions of time
Alternating current
Resistance in an AC circuit
Inductance in an AC circuit
Capacitance in an AC circuit
Series LRC circuit
Power loss
Resonance
by Dr. John Dayton
AC CIRCUITS
Every slide contains valuable and need-to-know information
that has to be understood and retained before proceeding.
Throughout this PowerPoint a color scheme has been
employed. Current is depicted in purple, voltage across a
resistor in orange, voltage across an inductor in green,
voltage across a capacitor in blue, and generator voltage
in red.
I
VL
VR
VC
E
GRAPHS OF SINUSOIDAL FUNCTIONS OF TIME.
Important Terms:
Period, T
Amplitude, A
Angular frequency, w
Frequency, f
Phase shift, f
The graph is just a small piece of y  Asin wt  . A is the
amplitude, w is the angular frequency in radians/s. Related to w is
f, the frequency in Hz = s-1. Adding a phase angle, f, shifts the
graph. y  Asin wt  f  shifts the graph to the left if f is positive.
RMS – Root Mean Square, the Effective Value
(purple graph)
The additional graph
shown in green is:
y  A sin wt 
2
T
2
RMS – Root Mean Square, the Effective Value
T
The average value of any function is the area under its graph divided by the
width of the graph.
Over one period the average value of y  Asin wt  is 0 since there is as
much positive area above the horizontal axis as there is negative area below
the horizontal axis.
RMS – Root Mean Square, the Effective Value
A2
1
2
A2
T
Area = 2A2T
T
Examine the graphs of y 2  A2 sin2 wt  shown above. The two
shaded areas in the first period lying above the midpoint can be
cut off and used to fill in the troughs as shown. This completes
a rectangular area equal to 12 A2T , further shaded in pink. Thus
the average value of A2 sin2 wt  is 12 A2 over one period. We
can now define what is called the root mean square of y to be
yrms 
1
2
A
A 
2
2
The pulsating dots show the original and modified distributions
of area.
EXAMPLE: Given: y  5sin 12t  , identify the following:
the amplitude, A
the angular frequency, w
the frequency, f
the period, T
the rms value of y, yrms
The basic sinusoidal equation is y  Asin wt  . Using this it can be
seen that
A = 5 (which may contain units)
w = 12 rad/s
Then:
w
1
f 
 1.91Hz
T   0.524 s
2
f
yrms
A
5


 3.54
2
2
PHASORS
A phasor can be likened to a rotating vector. It has a specific
and constant length equal to the maximum value, amplitude, of
a sinusoidally varying quantity such as a current or a voltage.
The tail end of the phasor is fixed on the origin of an x-y
coordinate system and is the pivot of the phasor’s rotation. As
the phasor rotates about this pivot, it sweeps out the angle
w t + f and its projection on the y-axis corresponds to the value
of the quantity represented by the phasor.
y
y
t
phasor
PHASORS
The diagram shows two phasors with the same frequency but
out of phase with each other. The blue phasor leads the red
phasor by an angular amount equal to their phase difference.
You will note that the blue phasor crosses the +y axis before or
ahead of the red phasor, indicating it leads. You can also say
the red phasor lags the blue phasor. On the graphs you can
see that the peak in the blue quantity occurs before the peak in
the red quantity.
y
y
t
graphic from web
Alternating Current
Instantaneous current:
i  I sin(wt )
AC Power Supply
Instantaneous emf:
E  Emax sin wt 
PHASOR
i
I
I
Graph of i
wt
t
The time varying current, i, in this
phasor diagram is represented by
the projection of I on the y-axis.
Resistance in an AC Circuit
Maximum Voltage:
VR
Instantaneous Voltage:
vR  VR sin  wt 
Peak-to-Peak Voltage:
2VR
This phasor diagram shows VR
and I rotating at w. vR and i are
represented as projections on
the y axis.
The Current and Voltage are in
Phase.
Maximum Current:
RMS Current:
I rms
VR
I
R
I

2
Resistance in an AC Circuit
I, V
Imax
Graphs of i and vR
p  i R  I R sin wt
2
Pavg  I R  I
1
2
VR
t
2
I
I rms 
2
V  IR
VR , rms  I rms R
Effective current, Irms
Effective voltage, Erms
2
VR , rms
VR

2
2
2
rms
R
Resistance in an AC Circuit
EXAMPLE: An AC generator has a maximum emf of 240
volts at a frequency of 90 Hz. Connected to the generator
is a 500 W resistor. What are the rms current through the
resistor and the average power loss?
I rms
Erms
E
240V



 0.339 A
R
R 2 500W  2
PAVG  I
2
rms
R   0.33941A (500W)  57.6W
2
REACTANCE VS RESISTANCE
As current flows through a resistor it interacts with the material
of the resistor and experiences an opposition to its flow known
as resistance. This is expressed by Ohm’s Law.
V  IR
As a changing current flows through an inductor or to and
from a capacitor there are voltage changes that occur due to
the fields in the respective devices. There is a relation
between this voltage and the current that involves an
opposition to current flow due to these fields called reactance.
Reactance is designated by X and Ohm’s Law can be written:
V  IX
Both resistance and reactance are measured in ohms, W, but
there is no power loss due to reactance as there is with
resistance.
Inductance in an AC Circuit
Reactance is an effective resistance to
the flow of current due to the presence
of electric or magnetic fields.
Inductance in an AC Circuit
i
vL  L
t
  I sin  wt  
vL  L
 L  wI cos  wt  
t
vL  wLI sin  wt  2   VL sin  wt  2 
VL  IX L
Inductive Reactance:
X L  wL
XL
w
Inductive Reactance
Inductive reactance increases linearly with frequency.
Inductance in an AC Circuit
This phasor diagram shows VL
and I rotating at w. vL and i are
represented as projections on
the y axis.
I, V
Graphs of i and vL
I
VL
t
VL leads I by 90o
The current through an inductor lags the voltage across it by
90o. The voltage peak across the inductor will be seen T/4
seconds before the current peak. (T = period = 1/f).
The flashing yellow dots on the graphs flash as the respective
phasor crosses the +y-axis indicating maximum value.
Inductance in an AC Circuit
EXAMPLE: A 300 peak voltage ac generator operating at 120
Hz is connected to a 15,000 turn solenoid of length 5.0 cm. and
radius 1.0 cm. (A) what is the inductance of the solenoid. (B)
What is the inductive reactance of the circuit? (C) What is the
rms current in the circuit? (D) At t = T/3 what is the stored
energy in the inductor?
L
0 N A
2
 4  107 TmA  15000  .01m 
2

2
 1.78H
l
.05m
X L  w L  2 fL  2 120 Hz 1.7765H   1.34  103 W
I rms 
Erms
E
300V


 0.158 A
3
X L X L 2 (1.33945  10 W) 2
E
 T E
 2 T  E
 2 
sin  w  
sin 

sin



XL
 3  XL
 T 3  XL
 3 
300V

  2 
i
 sin 
  0.19397 A
3
 1.33945  10 W   3 
i  I sin wt  
U  12 Li 2  12 (1.7765H )(0.19397 A)2  0.0334 J
Capacitance in an AC Circuit
Capacitance in an AC Circuit
q
vC 
C
q
i
t
q  it
q   q   it   I sin wt  t
q
 I cos wt 

I sin wt  2 
w
w
I sin wt  2 
vC 
 VC sin(wt  2 )
wC
Capacitance in an AC Circuit
I sin wt  2 
vC 
 VC sin(wt  2 )
wC
VC  IX C
XC
1
XC 
wC
Capacitive Reactance
w
Capacitive reactance decreases
with the reciprocal of the frequency.
I, V
I
Capacitance in an AC Circuit
Graphs of i and vC
t
VC
The current to/from a
capacitor leads the voltage
by 90o. The current peak
will be seen T/4 seconds
before the voltage peak.
This phasor diagram shows VC and I
rotating at w. vC and i are represented
as projections on the y axis.
VC lags I by 90o
Capacitance in an AC Circuit
EXAMPLE: An AC generator operating at 140 V and 75 Hz
is connected to a 12.7 F capacitor. (A) What is the
capacitive reactance? (B) What is the rms current in the
circuit? (C) What is the average power loss?
1
1
1
2
XC 



1.67

10
W
6
wC 2 fC 2  75Hz  12.7 10 F 
I rms
Erms
E
140V



 0.592 A
2
X C X C 2 (1.6709 10 W)  2
Since there is no resistance, R, there is no
power loss in the circuit.
Series LRC Circuits
The current i in the circuit
is the same at all points at
any instant of time and
varies sinusoidally.
i  I sin  wt 
The instantaneous voltage across the resistor is in phase with the
current.
The instantaneous voltage across the inductor leads the current
by 90o.
The instantaneous voltage across the capacitor lags the current
by 90o.
Series LRC Circuits
I, VR, VL, VC, and Emax are all maximum values.
i, vR, vL, vC , E are all time-varying values.
Color distinction has been used
throughout for convenience.
Graphs of i, v
I
VR
VC
Emax
VL
t
Series LRC Circuits
Phasors shown at t = 0. Click to see phasor rotation. View
time-varying voltages and current as projections on the yaxis.
Phasors
Emax
f
E  vR  vC  vL  Emax sin(wt  f )
Series LRC Circuits
Series LRC Circuits
Emax Phase:
f
VL-VC
VR
VL  VC IX L  IX C X L  X c
tan(f ) 


VR
IR
R
XL  Xc
tan(f ) 
R
1  X L  X c 
f  tan 

R


Series LRC Circuits
Impedance, Z:
VL-VC
f
IZ
Z
f
f
I(XL-XC)
R
IR
VR
XL-XC
Emax  V  (VL  VC )  I R   IX L  IX C 
2
R
2
2
2
2
Emax  I R  ( X L  X C )  IZ
2
2
Z  R  (X L  XC )
2
2
Emax
I
Z
Impedance, Z, is the effective resistance due to a combination
of resistance and reactance and is measured in ohms, W.
Series LRC Circuits
In an inductive circuit the generator voltage leads the current.
f
VL-VC
f is positive, XL > XC
VR
In a capacitive circuit the generator voltage lags the current.
VR
f
VL-VC
f is negative, XC > XL
EXAMPLE:
Series LRC Circuits
A series RLC circuit with R=300 W, L = .75 H, C = 4.6F, f = 60 Hz
and Emax = 120V.
w  2f  376.99 s 1
X L  wL  282.74W
1
XC 
 576.65W
wC
Emax
I
 0.286 A
Z
Z  R 2  ( X L  X C ) 2  419.98W
 X L  XC
f  tan 
R

Since the phase angle f is
negative the circuit is capacitive
and the generator emf peak
lags the current peak by 44.4o.
VR  IR  85.8V
VL  IX L  80.9V
VC  IX C  164.9V

o
  44.4

1
VL
VR
f
Emax
R
VC
f
Z
XL-XC
Series LRC Circuits
POWER LOSS
No power loss occurs at a pure inductor or a pure capacitor.
Power loss occurs at a resistor where energy is lost as thermal
energy.
P  I R  I rmsVR, rms  I rmsErms cos f 
2
rms
P  I rmsErms cos f 
cos(f is called the power factor.
Series LRC Circuits
RESONANCE
Generally:
I rms
At resonance:
Erms


Z
X L  XC
1
wo 
LC
A maximum value of the
current and power loss
occurs when XL = XC, w = wo.
Erms
R 2  ( X L  X C )2
1
 wo L 
woC
resonance angular
frequency
Z  R, f  0
Erms
I rms 
R
o
Series LRC Circuits
Resonance is related to the natural frequency of oscillation
between a fully charged capacitor and an inductor connected in
parallel. Initially the capacitor discharges through the inductor
transferring all its stored energy from its electric field to an
induced magnetic field inside the inductor. As the magnetic field
collapses, its energy flows back to the capacitor by recharging
the capacitor and reestablishing the electric field. This cycle
repeats with each cycle reversing the polarity of the fields. the
natural frequency is the same as the resonance frequency.
w0 
1
LC
image from Wikipedia
Series LRC Circuits
RESONANCE
A resonance:
w  wo 
1
LC
X L  XC
ZR
f 0
If there is no resistance,
then Z = 0
0
Emax
I peak 
R
Erms
I peak , rms 
R
Series LRC Circuits
C = 12.0 F
R = 100 W
L = 4.0 H
Emax = 200 V
Current, I, and power, P,
vary with w and reach
peak values at resonance
when w = wo.
wo 
1
= 144.338 rads
LC
Power
Current
wo
w
wo
w
IMPORTANT GRAPHS
R
At resonance XL = XC, Z = R, f = 0, I and P peak.
Series LRC Circuits
EXAMPLE: Given the RLC series circuit with R = 150 W, L = 1.25
H, C = 8.4 F and Emax = 150 V. (A) Calculate the resonance
frequency, wo; and the peak current and power loss at resonance.
(B) Calculate the impedance, current, power loss, and phase
angle at w = .5wo.
1

LC
1
A
wo 
I
Emax Emax 150V


 1.00 A
Z
R
150W
1.25H  8.4  106 F 
 308.607 rads
I 2 R 1.00 A (150W)
2
P  I rms R 

 75W
2
2
2
Continued on next slide.
B
w  12 wo  12 (308.607 rads )  154.303 rads
1 

Z  R  wL 
wC 

2
2


1
Z  (150W) 2   154.303 rads  (1.25H ) 

6
rad
(154.303
)(8.4

10
F
)
s


2
Z  597.764W
E
150V
I  max 
 0.251A
Z
597.764W
I 2 R  0.25094 A (150W)
P

 4.72W
2
2
1 

w
L



1
w
C
f  tan 

R




2
1


rad
154.303
(1.25
H
)



s
6


rad
154.303
8.4

10
F




s
  75.5o
f  tan 1 


150W




FILTER CIRCUITS
VIN
VIN
VOUT
VOUT
A low-pass filter:
VOUT IX C
XC
1



2
2
2
VIN
IZ
R  XC
1  w RC 
A high-pass filter:
VOUT IR
R



2
2
VIN
IZ
R  XC
1
1
1
w RC 
2
FILTER CIRCUITS
VIN
VOUT
VOUT IX C
XC
1
1



 2
VIN
IZ
X L  X C  X L  w LC  1


 X C 1 
1
2
 wo LC  1
At resonance: w  wo 
LC
VOUT

VIN
 w 2  wo2 
Exercise: show that Z  L 

 w 
END OF PRESENTATION