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AC Circuits
w
i m wL

im
wC

m

im R
iron


V1
V2
N1
(primary)
N2
(secondary)
Physics 1304: Lecture 18, Pg 1
Lecture Outline


Driven Series LCR Circuit:
• General solution
• Resonance condition
» Resonant frequency
» “Sharpness of resonance” = Q
• Power considerations
» Power factor depends on impedance
Transformers
• Voltage changes
• Faraday’s Law in action gives induced primary current.
• Power considerations
Text Reference: Chapter 33.4-6
Physics 1304: Lecture 18, Pg 2
Phasors


i R  m sin wt
R
Q
  m sin wt
C

i C  wC m cos wt
di L
  m sin wt
dt

• R: V in phase with i
VR  Ri R   m sin wt
• C: V lags i by 90
VC 
• L: V leads i

by 90
VL  L
iL  
m
cos wt
wL
A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I)
and which rotates counterclockwise in a 2-d plane with angular velocity w. Recall
uniform circular motion:
x  r coswt
y  r sinwt
The projections of r (on
the vertical y axis) execute
sinusoidal oscillation.
y
y
w
x
Physics 1304: Lecture 18, Pg 3
Suppose:
xx
00,,
r1
r1
nn
.... r1
r1 Phasors
for L,C,R
11

V
i
R
V
f(f(xx))000
VR  Ri m sinwt
xx
00
22
44
66
1
i m cos wt
wC
x
xx
f( x ) 00
V
r1
0 , .. r1
n 1
wt
C
V
2
4
V
i
f( x ) 0
0
V
0
C
66
x
1.01 1
w
i
0
1.01 1
VL  wLi m cos wt
R
t
i
VC  
w
wt
r1
00,, r1 .... r1
r1
n 11
11
i
L
w
i
wt
L
2
4
6
Physics 1304: Lecture 18, Pg 4
Series LCR
AC Circuit
R
•
Back to the original problem: the loop
equation gives:
L
d 2Q
dt 2

Q
dQ
R
  m sin wt
C
dt
C


L
•
Assume a solution of the form:

Here all unknowns, (im,) , must be found from the loop eqn; the initial conditions
have been taken care of by taking the emf to be:   m sinwt.
•
To solve this problem graphically, first write down expressions
for the voltages across R,C, and L and then plot the appropriate
phasor diagram.
i  i m sin(wt  )
Physics 1304: Lecture 18, Pg 5
Phasors: LCR
• Given:    m sin wt
• Assume:
R

C
L



w
i m wL

From these equations, we can draw the phasor
diagram to the right.

This picture corresponds to a snapshot at t=0. The
projections of these phasors along the vertical axis
are the actual values of the voltages at the given
time.

im
wC

m

im R
Phasors: LCR
w
i m XL
R
C


m

L


i m XC
•
im R
The phasor diagram has been relabeled in terms of the
reactances defined from:
X L  wL
XC 
1
wC
The unknowns (im,) can now be solved for
graphically since the vector sum of the voltages
VL + VC + VR must sum to the driving emf .
Physics 1304: Lecture 18, Pg 7
Lecture 20, ACT 3

A driven RLC circuit is connected as shown.
For what frequencies w of the voltage
source is the current through the resistor
largest?
(a) w small
(b) w large (c) w 

R

w
1
LC
L
C
Physics 1304: Lecture 18, Pg 8
Conceptual Question

A driven RLC circuit is connected as shown.
For what frequencies w of the voltage
source is the current through the resistor
largest?
(a) w small
(b) w large (c) w 

R

w
1
LC
C
• This is NOT a series RLC circuit. We cannot blindly apply our
techniques for solving the circuit. We must think a little bit.
• However, we can use the frequency dependence of the impedances
(reactances) to answer this question.
• The reactance of an inductor = XL = wL.
• The reactance of a capacitor = XC = 1/(wC).
• Therefore,
• in the low frequency limit, XL  0 and XC   .
• Therefore, as w  0, the current will flow mostly through the
inductor; the current through the capacitor approaches 0.
• in the high frequency limit, XL   and XC  0 .
• Therefore, as w  , the current will flow mostly through the
capacitor, approaching a maximum imax = /R.
L
Phasors:LCR
i m XL
m




i m XC
X L  wL
im R
i m (XL-X C)
m

X L  XC
tan  
R
im R
XC 
1
wC

 2m  i 2m R 2   X L  XC 
2


Z  R 2   X L  XC 
2
im 
m

 m
2
Z
R 2   X L  XC 
Physics 1304: Lecture 18, Pg 10
Phasors:Tips
i m XL
y
• This phasor diagram was drawn as a
snapshot of time t=0 with the voltages
being given as the projections along the
y-axis.
• Sometimes, in working problems, it is
easier to draw the diagram at a time when
the current is along the x-axis (when i=0).


i m XC
imXL
m

m

imR
imXC
“Full Phasor Diagram”
x
im R
From this diagram, we can also create a
triangle which allows us to calculate the
impedance Z:
Z
X L  XC

R
“ Impedance Triangle”
Phasors:LCR
We have found the general solution for the driven LCR circuit:
i  i m sin(wt  )
imXL
m

w
the loop
eqn
tan  


  i mZ

im  m
Z
imR
imXC
XL
Z
X L  wL

1
XC 
wC
Z  R 2   X L  XC 
X L  XC
R
2
XL - XC
R
XC
Physics 1304: Lecture 18, Pg 12
Lagging & Leading
The phase  between the current and the driving emf depends on the
relative magnitudes of the inductive and capacitive reactances.
X L  wL
1
XC 
wC
X  XC
tan   L
R

im  m
Z
XL
Z
XL
XL


R
XC
XL > XC
>0
current
LAGS
applied voltage
Z
R
Z
XC
XL < XC
<0
current
LEADS
applied voltage
R
XC
XL = XC
=0
current
IN PHASE
applied voltage
Conceptual Question

The series LCR circuit shown is driven by a
generator with voltage  =  msinwt. The time
dependence of the current i which flows in the
circuit is shown in the plot.
How should w be changed to
1A bring the current and driving
voltage into phase?
(b) decrease w
(a) increase w
(c) impossible
1B • Which of the following phasors represents the current i at t=0?
(c)
(b) i
(a)
i
i
Physics 1304: Lecture 18, Pg 14
Lecture 21, ACT 1

The series LCR circuit shown is driven by a
generator with voltage  =  msinwt. The time
dependence of the current i which flows in the
circuit is shown in the plot.
xx
r1
.... r1
nn
im
11
1.01
R
C
How should w be changed to
1A bring the current and driving
voltage into phase?
(a) increase w
00,


L
f(
f(ixx)) 00
0
1.01-i 1
m
1
00
.53
(b) decrease w

io
2
2
t
4 4
xx
6 6
6.81
(c) impossible
• From the plot, it is clear that the current is LEADING the applied voltage.
XL
Therefore, the
phasor diagram must
look like this:
i

XC
Therefore,
XC  X L
• To bring the current into phase
with the applied voltage, we need
to increase XL and decrease XC.
• Increasing w will do both!!
Physics 1304: Lecture 18, Pg 15
Lecture 21, ACT 1
x

The series LCR circuit shown is driven by a
generator with voltage  =  msinwt. The time
dependence of the current i which flows in the
circuit is shown in the plot.
How should w be changed to
1A bring the current and driving
voltage into phase?
(a) increase w
0,
r1
.. r1
n
i1
1.01 m
R
C


(b) decrease w
io
L
f(ix ) 00
1.01-i 1
m
0
.53
2
t
4
x
(c) impossible
1B • Which of the following phasors represents the current i at t=0?
w
(c)
(b) i
(a)
i
w
i
• The projection
of i along the
vertical axis is
negative here.
• no way jose
• The sign of i is
correct at t=0.
• However, it
soon will become
negative!
• nope
• This one looks
just right!!
•  = -30 
Physics 1304: Lecture 18, Pg 16
6
6.81
Resonance

For fixed R,C,L the current im will be a maximum at the resonant frequency w0
which makes the impedance Z purely resistive.
ie:
m
im 

Z
m
R 2   X L  XC 
reaches a maximum when:
2
X L  XC
the frequency at which this condition is obtained is given from:
1
1
woL 
 wo 
w oC
LC
•
Note that this resonant frequency is identical to the natural
frequency of the LC circuit by itself!
•
At this frequency, the current and the driving voltage are in
phase!
X  XC
tan   L
0
R
Physics 1304: Lecture 18, Pg 17
Resonance
The current in an LCR circuit depends on the values of the elements and
on the driving frequency through the relation
m
im 

Z
m
R 2   X L  XC 

1

im  m
 m cos 
R 1  tan 2 
R
Suppose you plot the current versus w, the
source voltage frequency, you would get:
2
x
0.0 ,
r1
.. r1
n
m
1
R0
R=Ro
f( x )
im0.5
g( x )
R=2Ro
00
00
1
wx
2w2o
Physics 1304: Lecture 18, Pg 18
Power in LCR Circuit

•
The power supplied by the emf in a series LCR circuit depends on the frequency w.
It will turn out that the maximum power is supplied at the resonant frequency w0.
The instantaneous power (for some frequency, w) delivered at
time t is given by:
Remember what
this stands for
•
The most useful quantity to consider here is not the
instantaneous power but rather the average power delivered in
a cycle.
•
To evaluate the average on the right, we first expand the sin(wt) term.
Physics 1304: Lecture 18, Pg 19
Power in LCR Circuit
•
Expanding,
•
Taking the averages,
1
1.01+1
sinwtcoswt
(Product of even and odd function = 0)
•
Generally:
h( x ) 0
0
1.01 1
-1
•
x
Putting it all back together again,
1/2
0
00
2
r1
0.0 ,0 .. r1
n
1
+1
wt
x
4
2p
6
6.28
sin2wt
h( x ) 0
0
-11
2p
wt
Physics 1304: Lecture
18,
Pg
20
x
00
2
4
6
Power in LCR Circuit

This result is often rewritten in terms of rms values:
 rms 
1
m
2
i rms 
1
im
2

 P( t )    rmsi rms cos 

Power delivered depends on the phase, , the “power factor”

phase depends on the values of L, C, R, and w

therefore...
Physics 1304: Lecture 18, Pg 21
Power in RLC
 P( t )    rmsi rms cos 

Power, as well as current, peaks at w = w 0. The sharpness of the resonance depends
on the values of the components.

Recall:
• Therefore,
We can write this in the following manner (you can do the algebra):
x
…introducing the curious factors Q and x...
w
wo
Q
Lw o
R
Physics 1304: Lecture 18, Pg 22
The Q factor
A parameter “Q” is often defined to describe the sharpness of
resonance peaks in both mechanical and electrical oscillating
systems. “Q” is defined as
where Umax is max energy stored in the system and DU is the
energy dissipated in one cycle
period
For RLC circuit, Umax is (e.g.)
And losses only in R, namely
This gives
And for completeness, note
Physics 1304: Lecture 18, Pg 23
Power in RLC
For Q > few,
x
0.0 ,
r1
 2 rms
1
R0
wres
Q
fwhm
.. r1
n
Q=3
FWHM
R=Ro
f( x )
<P>0.5
g( x )
R=2Ro
00
00
w
1
2wo2
x
Physics 1304: Lecture 18, Pg 24
Conceptual Question 2

Consider the two circuits shown where CII = 2 CI.
What is the relation between the quality
factors, QI and QII , of the two circuits?
2A
(a) QII < QI
(b) QII = QI
R
R
CI


L
CII


L
(c) QII > QI
2B • What is the relation between PI and PII , the power delivered by
the generator to the circuit when each circuit is operated at its
resonant frequency?
(a) PII < PI
(b) PII = PI
(c) PII > PI
Physics 1304: Lecture 18, Pg 25
Lecture 21, ACT 2

Consider the two circuits shown where CII = 2 CI.
What is the relation between the quality
factors, QI and QII , of the two circuits?
2A
(a) QII < QI
(b) QII = QI
• We know the definition of Q: Q 
R
R
CI


L
CII


L
(c) QII > QI
w 0L
R
• At first glance, it looks like Q is independent of C.
• At second glance, we see this cannot be true, since the resonant
frequency wo depends on C!
w0 
1
LC
Doubling C decreases wo by sqrt(2)!
\ Doubling C decreases Q by sqrt(2)!
Doubling C increases the width of the resonance!
Physics 1304: Lecture 18, Pg 26
Lecture 21, ACT 2

Consider the two circuits shown where CII = 2 CI.
What is the relation between the quality
factors, QI and QII , of the two circuits?
2A
(a) QII < QI
(b) QII = QI
R
R
CI


L
CII


L
(c) QII > QI
2B • What is the relation between PI and PII , the power delivered by
the generator to the circuit when each circuit is operated at its
resonant frequency?
(a) PII < PI
(b) PII = PI
(c) PII > PI
• At the resonant frequency, the impedance of the circuit is purely
resistive.
• Since the resistances in each circuit are the same, the impedances at
the resonant frequency for each circuit are equal.
• Therefore, The power delivered by the generator to each circuit is
identical.
Physics 1304: Lecture 18, Pg 27
Power Transmission

How can we transport power from power stations to homes? Why do we use “high tension” lines?
At home, the AC voltage obtained from outlets in this country is 120V at
60Hz.
Transmission of power is typically at very high voltages
( eg ~500 kV)
Transformers are used to raise the voltage for transmission and lower the
voltage for use. We’ll describe these next.

But why?
Calculate ohmic losses in the transmission lines:
Define efficiency of transmission:
Pout iVin  i 2 R
iR


 1
Pin
iVin
Vin
 Vin

 Vin

Pin R
  1  2
V in

Keep R
small
Make
Vin big
– Note for fixed input power and line resistance, the inefficiency  1/V2
Example: Quebec to Montreal
1000 km  R= 220W
suppose Pin = 500 MW
With Vin=735kV,  = 80%.
The efficiency goes to zero quickly if
Vin were lowered!
Physics 1304: Lecture 18, Pg 28
Physics 1304: Lecture 18, Pg 29
Transformers
•
AC voltages can be stepped up
or stepped down by the use of
transformers.

The AC current in the primary circuit creates a

time-varying magnetic field in the iron

This induces an emf on the secondary
windings due to the mutual inductance of the
two sets of coils.
•
iron

V1
V2
N1
(primary)
N2
(secondary)
The iron is used to maximize the mutual inductance. We
assume that the entire flux produced by each turn of the
primary is trapped in the iron.
Physics 1304: Lecture 18, Pg 30
Ideal
No resistance losses

Transformers (no load)
All flux contained in iron
Nothing connected on secondary
The primary circuit is just an AC voltage source in series with an
inductor. The change in flux produced in each turn is given by:
d turn V1

dt
N1
iron


• The change in flux per turn in the secondary coil is the
same as the change in flux per turn in the primary coil
(ideal case). The induced voltage appearing across the
secondary coil is given by:
V2  N 2
V1
V2
N1
(primary)
N2
(secondary)
d turn N 2

V1
dt
N1
• Therefore,
• N2 > N1  secondary V2 is larger than primary V1 (step-up)
• N1 > N2  secondary V2 is smaller than primary V1 (step-down)
• Note: “no load” means no current in secondary. The primary current,
termed “the magnetizing current” is small!
Physics 1304: Lecture 18, Pg 31
Ideal

Transformers
with a Load
iron
What happens when we connect a resistive load to
the secondary coil?
Flux produced by primary coil
induces an emf in secondary
emf in secondary produces current i2
V
i2  2
R


V1
V2
N1
(primary)
R
N2
(secondary)
This current produces a flux in the
secondary coil  N2i2, which opposes the
original flux -- Lenz’s law
This changing flux appears in the primary
circuit as well; the sense of it is to reduce
the emf in the primary...
However, V1 is a voltage source.
Therefore, there must be an increased
current i1 (supplied by the voltage source) in
the primary which produces a flux  N1i1
which exactly cancels the flux produced by
i2.
N2
i1 
i2
N1
Physics 1304: Lecture 18, Pg 32
Transformers with a Load

iron
With a resistive load in the secondary, the primary current
is given by:


V1
V2
N1
(primary)
R
N2
(secondary)
V1
N12
 Req  R 2
i1
N2
This is the equivalent resistance seen by the source.
Physics 1304: Lecture 18, Pg 33
Lecture 21, ACT 3
iron

The primary coil of an ideal transformer is connected to an AC
voltage source as shown. There are 50 turns in the primary
and 200 turns in the secondary.

3A
If V1 = 120 V, what is the potential drop across
the resistor R ?

V1
V2
N1
(primary)
(a) 30 V
3B
(b) 120 V
N2
(secondary)
(c) 480 V
If 960 W are dissipated in the resistor R,
what is the current in the primary ?
(a) 8 A
(b) 16 A
(c) 32 A
Physics 1304: Lecture 18, Pg 34
R
Lecture 21, ACT 3
iron

The primary coil of an ideal transformer is connected to an AC
voltage source as shown. There are 50 turns in the primary
and 200 turns in the secondary.

3A
If V1 = 120 V, what is the potential drop across
the resistor R ?

V1
V2
N1
(primary)
(a) 30 V
(b) 120 V
N2
(secondary)
(c) 480 V
The ratio of turns, (N2/N1) = (200/50) = 4
The ratio of secondary voltage to primary voltage is equal to the ratio
of turns, (V2/V1) = (N2/N1)
Therefore, (V2/V1) = 480 V
Physics 1304: Lecture 18, Pg 35
R
iron
Lecture 21, ACT 3

The primary coil of an ideal transformer is connected to an AC
voltage source as shown. There are 50 turns in the primary 
and 200 turns in the secondary.

If V1 = 120 V, what is the potential drop across
the resistor R ?
V1
V2
N1
3A
(primary)
(a) 30 V
N2
(secondary)
(c) 480 V
(b) 120 V
The ratio of turns, (N2/N1) = (200/50) = 4
The ratio (V2/V1) = (N2/N1). Therefore, (V2/V1) = 480 V
3B
If 960 W are dissipated in the resistor R, what
is the current in the primary ?
(a) 8 A
(c) 32 A
(b) 16 A
Gee, we didn’t talk about power yet….
But, let’s assume energy is conserved…since it usually is around here
Therefore, 960 W should be produced in the primary
P1 = V1 I1 implies that 960W/120V = 8 A
Physics 1304: Lecture 18, Pg 36
R
Transformers with a Load
iron
• To get that last ACT, you had to use a
general philosophy -- energy
conservation.
• An expression for the RMS power flow
looks like this:
Prms  V1rmsi1rms


V1
V2
N1
(primary)
R
N2
(secondary)
N1
N2

V2 rms
i 2 rms  V2 rmsi 2 rms
N2
N1
Note: This equation simply says that all power delivered by the
generator is dissipated in the resistor ! Energy conservation!!
Physics 1304: Lecture 18, Pg 37
Physics 1304: Lecture 18, Pg 38