physics 1304: lecture 18, pg 1 ac circuits physics 1304: lecture 18, pg 2 lecture outline l driven...
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Physics 1304: Lecture 18, Pg 1
AC CircuitsAC Circuits
imR
imL
imC
m
N2N1
(primary) (secondary)
iron
V2V1
x ..,0.0r1
nr1
0 1 20
0.5
1
f( )x
g( )x
x
im
00 o
m
R0
R=Ro
R=2Ro
Physics 1304: Lecture 18, Pg 2
Lecture OutlineLecture 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 3
PhasorsPhasors
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 . Recall uniform circular motion:
The projections of r (on the vertical y axis) execute sinusoidal oscillation.
iL
tLm
cos
i C tC m cos
iR
tRm sin
x r t cos
y r t sin
V Ri tR R m sin• R: V in phase with i
VQ
CtC m sin• C: V lags i by 90
V Ldi
dttL
Lm sin• L: V leads i by 90
x
y y
Physics 1304: Lecture 18, Pg 4
Phasors for Phasors for L,C,RL,C,R
i
t
V
R
i
t
VL
i
t
VC
Suppose:
V Ri tR m sin
VC
i tC m 1
cos
V Li tL m cos
t
iV
R
0
VC
0
i
i
0
VL
Physics 1304: Lecture 18, Pg 5
Series LCRSeries LCRAC CircuitAC Circuit
• Back to the original problem: the loop equation gives:
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 sint.
• To solve this problem graphically, first write down expressions for the voltages across R,C, and L and then plot the appropriate phasor diagram.
LC
R
Ld Q
dt
Q
CR
dQ
dttm
2
2 sin
• Assume a solution of the form: i i tm sin( )
Phasors: LCRPhasors: LCR
• Assume:
From these equations, we can draw the phasor diagram to the right.
imR
imL
imC
m
LC
R• Given: m tsin
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.
Physics 1304: Lecture 18, Pg 7
Phasors: LCRPhasors: LCR
• The phasor diagram has been relabeled in terms of the reactances defined from:
imR
m
imXC
imXL
LC
R
XCC 1
X LL
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 8
Lecture 20, ACT 3Lecture 20, ACT 3
A driven RLC circuit is connected as shown.
For what frequencies of the voltage source is the current through the resistor largest?
(a) small (b) large (c) 1
LC
L
C
R
Conceptual Conceptual QuestionQuestion
A driven RLC circuit is connected as shown.
For what frequencies of the voltage source is the current through the resistor largest?
(a) small (b) large (c)
L
C
R
1
LC
• 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 = L. • The reactance of a capacitor = XC = 1/(C).• Therefore,
• in the low frequency limit, XL 0 and XC . • Therefore, as 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 , the current will flow mostly through the capacitor, approaching a maximum imax = /R.
Physics 1304: Lecture 18, Pg 10
Phasors:LCRPhasors:LCR
imR
m
im(XL-X C)
imR
m
imXC
imXL
XCC 1
X LL
Z R X XL C 2 2
tan X X
RL C
m m L Ci R X X2 2 2 2
i
R X X Zmm
L C
m
2 2
Phasors:TipsPhasors:Tips
• This phasor diagram was drawn as a snapshot of time t=0 with the voltages being given as the projections along the y-axis.
imR
m
imXC
imXLy
x
imR
imXL
imXC
m
“Full Phasor Diagram”
From this diagram, we can also create a triangle which allows us to calculate the impedance Z:
X XL C
Z
R
“ Impedance Triangle”
• 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).
Physics 1304: Lecture 18, Pg 12
Phasors:LCRPhasors:LCR
We have found the general solution for the driven LCR circuit:
X LL
XCC 1
Z R X XL C 2 2
R
XL
XC
tan X X
RL C
iZmm
i Zm
the loop
eqn
XL - XC
i i tm sin( )
imR
imXL
imXC
m
Lagging & LeadingLagging & Leading
The phase between the current and the driving emf depends on the relative magnitudes of the inductive and capacitive reactances.
R
XL
XC
tan X X
RL Ci
Zmm X LL
XCC 1
XL > XC
> 0 current LAGS
applied voltage
R
XL
XC
XL < XC
< 0 current LEADS
applied voltage
XL = XC
= 0 current
IN PHASE applied voltage
R
XL
XC
Physics 1304: Lecture 18, Pg 14
Conceptual Conceptual QuestionQuestion
The series LCR circuit shown is driven by a generator with voltage = msint. The time dependence of the current i which flows in the circuit is shown in the plot.
How should be changed to bring the current and driving voltage into phase?
(a) increase (b) decrease (c) impossible
• Which of the following phasors represents the current i at t=0?1B
1A
LC
R
0
x ..,0r1
nr1
2 4 61
0
11.01
1.01
f( )x
6.81.53 x
io
im
-im0
i
t
(a) (b) (c) ii
i
Physics 1304: Lecture 18, Pg 15
Lecture 21, ACT 1Lecture 21, ACT 1
The series LCR circuit shown is driven by a generator with voltage = msint. The time dependence of the current i which flows in the circuit is shown in the plot.
How should be changed to bring the current and driving voltage into phase?
(a) increase (b) decrease (c) impossible
1A
LC
R
0
x ..,0r1
nr1
2 4 61
0
11.01
1.01
f( )x
6.81.53 x
io
im
-im0
i
t
• From the plot, it is clear that the current is LEADING the applied voltage.
i
XC
XL
Therefore, the phasor diagram must look like this:
Therefore, X XC L• To bring the current into phase with the applied voltage, we need to increase XL and decrease XC.• Increasing will do both!!
Physics 1304: Lecture 18, Pg 16
Lecture 21, ACT 1Lecture 21, ACT 1
The series LCR circuit shown is driven by a generator with voltage = msint. The time dependence of the current i which flows in the circuit is shown in the plot.
How should be changed to bring the current and driving voltage into phase?
(a) increase (b) decrease (c) impossible
• Which of the following phasors represents the current i at t=0?
(a) (b) (c)
1A
1B
LC
R
0
x ..,0r1
nr1
2 4 61
0
11.01
1.01
f( )x
6.81.53 x
io
im
-im0
i
t
ii
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 17
ResonanceResonance
For fixed R,C,L the current im will be a maximum at the resonant frequency 0 which makes the impedance Z purely resistive.
the frequency at which this condition is obtained is given from:
• 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!
ie:
iZ R X X
mm m
L C
2 2
reaches a maximum when: X XL C
o
oL
C 1
o LC 1
tan X X
RL C 0
Physics 1304: Lecture 18, Pg 18
ResonanceResonance
The current in an LCR circuit depends on the values of the elements and on the driving frequency through the relation
i
Z R X Xm
m m
L C
2 2
iR Rmm m
1
1 2tancos
im
00
o
m
R0
R=Ro
R=2Ro
Suppose you plot the current versus , the source voltage frequency, you would get:
Physics 1304: Lecture 18, Pg 19
Power in LCR CircuitPower in LCR Circuit
The power supplied by the emf in a series LCR circuit depends on the frequency . It will turn out that the maximum power is supplied at the resonant frequency 0.
• The instantaneous power (for some frequency, ) delivered at time t is given by:
• 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(t-) term.
Remember what this stands for
Physics 1304: Lecture 18, Pg 20
Power in LCR CircuitPower in LCR Circuit
• Expanding,
• Taking the averages,
• Generally:
sintcost
t0
0
+1
-1
sin2t
t0
0
+1
-1
• Putting it all back together again,
(Product of even and odd function = 0)
01/2
Physics 1304: Lecture 18, Pg 21
Power in LCR CircuitPower in LCR Circuit
This result is often rewritten in terms of rms values:
rms m 1
2i irms m 1
2 P t irms rms( ) cos
Power delivered depends on the phase, the“power factor”
phase depends on the values of L, C, R, and
therefore...
Physics 1304: Lecture 18, Pg 22
Power in RLCPower in RLC
Power, as well as current, peaks at = 0. The sharpness of the resonance depends on the values of the components.
Recall:
P t irms rms( ) cos
• Therefore,
We can write this in the following manner (you can do the algebra):
…introducing the curious factors Q and x...
o
x
R
LQ o
Physics 1304: Lecture 18, Pg 23
The Q factorThe 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 U is the energy dissipated in one cycle
For RLC circuit, Umax is (e.g.)
And losses only in R, namely
period
This gives
And for completeness, note
Physics 1304: Lecture 18, Pg 24
Power in RLCPower in RLC
<P>
00
o
2
0
rms
R
R=Ro
R=2Ro
Q=3
FWHM
For Q > few, fwhm
Q res
Physics 1304: Lecture 18, Pg 25
Conceptual Question Conceptual Question 22
Consider the two circuits shown where CII = 2 CI.
What is the relation between the quality factors, QI and QII , of the two circuits?
(a) QII < QI (b) QII = QI (c) QII > QI
(a) PII < PI (b) PII = PI (c) PII > PI
• 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?
2B
2A
LC
R
I LC
R
II
Physics 1304: Lecture 18, Pg 26
Lecture 21, ACT 2Lecture 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?
(a) QII < QI (b) QII = QI (c) QII > QI
2A
LC
R
I LC
R
II
• We know the definition of Q: QL
R0
• At first glance, it looks like Q is independent of C.• At second glance, we see this cannot be true, since the resonant frequency o depends on C!
01LC Doubling C decreases o by sqrt(2)!
Doubling C decreases Q by sqrt(2)!
Doubling C increases the width of the resonance!
Physics 1304: Lecture 18, Pg 27
Lecture 21, ACT 2Lecture 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?
(a) QII < QI (b) QII = QI (c) QII > QI
• 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
2A
2B
LC
R
I LC
R
II
• 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 28
Power TransmissionPower 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:
in2
in
in
in
inin
2in
in
out
V
RP1
V
V
V
iR1
iV
RiiV
P
P
– Note for fixed input power and line resistance, the inefficiency 1/V2
Example: Quebec to Montreal 1000 km R= 220suppose Pin = 500 MW
With Vin=735kV, = 80%.The efficiency goes to zero quickly if Vin were lowered!
Keep R small
Make Vin big
Physics 1304: Lecture 18, Pg 29
Physics 1304: Lecture 18, Pg 30
TransformersTransformers
N2N1
(primary) (secondary)
• AC voltages can be stepped up or stepped down by the use of transformers.
iron
The AC current in the primary circuit creates a time-varying magnetic field in the iron
• 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.
V2V1
This induces an emf on the secondary windings due to the mutual inductance of the two sets of coils.
Physics 1304: Lecture 18, Pg 31
IdealIdeal Transformers Transformers (no load)(no load)
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:
N2N1
(primary) (secondary)
iron
V2V1
1
1turn
N
V
dt
d
• 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:
• 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!
V Nd
dt
N
NVturn
2 22
11
No resistance losses All flux contained in iron Nothing connected on secondary
Physics 1304: Lecture 18, Pg 32
IdealIdeal Transformers Transformers
What happens when we connect a resistive load to the secondary coil?
Flux produced by primary coil induces an emf in secondary
N2N1
(primary) (secondary)
iron
V2V1 R
iV
R22
iN
Ni1
2
12
with a Loadwith a Load
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.
emf in secondary produces current i2
This current produces a flux in the secondary coil N2i2which opposes the original flux -- Lenz’s law
Physics 1304: Lecture 18, Pg 33
Transformers with a LoadTransformers with a Load
With a resistive load in the secondary, the primary current is given by:
N2N1
(primary) (secondary)
iron
V2V1 R
This is the equivalent resistance seen by the source.
22
21
1
1
N
NRR
i
Veq
Physics 1304: Lecture 18, Pg 34
Lecture 21, ACT 3Lecture 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 ?
(a) 30 V (b) 120 V (c) 480 V
3A
N2N1
(primary) (secondary)
iron
V2V1 R
3B 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 35
Lecture 21, ACT 3Lecture 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 ?
(a) 30 V (b) 120 V (c) 480 V
3A
N2N1
(primary) (secondary)
iron
V2V1 R
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 36
Lecture 21, ACT 3Lecture 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 ?
(a) 30 V (b) 120 V (c) 480 V3A
N2N1
(primary) (secondary)
iron
V2V1 R
The ratio of turns, (N2/N1) = (200/50) = 4 The ratio (V2/V1) = (N2/N1). Therefore, (V2/V1) = 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
3B
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 37
Transformers with a LoadTransformers with a Load
N2N1
(primary) (secondary)
iron
V2V1 R• To get that last ACT, you had to use a general philosophy -- energy conservation.
• An expression for the RMS power flow looks like this:
P V iN
NV
N
Ni V irms rms rms rms rms rms rms 1 1
1
22
2
12 2 2
Note: This equation simply says that all power delivered by the generator is dissipated in the resistor ! Energy conservation!!
Physics 1304: Lecture 18, Pg 38