Download AC Circuits and Resonance Conclusion

AC Circuits and Resonance
Conclusion
April 20, 2005
Alternating Current Circuits
An “AC” circuit is one in which the driving voltage and
hence the current are sinusoidal in time.
V(t)
Vp
fv
p
2p
wt
V = VP sin (wt - fv )
I = IP sin (wt - fI )
-Vp
w is the angular frequency (angular speed) [radians per second].
Sometimes instead of w we use the frequency f [cycles per second]
Frequency  f [cycles per second, or Hertz (Hz)]
w = 2p f
Phase Term
V = VP sin (wt - fv )
V(t)
Vp
p
fv
-Vp
2p
wt
Resistors in AC Circuits
R
E
~
EMF (and also voltage across resistor):
V = VP sin (wt)
Hence by Ohm’s law, I=V/R:
I = (VP /R) sin(wt) = IP sin(wt)
(with IP=VP/R)
V
I
p
2p
wt
V and I
“In-phase”
Capacitors in AC Circuits
C
Start from:
q = C V [V=Vpsin(wt)]
Take derivative: dq/dt = C dV/dt
So
I = C dV/dt = C VP w cos (wt)
E
~
I = C w VP sin (wt + p/2)
V
I
p
2p wt
This looks like IP=VP/R for a resistor
(except for the phase change).
So we call
Xc = 1/(wC)
the Capacitive Reactance
The reactance is sort of like resistance in
that IP=VP/Xc. Also, the current leads
the voltage by 90o (phase difference).
V and I “out of phase” by 90º. I leads V by 90º.
I Leads V???
What the **(&@ does that mean??
2
V
f
1
I
I = C w VP sin (wt + p/2)
Current reaches it’s maximum at
an earlier time than the voltage!
Inductors in AC Circuits
~
L
V = VP sin (wt)
Loop law: V +VL= 0 where VL = -L dI/dt
Hence:
dI/dt = (VP/L) sin(wt).
Integrate: I = - (VP / Lw) cos (wt)
or
V
Again this looks like IP=VP/R for a
resistor (except for the phase change).
I
p
I = [VP /(wL)] sin (wt - p/2)
2p
wt So we call
the
XL = w L
Inductive Reactance
Here the current lags the voltage by 90o.
V and I “out of phase” by 90º. I lags V by 90º.
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Vp
Ip
wt
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Vp
Ip
Ip
wt
wt
Vp
Phasor Diagrams
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
The “y component” is the actual voltage or current.
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
wt
Ip
wt
wt
Vp
i
i
+
+
+
time
i
i
LC Circuit
i
i
+
+
+
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U = UB  UE = LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
= ( LI 
) = 0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
= 0 = L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q = 0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U = UB  UE = LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
= ( LI 
) = 0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
= 0 = L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q = 0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U = UB  UE = LI 
2
2 C
2
Differentiate :
dU d 1 2 1 q
= ( LI 
) = 0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
= 0 = L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q = 0
dt
C
Analyzing the L-C Circuit
Total energy in the circuit:
1 2 1 q2
U = UB  UE = LI 
2
2 C
2
Differentiate :
d 2q
2

w
q=0
2
dt
q = q p cos wt
dU d 1 2 1 q
= ( LI 
) = 0 N o change
in energy
dt dt 2
2 C
dI q dq
dq d 2 q q dq
 LI 
= 0 = L( ) 2 
dt C dt
dt dt
C dt
d 2q 1
L 2  q = 0
dt
C
The charge sloshes back and
forth with frequency w = (LC)-1/2
LC Oscillations
• Work out equation for LC circuit (loop rule)
C
q
di
 L =0
C
dt
• Rewrite using i = dq/dt
d 2q
q
d 2q
2
L 2  =0 

w
q=0
2
C
dt
dt
– w (angular frequency) has dimensions of 1/t
w=
1
LC
• Identical to equation of mass on spring
m
d 2x
dt
2
 kx = 0 
d 2x
dt
2
w x = 0
2
k
w=
m
L
LC Oscillations (2)
• Solution is same as mass on spring 
oscillations
q = qmax cos w t   )
i = w qmax sin w t   ) = imax sin w t   )
– qmax is the maximum charge on capacitor
–  is an unknown phase (depends on initial
conditions)
• Calculate current: i = dq/dt
Plot Charge and Current vs t
w = 1 T = 2p
q t )
i t )
Energy Oscillations
Equation of LC circuit
di q
L  =0
dt C
• Total energy in circuit is conserved. Let’s
see why
di
q dq
L i
=0
dt C dt
 )
(Multiply by i = dq/dt)
 )
Ld 2
1 d 2
i 
q =0
2 dt
2C dt
d  1 2 1 q2 
 2 Li  2  = 0
dt 
C
dx 2
dx
Use
= 2x
dt
dt
2
q
1 Li 2  1
= const
2
2 C
UL + UC = const
Oscillation of Energies
• Energies can be written as (using w2 = 1/LC)
2
q 2 qmax
UC =
=
cos 2 w t   )
2C
2C
2
q
2
U L = 12 Li 2 = 12 Lw 2 qmax
sin 2 w t   ) = max sin 2 w t   )
2C
• Conservation of energy:
2
qmax
UC  U L =
= const
2C
• Energy oscillates between capacitor and inductor
– Endless oscillation between electrical and magnetic energy
– Just like oscillation between potential energy and kinetic energy
for mass on spring
Plot Energies vs t
UC t ) U L t )
Sum
UNDRIVEN RLC Circuit
• Work out equation using loop rule
di
q
L  Ri  = 0
dt
C
• Rewrite using i = dq/dt
d 2q
R dq q


=0
2
L dt LC
dt
2
tR / 2 L


q = qmax e
cos w t   ) w = 1/ LC   R / 2 L )
• Solution slightly more complicated than LC
case
Charge and Current vs t in RLC Circuit
q t )
i t )
e
tR / 2 L
RLC Circuit (Energy)
di
q
L  Ri  = 0
dt
C
Basic RLC equation
di
q dq
2
L i  Ri 
=0
dt
C dt
Multiply by i = dq/dt
d  1 2 1 q2 
2
Li

=

i
R
 2

2
dt 
C
Collect terms
(similar to LC circuit)
d
U L  U C ) = i 2 R
dt
Total energy in circuit
decreases at rate of i2R
(dissipation of energy)
U tot
etR / L
Energy in RLC Circuit
UC t )
U L t )
Sum
e
tR / L
AC Circuits
• Enormous impact of AC circuits
–
–
–
–
–
Power delivery
Radio transmitters and receivers
Tuners
Filters
Transformers
• Basic components
–
–
–
–
R
L
C
Driving emf
AC Circuits and Forced
Oscillations
• RLC + “driving” EMF with angular frequency wd
 =  m sin wd t
L
di
q
 Ri  =  m sin w d t
dt
C
• General solution for current is sum of two terms
“Transient”: Falls
exponentially & disappears
Ignore
i
etR / 2 L cos w t
“Steady state”:
Constant amplitude
Steady State Solution
• Assume steady state solution of form
i = I m sin w d t  f )
– Im is current amplitude
– f is phase by which current “lags” the driving EMF
– Must determine Im and f
• Plug in solution: differentiate & integrate sin(wt-f)
i = I m sin w d t  f )
di
= w d I m cos w d t  f )
dt
Im
q=
cos w d t  f )
Substitute
di
q
L  Ri  =  m sin w t
dt
C
wd
Im
I mw d L cos w d  f )  I m R sin w d t  f ) 
cos w d t  f ) =  m sin w d t
wd C
Steady State Solution for AC
Im
Current
(2)
I mw d L cos w d  f )  I m R sin w d t  f ) 
cos w d t  f ) =  m sin w d t
wd C
• Expand sin & cos expressions
sin wd t  f ) = sin wd t cos f  cos w d t sin f
cos wd t  f ) = cos wd t cos f  sin w d t sin f
High school trig!
• Collect sinwdt & coswdt terms separately
coswdt terms
wd L  1/ wd C ) cos f  R sin f = 0
sinwdt terms
I m wd L  1/ wd C ) sin f  I m R cos f =  m
• These equations can be solved for Im and f
(next slide)
Steady State Solution for AC Current (3)
wd L  1/ wd C ) cos f  R sin f = 0
I m wd L  1/ wd C ) sin f  I m R cos f =  m
• Solve for f and Im in terms of
tan f =
wd L  1/ wd C
R
X  XC
 L
R
Im =
m
Z
• R, XL, XC and Z have dimensions of resistance
X L = wd L
Inductive “reactance”
X C = 1/ wd C
Capacitive “reactance”
Z = R2   X L  X C )
2
Total “impedance”
• Let’s try to understand this solution using
“phasors”
REMEMBER Phasor Diagrams?
A phasor is an arrow whose length represents the amplitude of
an AC voltage or current.
The phasor rotates counterclockwise about the origin with the
angular frequency of the AC quantity.
Phasor diagrams are useful in solving complex AC circuits.
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
wt
Ip
wt
wt
Vp
Reactance - Phasor Diagrams
Resistor
Capacitor
Inductor
Vp
Ip
Vp
Ip
wt
Ip
wt
wt
Vp
“Impedance” of an AC Circuit
R
L
~
C
The impedance, Z, of a circuit relates peak
current to peak voltage:
Ip =
Vp
Z
(Units: OHMS)
“Impedance” of an AC Circuit
R
L
~
C
The impedance, Z, of a circuit relates peak
current to peak voltage:
Ip =
Vp
Z
(Units: OHMS)
(This is the AC equivalent of Ohm’s law.)
Impedance of an RLC Circuit
R
E
~
L
C
As in DC circuits, we can use the loop method:
E - V R - VC - VL = 0
I is same through all components.
Impedance of an RLC Circuit
R
E
~
L
C
As in DC circuits, we can use the loop method:
E - V R - VC - VL = 0
I is same through all components.
BUT: Voltages have different PHASES
 they add as PHASORS.
Phasors for a Series RLC Circuit
Ip
VLp
VRp
f
(VCp- VLp)
VP
VCp
Phasors for a Series RLC Circuit
Ip
VLp
VRp
f
(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
Phasors for a Series RLC Circuit
Ip
VLp
VRp
f
(VCp- VLp)
VP
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
= Ip2 R2 + (Ip XC - Ip XL) 2
Impedance of an RLC Circuit
R
Solve for the current:
Ip =
Vp
Vp
=
Z
R2  (X c  X L )2
~
L
C
Impedance of an RLC Circuit
R
Solve for the current:
Ip =
~
L
C
Vp
=
Z
R2  (X c  X L )2
Impedance:
Vp
Z=
 1

R 
 wL
wC
2
2
Impedance of an RLC Circuit
Vp
Ip =
Z
 1
R 
 wL
wC
Z=
The current’s magnitude depends on
the driving frequency. When Z is a
minimum, the current is a maximum.
This happens at a resonance frequency:
2
2
The circuit hits resonance when 1/wC-wL=0: w r=1/ LC
When this happens the capacitor and inductor cancel each other
and the circuit behaves purely resistively: IP=VP/R.
IP
R =10W
L=1mH
C=10mF
R = 1 0 0 W
0
1 0
wr
2
1 0
3
1 0
4
1 0
5
w
The current dies away
at both low and high
frequencies.
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
(VCp- VLp)
f
VP
tan f = (VCp - VLp)/ VRp
or;
or
VCp
tan f = (XC-XL)/R.
tan f = (1/wC - wL) / R
Phase in an RLC Circuit
Ip
VLp
We can also find the phase:
VRp
(VCp- VLp)
f
VP
tan f = (VCp - VLp)/ VRp
or;
or
VCp
tan f = (XC-XL)/R.
tan f = (1/wC - wL) / R
More generally, in terms of impedance:
cos f = R/Z
At resonance the phase goes to zero (when the circuit becomes
purely resistive, the current and voltage are in phase).
Power in an AC Circuit
V
f= 0
p
I
2p
wt
V(t) = VP sin (wt)
I(t) = IP sin (wt)
(This is for a purely
resistive circuit.)
P
P(t) = IV = IP VP sin 2(wt)
Note this oscillates
twice as fast.
p
2p
wt
Power in an AC Circuit
The power is P=IV. Since both I and V vary in time, so
does the power: P is a function of time.
Use, V = VP sin (wt) and I = IP sin (w t+f ) :
P(t) = IpVpsin(wt) sin (w t+f )
This wiggles in time, usually very fast. What we usually
care about is the time average of this:
1 T
P = 0 P( t )dt
T
(T=1/f )
Power in an AC Circuit
Now: sin( wt  f ) = sin( wt )cos f  cos(wt )sin f
Power in an AC Circuit
Now: sin( wt  f ) = sin( wt )cos f  cos(wt )sin f
P( t ) = I PVP sin( w t )sin( w t  f )
= I PVP sin 2( w t )cos f  sin( w t )cos( w t )sin f
Power in an AC Circuit
Now: sin( wt  f ) = sin( wt )cos f  cos(wt )sin f
P( t ) = I PVP sin( w t )sin( w t  f )
= I PVP sin 2( w t )cos f  sin( w t )cos( w t )sin f
Use:
and:
So
sin (w t ) =
2
1
2
sin(w t ) cos(w t ) = 0
P =
1
2
I PV P cos f
Power in an AC Circuit
Now: sin( wt  f ) = sin( wt )cos f  cos(wt )sin f
P( t ) = I PVP sin( w t )sin( w t  f )
= I PVP sin 2( w t )cos f  sin( w t )cos( w t )sin f
Use:
and:
So
sin (w t ) =
2
1
2
sin(w t ) cos(w t ) = 0
P =
1
2
I PV P cos f
which we usually write as
P = IrmsVrms cos f
Power in an AC Circuit
P = IrmsVrms cos f
f goes from -900 to 900, so the average power is positive)
cos(f) is called the power factor.
For a purely resistive circuit the power factor is 1.
When R=0, cos(f)=0 (energy is traded but not dissipated).
Usually the power factor depends on frequency.
Power in an AC Circuit
P = IrmsVrms cos f
What if f is not zero?
I
P
V
Here I and V are 900
out of phase. (f= 900)
wt (It is purely reactive)
The time average of
P is zero.
STOP!
Transformers
Transformers use mutual inductance to change voltages:
N2
V2 =
V1
N1
N1 turns
Iron Core
V1
Primary
Power is conserved, though:
(if 100% efficient.)
N2 turns
V2
Secondary
I1V1 = I 2V2
Transformers & Power Transmission
Transformers can be used to “step up” and “step
down” voltages for power transmission.
110 turns
Power
=I1 V1
V1=110V
20,000 turns
V2=20kV Power
=I2 V2
We use high voltage (e.g. 365 kV) to transmit electrical
power over long distances.
Why do we want to do this?
Transformers & Power Transmission
Transformers can be used to “step up” and “step down”
voltages, for power transmission and other applications.
110 turns
Power
=I1 V1
V1=110V
20,000 turns
V2=20kV Power
=I2 V2
We use high voltage (e.g. 365 kV) to transmit electrical
power over long distances.
Why do we want to do this?
P = I2R
(P = power dissipation in the line - I is smaller at high voltages)