Download Lecture 29 Chapter 33 EM Oscillations and AC

Lecture 29
Chapter 33
EM Oscillations and AC
Review
• RL and RC circuits
– Charge, current, and potential
grow and decay exponentially
• LC circuit
– Charge, current, and potential
grow and decay sinusoidally –
electromagnetic oscillations
– Total electromagnetic energy is
Li 2 q 2
U = UB + UE =
+
2
2C
Review
dU
=0
– Total energy conserved dt
• Ideal LC circuit
– Solved differential equation to find
q = Q cos(ωt + φ )
i = − I sin(ωt + φ )
1
LC
ω=
– Substituting q and i into energy equations
U
E
Q2
=
cos
2C
2
(ω t + φ )
U
B
Q2
=
sin
2C
U = U B + U E = Q 2 / 2C
2
(ω t + φ )
Review
• RLC circuit
– Energy is no longer conserved,
becomes thermal energy in resistor
– Oscillations are damped
– Solved differential equation to find
− Rt / 2 L
q = Qe
dU
= −i 2 R
dt
cos(ω ′t + φ )
2
2
′
ω = ω − ( R / 2 L)
ω=
If R is very small
1
LC
ω′ = ω
– Energy decreases as
Q 2 − Rt / L
UE =
e
cos2 (ω ′t + φ )
2C
Q 2 − Rt / L
Utot =
e
2C
EM Oscillations (23)
• LC and RLC circuits with no external emf
– Free oscillations with natural
frequency, ω
• Add external emf (e.g. ac
to RLC circuit
– Oscillations said to be
or forced
– Oscillations occur at driving
frequency, ωd
– When ωd = ω, called resonance,
current amplitude, I, is maximum
angular
ω=
1
LC
generator)
driven
angular
the
EM Oscillations (24)
• ac generator – mechanically
turn loop in B field, induces a
current and therefore an emf
• Used Faraday’s law to find emf
dΦ B
E = −N
= NBA ω sin ωt
dt
E max = NBA ω
E = E max sin ωt
E
t
EM Oscillations (25)
E = E m sin ω d t
– Driving angular frequency, ωd
is equal to angular speed that
loop rotates in B field
– The phase of the emf is ωd t
– Amplitude is Em where m stands
for maximum
• Current of ac generator where φ
corresponds to phase difference
between the current and emf
i = I sin (ω d t − φ )
ω d = 2πf d
• I is amplitude
• Minus sign historical
EM Oscillations (26)
• Purely resistive load
• Apply loop rule
E − vR = 0
• Using
E = E m sin ω d t
• Substitute
vR = Em sin ω d t
• Amplitude across
resistor is same as
across emf
E m = VR
• Rewrite vR as
vR = VR sin ω d t
EM Oscillations (27)
• Use definition of
resistance to find iR
v R = V R sin ω d t
VR
R=
IR
vR VR
iR =
=
sin ω d t
R
R
• Voltage and current are
functions of sin(ωdt) with
φ = 0 so are in phase
• No damping of vR and iR ,
generator supplies energy
EM Oscillations (28)
• Compare iR to i for emf
vR VR
iR =
= sin ω d t
R
R
iemf = iR = I R sin(ωd t − φ )
• For purely resistive load the
phase constant φ = 0
• Voltage amplitude is related
to current amplitude
VR = I R R
VR
IR =
R
VR = I R R
EM Oscillations (29)
• Purely capacitive load
• Apply loop rule
E − vC = 0
• Using
E = E m sin ω d t
• Substitute
vC = Em sin ω d t
• Amplitude across
capacitor is same
as across emf
E m = VC
• Rewrite vC as
vC = VC sin ω d t
EM Oscillations (30)
• Use definition of
capacitance
qC = CvC
qC = Cv C = CV C sin ω d t
• Use definition of current
and differentiate
dq
i=
dt
dq C
iC =
= ω d CV C cos ω d t
dt
• Replace cosine term
with a phase-shifted
sine term
cosωd t = sin(ωd t + 90°)
EM Oscillations (31)
• Compare vC and iC
of capacitor
vC = VC sin ω d t
iC = ωdCVC sin(ωd t + 90°)
• Voltage and current are
out of phase by 90°
• Current leads voltage
– Current reaches its max
before voltage does by a
quarter cycle or T/4
EM Oscillations (32)
• Now compare currents
iC = ωd CVC sin(ωd t + 90°) iemf = iC = I C sin(ωd t − φ )
• For purely capacitive load phase φ = -90°
• VC amplitude is related to IC amplitude
1
I C = ω d CVC VC = I C ω C = I C X C
d
• XC is the capacitive reactance and has SI
unit of ohm, Ω just like resistance R
VC = I C X C
1
XC =
ωd C
EM Oscillations (33)
• Purely inductive load
• Apply loop rule
E − vL = 0
• Using
E = E m sin ω d t
• Substitute
vL = Em sin ω d t
• Amplitude across
inductor is same
as across emf
E m = VL
• Rewrite vL as
vL = VL sin ω d t
EM Oscillations (34)
• Self-induced emf across
an inductor is
di
• Relate
E L = vL = L
dt
di
v L = VL sin ω d t = L
dt
di
VL
=
sin ω d t
dt
L
• Want current so integrate
 VL 
VL
 cosωd t
iL = ∫ diL = ∫ sinωd tdt = −
L
 ωd L 
EM Oscillations (35)
• Replace -cosine with a
phase-shifted sine term
− cos ω d t = sin(ω d t − 90°)
 VL 
VL


iL = − 
cos ω d t =
sin(ω d t − 90°)

ωd L
 ωd L 
• Compare iL to vL
• iL and vL are 90° out
of phase
• Current lags voltage
– iL reaches max after vL
by a quarter cycle or T/4
vL = VL sin ω d t
EM Oscillations (36)
• Now compare currents
VL
iL =
sin( ω d t − 90 ° )
ωd L
iemf = iL = I L sin (ωd t − φ )
• For purely inductive load phase φ = +90°
• VL amplitude is related to IL amplitude
VL
IL =
V
=
I
ω
L
=
I
X
L
L
d
L
L
ωd L
• XL is the inductive reactance and has SI unit
of ohm, Ω just like resistance R
VL = I L X L
X L = ωd L
EM Oscillations (37)
Element
Resistor
Capacito
r
Inductor
Reactance Phase of Phase Amplitude
/
Current angle φ Relation
Resistance
VR=IRR
R
In phase
0°
XC= 1/ωdC Leads
vC (ICE)
XL=ωdL Lags vL
(ELI)
-90°
VC=ICXC
+90°
VL=ILXL
• ELI the ICE man
– Voltage or emf (E) before current (I) in an inductor (L)
– Current (I) before voltage or emf (E) in capacitor (C)
EM Oscillations (38)
• Checkpoints #3, 5, & 6 – If the driving
frequency, ωd , in a circuit is increased does the
amplitude voltage and amplitude current
increase, decrease or remain the same?
• For purely resistive circuit
• From loop rule VR = Em
• So amplitude voltage, VL stays the same
• IR also stays the same VR
IR =
only depends on R
R
IR
EM Oscillations (39)
• Checkpoints #3, 5, & 6 – If the driving
frequency, ωd , in a circuit is increased does the
amplitude voltage and amplitude current
increase, decrease or remain the same?
• For purely capacitive circuit
• From loop rule Vc = Em
• So amplitude voltage, VC stays the same
• IC depends on XC which depends on ωd by
• So IC increases
V
IC =
C
XC
= ω d CVC
EM Oscillations (40)
• Checkpoints #3, 5, & 6 – If the driving
frequency, ωd , in a circuit is increased does the
amplitude voltage and amplitude current
increase, decrease or remain the same?
• For purely inductive circuit
• From loop rule VL = Em
• So amplitude voltage, VL stays the same
• IL depends on XL which depends on ωd by
• So IL decreases
V
V
IL =
L
XL
=
L
ωd L