8.022 (E&M) – Lecture 17
Topics:
„
Discussion of Exam 2 and make-up exam
„
Back to E&M:
„
RCL circuits: recap undriven RCLs, driven RCLs, inductance
Last time
„
„
What happens when we put inductors in circuits?
„
RL circuits: exponential solutions
„
LC circuits: oscillatory solution
„
RCL circuits: damped oscillation
RCL circuits are particularly interesting
„
Let’s see them in some more detail…
G. Sciolla – MIT
8.022 – Lecture 17
4
1
Undriven RCL circuits: recap
Kirchoff’s second rule:
d 2Q
dQ
1
L
+ R
+ Q =0
dt 2
dt
C
„
Does it look familiar?
d 2x
dx
m
+ kf
+ ke x = 0
2
dt
dt
„
„
Mechanics: harmonic oscillator!
RCL
L
d2Q/dt2
Mechanics
ma=m
Interpretation
d2x/dt2
L ~ m: inertia term
R dQ/dt
kf v = kf dx/dt
R ~ kf Æ friction (damping) term
1/C Q
ke x
1/C ~ke Æ elastic term due to spring
G. Sciolla – MIT
8.022 – Lecture 17
5
Undriven RCLs: solution
„
„
Differential equation governing loop:
d 2Q
R dQ
1
Q =0
+
+
2
dt
L dt
LC
Solve using complex number notation:
Q (t ) = e βt = e −αt e i ωt
NB: β = −α + i ω is a complex number, with α and ω real
e −αt = damping term, e i ωt = oscillatory term
Throw this into the equation and we get a quadratic equation in β :
β2 + β
R
1
R
R2
1
+
=0 ⇒ β =−
±
−
4L2 LC
L LC
2L
G. Sciolla – MIT
8.022 – Lecture 17
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3
RCL circuits: solution
⎧
⎪
2
⎪• β purely real: R − 1 > 0 ⇒ R>2 L ⇒
⎪
4L2 LC
C
⎪
⎪
⎪• β purely imaginary: ⇒ R = 0 ⇒ undamped LC ⇒
R
R2
1 ⎪
β =− ±
:⎨
−
2L
4L2 LC ⎪
2
⎪• β truly complex: R>0 and R 2 − 1 < 0 ⇒
4L LC
⎪
Q (t ) = e βt = e −αt e i ωt ⎪
2
⎪α= R and ω= 1 − R
⎪ 2L
LC 4L2
⎪
⎩
2
R
1
When 2 −
= 0 critical damping (fastest way to damp an oscillator).
4L LC
G. Sciolla – MIT
8.022 – Lecture 17
7
RCL in weak damping limit
„
Initial conditions: Q(0)=Q0 =Acos(φ0 ) and I(0)=0=Aω0 sinφ0 ⇒ A = Q 0 ; φ0 = 0
R
− t
⎧
2L
cos(ω 0t )
⎪Q (t ) ~ Q 0e
⇒ ⎨
R
− t
⎪ I (t ) ~ ω Q e 2 L sin(ω t )
0 0
0
⎩
„
Graphical representation of solution:
I(t)
t
G. Sciolla – MIT
8.022 – Lecture 17
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4
Energy
„
Energy of the circuit in the weak damping limit:
U C (t ) =
U L (t ) =
Q 2 (t ) Q 02 −Rt / L
e
cos 2 ω0t
=
2C
2C
Q2
1
1
LI (t )2 = ω02LQ 20 e −Rt / L sin2 ω0t = 0 e −Rt / L sin2 ω0t
2
2
2C
⇒ U (t ) = U L (t ) + U C (t ) =
„
Q 02 −Rt / L
Q2
(sin2 ω0t + cos 2 ω0t ) = 0 e −Rt / L
e
2C
2C
Since Q20/2C=total energy stored initially in the system
Æ U decreases exponentially over time: as expected!
G. Sciolla – MIT
8.022 – Lecture 17
9
Quality Factor
„
Definition 1: the quality factor measures how many times the circuit oscillates
before it loses a certain amount of energy
Q2
U (t )= 0 e −Rt / L
In the time τ =L/R the energy decreases by ∆U(t)=1/e
2C
The oscillation is ωτ radians ⇒ Q = ωτ =
„
Definition 2: the quality factor measures the ratio between energy stored (in
C and L) and average power dissipated (in R)
For an oscillation with frequency ω ⇒ Q = ω
„
ωL
R
LI 2 / 2 ωL
Energy stored
= ω 02
=
<Power>
RI 0 / 2 R
Q factor can be defined for any system that creates vibrations.
„
Acoustics: Q of a tuning fork is much higher than the Q of a table…
G. Sciolla – MIT
8.022 – Lecture 17
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5
Today’s goal:
Driven RCL circuits
„
~ is an AC e.m.f.
AC voltage supplied to the circuit:
„
Convenient assumption:
„
emf (t ) = V 0 cos ωt
V (t ) = Re ⎡⎣V (t ) ⎤⎦
„
with V (t ) = V 0e i ωt
„ NB: V0 is purely real!
How to solve this? Just generalize what we used for DC!
„ Sum of voltage drops in loop is equal to emf (Kirchoff #2)
Vemf (t ) = VR (t ) +VC (t ) +V L (t )
Vemf (t ) = VR (t ) +VC (t ) +V L (t )
The same current must pass through every circuit element
„
I (t ) = I R (t ) = I C (t ) = I L (t )
G. Sciolla – MIT
I (t ) = I R (t ) =8.022
I C (t –) Lecture
= I L (t 17
)
11
AC current
„
Consider a B constant in magnitude and a loop rotating
around its axis with angular velocity ω
ω
B
θ
∫
„
If S is the area of the loop: B ida = BS cos θ = BS cos ωt
S
Faraday:
1 ∂
ω
e .m .f . =
(BS cos ωt ) = BS sin ωt
c ∂t
c
„
This is how AC power is generated. In U.S.: ν=60 Hz Æ ω=377
„
G. Sciolla – MIT
8.022 – Lecture 17
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6
AC emf + resistor R
„
Ohm’s law holds for AC too:
I
V (t ) = VR (t ) = I (t )R
„
V
Let’s plot I(t) and V(t) on the same graph:
~
R
V(t)
--- I(t)
__ V(t)
t
Æ In a resistor the voltage and the current are in phase
(peak voltage occurs at the same time as peak current)
G. Sciolla – MIT
8.022 – Lecture 17
13
Reminder: phasor notation
Any complex number z = x + i y
with i= -1
can always be represented as the product of a real number (magnitude)
and a complex exponential:
⇒ z = re i θ
(Phasor representation)
y
where magnitude r= x 2 +y 2 and phase θ =arctg
x
⇒ z = r (cos θ + i sin θ )
and given Euler’s relation:
e i θ = cos θ + i sin θ
which can be easily proved using
Maclaurin expansion
G. Sciolla – MIT
y
8.022 – Lecture 17
x
z = x +i y
y
x
14
7
AC emf + R with phasors
„
The same information can be represented with phasors in the
complex plane:
V (t ) = RI (t )
I
~
V
R
Æ In a resistor the voltage and the current are in phase
In phase means that both phasors are at the same angle
G. Sciolla – MIT
8.022 – Lecture 17
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AC emf + capacitor C
„
Connect AC emf across a capacitor C:
Q (t )
V (t ) = VC (t ) =
C
~
V
C
Since V(t)=V0cosωt and I(t)= dQ/dt:
dQ (t )
π
= −ωCV 0 sin ωt = ωCV 0 cos(ωt + )
I (t ) =
dt
2
Æ I(t) LEADS V(t) by 90 deg / V(t) lags I(t) by 90 deg
(maxima in I(t) occur before maxima in V(t))
„
V(t)
t
G. Sciolla – MIT
8.022 – Lecture 17
--- I(t)
__ V(t)
16
8
Ohm’s law revisited and Impedance
„
Relation between I(t) and V(t) becomes more obvious when using
phasor notation:
VC (t ) = V 0 cos ωt = Re ⎡⎣VC (t ) ⎤⎦
„
with
V (t ) = V 0e i ωt
For the current:
π
I (t ) = ωCV 0 cos(ωt + ) = Re ⎡⎣I C (t ) ⎤⎦
2
with I (t ) = ωCV 0e
„
π⎞
⎛
i ⎜ ωt + ⎟
2
⎝
⎠
i
π
= i ωCV 0e i ωt (remember: e 2 = i )
Combining complex currents and voltages we can write:
V (t ) = I (t )Z C (complex equivalent of Ohm's law)
where Z C is the impedance of a capacitor: Z C =
G. Sciolla – MIT
8.022 – Lecture 17
1
iω C
17
AC emf + C: phasor representation
„
Given
V (t ) = V 0e i ωt
and
I (t ) = Z CV 0e i ωt =i ωCV 0e i ωt
V(t) and I(t) can easily be represented in the complex plane:
NB: I(t) is ahead of V(t) by 90 degrees: I(t) leads V(t) by 90 degrees
G. Sciolla – MIT
8.022 – Lecture 17
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9
AC emf + inductor L
„
„
Connect AC emf across an inductor L:
dI
V (t ) = VL (t ) = L
dt
Since V(t)=V0cosωt:
dI V 0
=
cos ωt
dt
L
⇒
I (t ) =
~
L
V
V0
V
π⎞
⎛
sin ωt = 0 cos ⎜ ωt − ⎟
ωL
ωL
2⎠
⎝
V(t)
t
--- I(t)
__ V(t)
Æ I(t) LAGS V(t) by 90 degrees, or V(t) LEADS I(t) by 90 degrees
(maxima in I(t) occur before maxima in V(t))
G. Sciolla – MIT
8.022 – Lecture 17
19
Impedance of inductors
„
Using phasor notation:
VC (t ) = V 0 cos ωt = Re ⎡⎣VL (t ) ⎤⎦
„
V (t ) = V 0e i ωt
The current is:
I (t ) =
V0
π
cos(ωt − ) = Re ⎡⎣I (t ) ⎤⎦
ωL
2
⎛
with I (t ) =
„
with
π⎞
π
-i
V 0 i ⎜⎝ ωt − 2 ⎟⎠ V 0 i ωt
−1
=
e
e
(remember: e 2 = ( i ) = −i )
ωL
i ωL
Combining complex currents and voltages we can write:
V (t ) = I (t )Z L (complex equivalent of Ohm's law)
where ZL is the impedance of an inductor: ZL =iωL
G. Sciolla – MIT
8.022 – Lecture 17
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10
AC emf + L: phasor representation
„
V 0 i ωt
e
i ωL
V(t) and I(t) can easily be represented in the complex plane:
Given V (t ) = V 0e i ωt
and
I (t ) = Z LV 0e i ωt =
NB: I(t) is 90 degrees behind V(t): I(t) lags V(t) by 90 degrees
G. Sciolla – MIT
8.022 – Lecture 17
21
Driven RCLs using inductance
„
„
Inductance simplifies the study of driven
RCL circuits
Let’s work with complex numbers and use
Ohm’s and Kirchoff’s extensions
Vemf (t ) = VR (t ) +VC (t ) +V L (t )
⎧
V R (t ) = RI (t )
⎪
1
⎪
I (t )
Since ⎨VC (t ) = Z C I (t ) =
i
ω
C
⎪
⎪ V L (t ) = Z L I (t ) = i ωLI (t )
⎩
⎛
1
⎛
⇒ V emf (t ) = I (t ) ⎜ R + i ⎜ ωL −
ω
C
⎝
⎝
1 ⎞
⎛
where total impedance of the circuit is Z tot ≡ R + i ⎜ ωL −
ω
C ⎟⎠
⎝
G. Sciolla – MIT
8.022 – Lecture 17
⎞⎞
⎟ ⎟ = I (t )Z tot
⎠⎠
22
11
Driven RCLs: phasor notation
„
The complex current can be written as
I (t ) =
„
V emf (t )
=
Z tot
V 0e i ωt
This can be written as:
I (t ) =
⎛
⎝
V 0e i ωt
V e i ωt
= 0 * Z *tot =
Z tot
Z tot Z tot
Remembering that e-iθ
G. Sciolla – MIT
1 ⎞
⎠
R + i ⎜ ωL −
ωC ⎟
V 0e i ωt
⎡
1 ⎞⎤
⎛
i ωt − i φ
⎢R − i ⎜ ωL − ωC ⎟ ⎥ = I 0e e
⎝
⎠
1
⎣
⎦
⎛
⎞
R 2 + ⎜ ωL −
ωC ⎟⎠
⎝
V0
⎧
⎪ I0 =
2
1 ⎞
⎛
⎪
2
R
L
ω
+
−
⎜
⎟
⎪
ωC ⎠
⎝
= cos θ − i sin θ ⇒ ⎨
⎪
1
ωL −
⎪
C = ωL − 1
ω
⎪ tgφ =
8.022 – Lecture
17
23
R
R ωRC
⎩
2
Dependence of φ from ω
φ(ω)
ω
tgφ =
1
ωL
−
R ωRC
NB: I (t ) = I 0e i ωt e −i φ
→ high ω: I lags voltage by 90 o
→ low ω: I leads voltage by 90 o
G. Sciolla – MIT
8.022 – Lecture 17
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12
AC motor (H26)
Coil 1
„
2 RL circuits driven by 60 Hz AC voltage
„
„
„
Coil 2
What is the ∆φ between the 2 currents?
„
„
„
~
Coil 1: R=2.3 Ω, L=1.5mH
Coil 2: R=2.5 Ω, L=31 mH
10-3
Z1=R1+iωL1=2.3+i 377 1.5
Z2=R2+iωL2=2.5+i 377 31 10-3
Æ ∆φ=64 degrees
~
The difference in phase will create a rotating B field Æ
Eddie currents in the metal can will make it rotate!
G. Sciolla – MIT
8.022 – Lecture 17
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Dependence of I0 from ω
I0
V0
I0 =
⎛
⎝
R 2 + ⎜ ωL −
1 ⎞
ωC ⎟⎠
2
Maximum current when ωL =
⇒ ω0 =
ω0
G. Sciolla – MIT
1
LC
1
ωC
resonance frequency
ω
8.022 – Lecture 17
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13
RCL resonance (Demo L8)
RCL circuit driven with variable frequency ω
„
„
„
L=50 mH
C=0.3 µF
scope
Measure VR on scope and tune frequency to maximize VR
„
„
What is the expect resonance frequency?
ω0 =
G. Sciolla – MIT
1
LC
= 8.2 × 103 ⇒ ν = 1.3 kHz
8.022 – Lecture 17
27
Demo L8: part 2
„
Same RCL circuit driven with variable frequency ω
„
„
„
„
Frequency is driven by a voltage Vin
L=50 mH
C=0.3 µF
scope
Display VR vs on the scope while sweeping Vin
„
What do you expect to see?
G. Sciolla – MIT
8.022 – Lecture 17
ω0=1.3 kHz
28
14
Resonant RCL with light bulb (L6)
„
RCL circuit driven by AC voltage
„
„
„
C can be adjusted using set of switches
L can be adjusted moving the Fe core
inside a solenoid
For each setting of C we can find an L that turn on the
light bulb
„
What is that L?
G. Sciolla – MIT
L=
1
C ω2
8.022 – Lecture 17
29
Summary and outlook
„
Today:
„
Undriven RCL circuits
„
„
Driven RCL AC circuits
„
„
Energy stored and quality factor in weak damping limit
Simple solution when introducing complex impedance Z
„
ZR = R
„
ZC = 1/(iωC)
„
ZL = iωL
Next Tuesday:
„
More on driven RCLs: power, resonances, filters…
G. Sciolla – MIT
8.022 – Lecture 17
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