8.022 (E&M) – Lecture 18
Topics:
„
RCL circuits: the hardest of the easiest part of the course?
„
More on complex impedance
„
Power and energy
„
Filters
Last time: AC driven RCLs
Simple solution when introducing following rules:
„
Work with complex V and I
„
Generalization of Ohm’s law to complex V and I:
„
Real currents and voltages are just the real part of the V and I.
V (t ) = I (t )Z X
„
⎧ ZR = R
⎪
1
⎪
where Z X is the impedance of component X: ⎨ Z C =
i
C
ω
⎪
⎪⎩ ZL =iωL
Analyze circuit as if it were DC with only resistors
„
Take the real part of I(t) and V(t)
„
The End.
G. Sciolla – MIT
8.022 – Lecture 17
2
1
“Analyze as DC with only resistors”
What do I mean with this statement?
„
Impedances in series
„
Z2
Z1
Same current flowing in each element
I1Z1=V1; I2Z2=V2; V1+V2=V; V=ZI
~
Æ Zeq=Z1+Z2
„
Impedances in parallel
„
Same voltage drop across each element
„
V1/Z1=V2/Z2=V/Zeq; V1=V2=V
Z1
Z2
~
Æ 1/Zeq=1/Z1+1/Z2
Æ Same rules as resistors in series and parallel!
G. Sciolla – MIT
8.022 – Lecture 17
3
Is the current leading or lagging?
Instead of thinking of the problems in terms of complex
currents, think in terms of complex impedance!
„
Generalized Ohm’s law: V (t ) = I (t )Z C
„
„
All what we really care about is amplitude of I and relative
phase between I and V
Trick: let’s choose V real (no law against it!) and draw the
complex I, V and Z in the complex plane
Z
Z
ωL
V
R
1/ωC
G. Sciolla – MIT
I=|V|/|Z|
φ=-φΖ
Ι
8.022 – Lecture 17
4
2
Is current leading or lagging? (2)
Consider the complex impedance:
„
Real part: only R contributes
„
Imaginary part: ZL “pulls up” by ωL and ZC pulls down by 1/ωC
Æ The phase of Z will depend on who prevails:
Z
ωL
Z
„
ZL>ZC Æ φΖ>0
„
ZL<ZC Æ φΖ<0;
| Z |= Re2 (Z ) + Im2 (Z ) =
2
1 ⎞
⎛
= R 2 + ⎜ ωL −
ωC ⎟⎠
⎝
1
ωL −
Im(Z )
ωC
tg φZ =
=
Re(Z )
R
φZ
R
1/ωC
G. Sciolla – MIT
8.022 – Lecture 17
5
Is current leading or lagging? (3)
Now remember that V (t ) = I (t )Z C
I (t ) =
V (t ) V (t ) −i φZ
=
e
ZC
ZC
Z
and that we chose a real V:
⇒ if φZ >0, I will be lagging V
if φZ <0, I will be leading V
Z
I=|V|/|Z|
φ=-φΖ
ωL
Z
V
φZ
R
1/ωC
Ι
G. Sciolla – MIT
8.022 – Lecture 17
6
3
Power in RCL circuits
„
Power delivered in a circuit is
P (t ) = V (t ) I (t )
⎧V (t ) = V 0 cos ω t
⎨
⎩ I (t ) = I 0 cos(ω t − φ )
„
Given
„
The average power over a period T will be
P
=
„
„
=
1
T
∫V (t ) I (t )dt
T
=
ω
2π
∫V
0
co s ω tI 0 co s(ω t − φ )dt =
T
ω V 20
co s ω t co s( ω t − φ )dt
2 π | Z | T∫
NB: when we say light bulb has a P of 100W we are referring to <P>
Using the identity: cos(α−β)=cosαcosβ+sinαsinβ we obtain:
ω
⎤
ω V 0 2 ⎡ 2ωπ
2π
P =
⎢ ∫0 cos ω t cos ω t cos φ dt + ∫0 cos ω t sin ω t sin φ dt ⎥
2π | Z | ⎣
⎦
G. Sciolla – MIT
8.022 – Lecture 17
7
Power in RCL circuits (2)
„
„
⎧ ω
⎪⎪
2π
Since: ⎨ ω
⎪
⎪⎩ 2 π
2π
0
∫
ω
2π
0
co s 2 ω t dt =
1
2
⇒
co s ω t sin ω t dt = 0
„
=
1 V 02
cos φ
2 | Z (ω ) |
cosφ=0 Æ no power dissipated in the circuit Æ no work done!
cosφ=0 when φ= 90o Æ when Z is purely imaginary: R needed!
Introducing: RMS (root mean squared) voltage and currents:
I
V
V RMS = 0 and I RMS = 0
2
„
Æ
P
NB: Power depends on relative phase between I and V
„
„
ω
∫
2
NB: in the US: outlet voltage is 120 V. This is the RMS voltage: Vmax=170
P =
V RMS 2
cos φ = RI RMS (ω ) 2
| Z (ω ) |
G. Sciolla – MIT
remembering that cos φ =
8.022 – Lecture 17
R
| Z (ω ) |
8
4
Power vs. frequency
NB: Z depends on ω Æ power dissipated depends on driving frequency!
P =
„
V RMS 2
R =
| Z (ω ) |2
„
1 ⎞
⎛
ωL −
1
ωC
2
R
P(ω)
Resonant
behavior!
R 2 + ⎜ ωL −
ω C ⎟⎠
⎝
At what ω is P is max?
„
„
V RMS 2
=0⇒ω =
1
LC
= ω0
What ω is the max P?
V 2
„
Pm ax = RMS
R
ω
What is the corresponding phase?
„
Zero: the imaginary part due to C and L exactly cancel out!
G. Sciolla – MIT
8.022 – Lecture 17
9
ω0 in term of L and C
What does ω=ω0 mean in terms of L and C?
„ Remember:
1
1
⇔ ωL =
ωC
LC
ω0 =
„
Back to the phasor representation for Z
Z
ωL
The imaginary part due to C exactly
compensates the one due to L
Æ Z is purely real!
1/ωC
G. Sciolla – MIT
R
8.022 – Lecture 17
10
5
How good is the resonant system?
„
„
Definition: width of resonance wrt the height
Width: ∆ω between the points where the
P(ω)
power goes to Pmax/2: ω1 and ω2
V RMS 2
⎛
⎝
R 2 + ⎜ ωL −
1 ⎞
ω C ⎟⎠
2
R =
V RMS 2
1
⇒ ωL −
= ±R
ωC
2R
1
⎧
ω1ω2
2
⎪ ω 1L − ω C = − R
⎪
⎪⎧ ω 1 L C + R C ω 1 − 1 = 0
1
⇒
⎨
⎨ 2
⎪⎩ ω 2 L C − R C ω 2 − 1 = 0
⎪ω L − 1 = R
2
⎪⎩
ω 2C
⎧
− RC ± R 2C 2 + 4 LC
=
⎪ω1 =
ω
Lω0
R
⎪
2LC
⇒ ∆ ω = ω 2 − ω1 =
⇒ Q = res =
⎨
2
2
ω
L
R
∆
RC ± R C + 4 LC
⎪
⎪⎩ω 2 =
LC
unphysica
G. Sciolla –2MIT
8.022 – Lecture
17l
ω
11
Application: FM antenna
Consider the following circuit:
„ L=8.22 µH
-12 F
„ C=0.27 pF=0.27x10
„ R=75 Ω
R
L
C
The radio signal in the air induces an alternated emf in the antenna:
VRMS=9.13 µV
„
Find frequency of incoming wave for which antenna is in tune
Resonance frequency: ω 0 =
ω 0 = 2πν ⇒ ν 0 =
G. Sciolla – MIT
ω0
2π
1
LC
= 6.7 x 10 8
= 106 MHz YES, FM rad i o!
8.022 – Lecture 17
12
6
Application: FM antenna (cont)
„
„
„
„
„
L=8.22 µH
C=0.27 pF=0.27x10-12 F
R=75 Ω
VRMS=9.13 µV
L
C
Calculate IRMS
I RMS = =
„
R
I0
2
=
V RMS
|Z0 |
=
V RMS
(NB : at resonance |Z 0 |=R )
R
∆VRMS across C
VC = I RM S Z C =
1 V RMS
= 0.66 m V
ωC R
Quest ion: VC = 0.66 m V w h il e VR MS = 9 µ V. How can th i s happen?
L and C cancel a l m ost perfect ly ⇒ Z can be sm a ll w h ile C and L
are large and Z~ rea l . NB: a ll ci rcu i ts w i th good Q value have this feature!
G. Sciolla – MIT
8.022 – Lecture 17
13
Application: FM antenna (cont)
„
Calculate width of resonance
R
∆ω
= 9 i10 6 ⇒∆ ν =
= 1.4 MHz
L
2π
Q: is this a good antenna?
∆ω=
R
L
C
No, since separat i on betw een stat i ons is ~ 0.2 MHz
„
Q factor
Q =
ω res
Lω0
=
= 73 good but not enough for a radio.
R
∆ω
How can this be im proved?
Can w e i ncrease L? No, i t w ou l d change frequency
⇒ decreasing R i s the so l ut i on
G. Sciolla – MIT
8.022 – Lecture 17
14
7
Low pass RL filter
„
„
RCL circuits have a frequency dependent response: they can act as
filters (select only certain frequencies)
Example: RL circuit
„
Calculate the complex current
V
V
=
⇒
Z R + i ωL
V
V0
=
| I |=
2
R + i ωL
R + ω 2L2
I =
VR = I R =
V0R
R + ω 2L2
2
⎧ω → 0 : VR → V0
⇒⎨
⇒ low pass filter
⎩ ω → ∞ : VR → 0
G. Sciolla – MIT
8.022 – Lecture 17
15
High pass RL filter
„
What if we take the voltage VL across the inductor?
„
Same complex current
V
V
=
⇒
Z R + i ωL
V
V0
=
| I |=
2
R + i ωL
R + ω 2L2
ωLV0
VL = ωL I =
R 2 + ω 2L2
ωLV0
ωLV0
⎧
→
→0
⎪ω → 0 : VL →
2
2 2
R
R +ω L
⎪⎪
⇒⎨
⇒ high pass filter
ωLV0
LV
→ 0 →V 0
⎪ ω → ∞ : VL →
2
L
R
⎪
ω 2 + L2
⎪
ω
⎩ – MIT
G. Sciolla
8.022 – Lecture 17
I =
16
8
Demo on filters
Low pass RC filter
„
Let’s now study the voltage across a capacitor of a driven RC circuit
„
The complex current is now:
I =
| I |=
V
V
=
i
Z
R −
ωC
V
V0
=
i
R −
ωC
I
VC =
=
ωC
⇒
1
R +
2
V0
ωC
R2 +
1
ω 2C 2
⎧ω → 0 : VR →V0
⇒⎨
⇒ low pass filter
ω C R + 1 ⎩ω → ∞ : VR → 0
V0
=
2
ω 2C 2
G. Sciolla – MIT
2
2
8.022 – Lecture 17
17
High pass RC filter
„
What if we take the voltage VR across the resistor?
„
Same complex current
I =
| I |=
V
V
=
i
Z
R −
ωC
V
R +
G. Sciolla – MIT
1
2
V0
R2 +
V0
=
i
R −
ωC
VR = R I =
⇒
1
ωC2
2
=
ω 2C 2
⎧ ω → 0 : VR → 0
⇒⎨
⇒ high pass filter
ω C R + 1 ⎩ω → ∞ : VR →V0
ωCRV0
2
2
2
8.022 – Lecture 17
18
9
Summary and outlook
„
Today:
„
„
End of RCL circuits
„
Some tricks to make RCL calculations easier
„
Power dissipated in RCL circuits
„
Antennas and high and low pass filters
Next time:
„
Back to Maxwell’s equation:
„
The missing ingredient!
G. Sciolla – MIT
8.022 – Lecture 17
19
10