Chapter 31: RLC Circuits
PHY2049: Chapter 31
1
Topics
ÎLC
Oscillations
‹ Conservation
ÎDamped
‹ Energy
ÎAC
of energy
oscillations in RLC circuits
loss
current
‹ RMS
ÎForced
quantities
oscillations
‹ Resistance,
reactance, impedance
‹ Phase
shift
‹ Resonant frequency
‹ Power
ÎTransformers
‹ Impedance
matching
PHY2049: Chapter 31
2
LC Oscillations
ÎWork
out equation for LC circuit (loop rule)
q
di
− −L =0
C
dt
ÎRewrite
C
using i = dq/dt
d 2q
q
d 2q
2
+
ω
L 2 + =0 ⇒
q=0
2
C
dt
dt
‹ω
L
ω=
1
LC
(angular frequency) has dimensions of 1/t
ÎIdentical
m
d 2x
dt
2
to equation of mass on spring
+ kx = 0 ⇒
d 2x
dt
2
2
+ω x = 0
PHY2049: Chapter 31
k
ω=
m
3
LC Oscillations (2)
is same as mass on spring ⇒ oscillations
k
q = qmax cos (ω t + θ )
ω=
m
ÎSolution
‹ qmax
is the maximum charge on capacitor
‹ θ is an unknown phase (depends on initial conditions)
ÎCalculate
current: i = dq/dt
i = −ω qmax sin (ω t + θ ) = −imax sin (ω t + θ )
ÎThus
both charge and current oscillate
frequency ω, frequency f = ω/2π
‹ Period: T = 2π/ω
‹ Angular
PHY2049: Chapter 31
4
Plot Charge and Current vs t
ω = 1 T = 2π
q (t )
i (t )
PHY2049: Chapter 31
5
Energy Oscillations
ÎTotal
energy in circuit is conserved. Let’s see why
di q
L + =0
Equation of LC circuit
dt C
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
PHY2049: Chapter 31
6
Oscillation of Energies
ÎEnergies
can be written as (using ω2 = 1/LC)
2
q 2 qmax
=
cos 2 (ω t + θ )
UC =
2C
2C
2
q
2
sin 2 (ω t + θ ) = max sin 2 (ω t + θ )
U L = 12 Li 2 = 12 Lω 2 qmax
2C
2
qmax
ÎConservation of energy: U C + U L =
= const
2C
ÎEnergy
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
PHY2049: Chapter 31
7
UC (t )
Plot Energies vs t
U L (t )
PHY2049: Chapter 31
Sum
8
LC Circuit Example
ÎParameters
‹C
= 20μF
‹ L = 200 mH
‹ Capacitor initially charged to 40V, no current initially
ÎCalculate
‹ω
ω, f and T
= 500 rad/s
‹ f = ω/2π = 79.6 Hz
‹ T = 1/f = 0.0126 sec
ÎCalculate
ω = 1/ LC = 1/
( 2 ×10−5 ) ( 0.2 ) = 500
qmax and imax
= CV = 800 μC = 8 × 10-4 C
= ωqmax = 500 × 8 × 10-4 = 0.4 A
‹ qmax
‹ imax
ÎCalculate
‹ UC
maximum energies
= q2max/2C = 0.016J UL = Li2max/2 = 0.016J
PHY2049: Chapter 31
9
LC Circuit Example (2)
ÎCharge
and current
q = 0.0008cos ( 500t ) i = −0.4sin ( 500t )
ÎEnergies
U C = 0.016cos 2 ( 500t )
U L = 0.016sin 2 ( 500t )
ÎVoltages
VC = q / C = 40cos ( 500t )
VL = Ldi / dt = − Lω imax cos ( 500t ) = −40cos ( 500t )
ÎNote
how voltages sum to zero, as they must!
PHY2049: Chapter 31
10
RLC Circuit
ÎWork
out equation using loop rule
di
q
L + Ri + = 0
dt
C
ÎRewrite
using i = dq/dt
d 2q
R dq q
+
+
=0
2
L dt LC
dt
ÎSolution
q = qmax e
ÎThis
slightly more complicated than LC case
−tR / 2 L
cos (ω ′t + θ ) ω ′ = 1/ LC − ( R / 2 L )
2
is a damped oscillator (similar to mechanical case)
‹ Amplitude
of oscillations falls exponentially
PHY2049: Chapter 31
11
Charge and Current vs t in RLC Circuit
q (t )
i (t )
e
PHY2049: Chapter 31
−tR / 2 L
12
RLC Circuit Example
ÎCircuit
‹L
parameters
= 12mL, C = 1.6μF, R = 1.5Ω
ÎCalculate
‹ω
ω, ω’, f and T
= 7220 rad/s
‹ ω’ = 7220 rad/s
‹ f = ω/2π = 1150 Hz
‹ T = 1/f = 0.00087 sec
ÎTime
ω = 1/
( 0.012 ) (1.6 ×10−6 ) = 7220
ω ′ = 72202 − (1.5/ 0.024 )
2
ω
for qmax to fall to ½ its initial value e−tR / 2 L = 1/ 2
‹t
= (2L/R) * ln2 = 0.0111s = 11.1 ms
‹ # periods = 0.0111/.00087 ≈ 13
PHY2049: Chapter 31
13
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
PHY2049: Chapter 31
14
Energy in RLC Circuit
UC (t )
U L (t )
Sum
e
−tR / L
PHY2049: Chapter 31
15
Quiz
ÎBelow
are shown 3 LC circuits. Which one takes the least
time to fully discharge the capacitors during the
oscillations?
‹ (1)
A
‹ (2) B
‹ (3) C
C
A
ω = 1/ LC
C
C
C
B
C
C
C has smallest capacitance, therefore highest
frequency, therefore shortest period
PHY2049: Chapter 31
16
AC Circuits
ÎEnormous
impact of AC circuits
‹ Power
delivery
‹ Radio transmitters and receivers
‹ Tuners
‹ Filters
‹ Transformers
ÎBasic
components
‹R
‹L
‹C
‹ Driving
ÎNow
emf
we will study the basic principles
PHY2049: Chapter 31
17
AC Circuits and Forced Oscillations
+ “driving” EMF with angular frequency ωd
ε = ε m sin ω d t
ÎRLC
L
di
q
+ Ri + = ε m sin ω d t
dt
C
ÎGeneral
solution for current is sum of two terms
“Transient”: Falls
exponentially & disappears
“Steady state”:
Constant amplitude
Ignore
i ∼ e−tR / 2 L cos ω ′t
PHY2049: Chapter 31
18
Steady State Solution
ÎAssume
steady state solution of form i = I m sin (ω d t − φ )
‹ Im
is current amplitude
‹ φ is phase by which current “lags” the driving EMF
‹ Must determine Im and φ
in solution: differentiate & integrate sin(ωt-φ)
i = I m sin (ω d t − φ )
ÎPlug
di
= ω d I m cos (ω d t − φ )
dt
Im
q=−
cos (ω d t − φ )
Substitute
L
di
q
+ Ri + = ε m sin ω t
dt
C
ωd
Im
I mω d L cos (ω dτ − φ ) + I m R sin (ω d t − φ ) −
cos (ω d t − φ ) = ε m sin ω d t
ωd C
PHY2049: Chapter 31
19
Steady State Solution for AC Current (2)
Im
I mω d L cos (ω dτ − φ ) + I m R sin (ω d t − φ ) −
cos (ω d t − φ ) = ε m sin ω d t
ωd C
ÎExpand
sin & cos expressions
sin (ω d t − φ ) = sin ω d t cos φ − cos ω d t sin φ
cos (ω d t − φ ) = cos ω d t cos φ + sin ω d t sin φ
ÎCollect
sinωdt & cosωdt terms separately
(ω d L − 1/ ω d C ) cos φ − R sin φ = 0
I m (ω d L − 1/ ω d C ) sin φ + I m R cos φ = ε m
ÎThese
High school trig!
cosωdt terms
sinωdt terms
equations can be solved for Im and φ (next slide)
PHY2049: Chapter 31
20
Steady State Solution for AC Current (3)
(ω d L − 1/ ω d C ) cos φ − R sin φ = 0
Same equations
I m (ω d L − 1/ ω d C ) sin φ + I m R cos φ = ε m
for φ and Im in terms of
ω L − 1/ ωd C X L − X C
εm
tan φ = d
≡
Im =
R
R
Z
ÎR, XL, XC and Z have dimensions of resistance
ÎSolve
X L = ωd L
Inductive “reactance”
X C = 1/ ωd C
Capacitive “reactance”
Z = R2 + ( X L − X C )
ÎLet’s
2
Total “impedance”
try to understand this solution using “phasors”
PHY2049: Chapter 31
21
Understanding AC Circuits Using Phasors
ÎPhasor
‹ Voltage
or current represented by “phasor”
‹ Phasor rotates counterclockwise with angular velocity = ωd
‹ Length of phasor is amplitude of voltage (V) or current (I)
‹ y component is instantaneous value of voltage (v) or current (i)
ε
ε = ε m sin ω d t
i
i = I m sin (ω d t − φ )
εm
Im
ωdt − φ
Current “lags” voltage by φ
PHY2049: Chapter 31
22
AC Source and Resistor Only
ÎVoltage
i
is vR = iR = VR sin ω d t
ÎRelation
of current and voltage
i = I R sin ω d t I R = VR / R
‹ Current
ε ~
R
is in phase with voltage (φ = 0)
IR
VR
PHY2049: Chapter 31
ωdt
23
AC Source and Capacitor Only
ÎVoltage
vC = q / C = VC sin ω d t
is
ÎDifferentiate
to find current
q = CVC sin ωd t
i = dq / dt = ωd CVC cos ωd t
ÎRewrite
i
ε ~
C
using phase
i = ωd CVC sin (ωd t + 90° )
ÎRelation
of current and voltage
i = I C sin (ω d t + 90° ) I C = VC / X C
ΓCapacitive
‹
ωdt
IC
ωdt + 90
reactance”: X C = 1/ ωd C
Current “leads” voltage by 90°
PHY2049: Chapter 31
VC
24
AC Source and Inductor Only
ÎVoltage
vL = Ldi / dt = VL sin ω d t
is
ÎIntegrate
di/dt to find current:
di / dt = (VL / L ) sin ωd t
i = − (VL / ω d L ) cos ω d t
ÎRewrite
i
ε ~
L
using phase
i = (VL / ω d L ) sin (ω d t − 90° )
ÎRelation
of current and voltage
i = I L sin (ω d t − 90° ) I L = VL / X L
ΓInductive
‹
reactance”: X L
= ωd L
VL
ωdt
IL
ωdt − 90
Current “lags” voltage by 90°
PHY2049: Chapter 31
25
What is Reactance?
Think of it as a frequency-dependent resistance
1
XC =
ωd C
X L = ωd L
ωd → 0, XC → ∞
- Capacitor looks like a break
ωd → ∞, XC → 0
- Capacitor looks like a wire (“short”)
ωd → 0, XL → 0
- Inductor looks like a wire (“short”)
ωd → ∞, XL → ∞
- Inductor looks like a break
( "XR " = R )
Independent of ωd
PHY2049: Chapter 31
26
Quiz
ÎThree
identical EMF sources are hooked to a single circuit
element, a resistor, a capacitor, or an inductor. The
current amplitude is then measured as a function of
frequency. Which one of the following curves corresponds
to an inductive circuit?
‹ (1)
a
‹ (2) b
‹ (3) c
‹ (4) Can’t tell without more info
a
b
Im
c
fd
X L = ωd L
For inductor, higher frequency gives higher
reactance, therefore lower current
PHY2049: Chapter 31
27
AC Source and RLC Circuit
Im
ÎVoltage is ε = ε m sin ω d t
ÎRelation
of current and voltage
i = I m sin (ω d t − φ )
“lags” voltage by φ
‹ Impedance: Due to R, XC and XL
‹ Current
ÎCalculate Im
‹ See
and φ using geometry
VL
next slide
εm V
R
ωdt − φ
VC
PHY2049: Chapter 31
28
AC Source and RLC Circuit (2)
triangle with sides VR, VL-VC and εm
VL − VC
VR = I m R
tan φ =
VR
VL = I m X L
2
VL−VC
ε m2 = VR2 + (VL − VC )
V =I X
ÎRight
C
ÎSolve
m
C
εm
φ
VR
for current: i = I m sin (ω d t − φ )
‹ (Magnitude
= Im, lags emf by phase φ)
Im = εm / Z
X L − X C ωd L − 1/ ωd C
tan φ =
=
R
R
Z = R + ( X L − X C ) = R + (ωd L − 1/ ωd C )
2
2
2
PHY2049: Chapter 31
2
29
AC Source and RLC Circuit (3)
X L − XC
tan φ =
R
ÎOnly
Z = R2 + ( X L − X C )
2
XL − XC is relevant, reactances cancel each other
ÎWhen
XL = XC, then φ = 0
in phase with emf, “Resonant circuit”: ω d
‹ Z = R (minimum impedance, maximum current)
‹ Current
ÎWhen
XL < XC, then φ < 0
‹ Current
ÎWhen
= ω0 = 1/ LC
leads emf, “Capacitive circuit”: ωd < ω0
XL > XC, then φ > 0
‹ Current
lags emf, “Inductive circuit”: ωd > ω0
PHY2049: Chapter 31
30
RLC Example 1
ÎBelow
are shown the driving emf and current vs time of
an RLC circuit. We can conclude the following
‹ Current
“leads” the driving emf (φ<0)
‹ Circuit is capacitive (XC > XL)
‹ ωd < ω0
I
ε
t
PHY2049: Chapter 31
31
Quiz
ÎWhich
one of these phasor diagrams corresponds to an
RLC circuit dominated by inductance?
‹ (1)
Circuit 1
‹ (2) Circuit 2
‹ (3) Circuit 3
Inductive: Current lags emf, φ>0
εm
1
εm
2
εm
3
PHY2049: Chapter 31
32
Quiz
ÎWhich
one of these phasor diagrams corresponds to an
RLC circuit dominated by capacitance?
‹ (1)
Circuit 1
‹ (2) Circuit 2
‹ (3) Circuit 3
Capacitive: Current leads emf, φ<0
εm
1
εm
2
εm
3
PHY2049: Chapter 31
33
RLC Example 2
ÎR
= 200Ω, C = 15μF, L = 230mH, εm = 36v, fd = 60 Hz
‹ ωd
= 120π = 377 rad/s
‹ Natural
‹ XL
frequency ω 0
= 1/
( 0.230 ) (15 ×10−6 ) = 538rad/s
= 377 × 0.23 = 86.7Ω
(
)
‹
X C = 1/ 377 ×15 × 10−6 = 177Ω
‹
Z = 2002 + ( 86.7 − 177 ) = 219Ω
‹
I m = ε m / Z = 36 / 219 = 0.164 A
‹
XL < XC
Capacitive circuit
2
−1 ⎛ 86.7 − 177 ⎞
φ = tan ⎜
⎝
200
⎟ = −24.3°
⎠
PHY2049: Chapter 31
Current leads emf
(as expected)
34
RLC Example 3
ÎCircuit
parameters: C = 2.5μF, L = 4mH, εm = 10v
‹ ω0
= (1/LC)1/2 = 104 rad/s
‹ Plot Im vs ωd / ω0
R = 5Ω
R = 10Ω
R = 20Ω
Im
Resonance
ω d = ω0
PHY2049: Chapter 31
35
Power in AC Circuits
ÎInstantaneous power
P = I m2 sin 2 ω d t − φ
(
ÎMore
emitted by circuit: P = i2R
)
Instantaneous power oscillates
useful to calculate power averaged over a cycle
‹ Use
<…> to indicate average over a cycle
P = I m2 R sin 2 (ω d t − φ ) = 12 I m2 R
ÎDefine
RMS quantities to avoid ½ factors in AC circuits
I rms =
ÎHouse
‹ Vrms
Im
2
ε rms =
εm
2
2
Pave = I rms
R
current
= 110V ⇒ Vpeak = 156V
PHY2049: Chapter 31
36
Power in AC Circuits (2)
2
P
=
I
ÎRecall power formula ave
rms R
ÎRewrite Pave =
‹ Based
on
ε rms
Z
cos φ =
I rms R = ε rms I rms cos φ
VR
εm
Im R R
=
=
ImZ Z
R
cos φ =
Z
Pave = ε rms I rms cos φ
Îcosφ
is the “power factor”
maximize power delivered to circuit ⇒ make φ close to zero
‹ Most power delivered to load happens near resonance
‹ E.g., too much inductive reactance (XL) can be cancelled by
increasing XC (decreasing C)
‹ To
PHY2049: Chapter 31
37
Power Example 1
ÎR
= 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, fd = 60 Hz
‹ ωd
= 120π = 377 rad/s
‹
Z = 200 + ( 80 − 120 ) = 211.9Ω
‹
I rms = ε rms / Z = 120 / 211.9 = 0.566 A
‹
2
2
−1 ⎛ 80 − 150 ⎞
φ = tan ⎜
⎟ = −19.3°
⎠
Current leads emf
Capacitive circuit
‹
⎝ 200
cos φ = 0.944
‹
Pave = ε rms I rms cos φ = 120 × 0.566 × 0.944 = 64.1W
‹
Same
2
Pave = I rms
R = 0.5662 × 200 = 64.1W
PHY2049: Chapter 31
38
Power Example 1 (cont)
ÎR
= 200Ω, XC = 150Ω, XL = 80Ω, εrms = 120v, fd = 60 Hz
ÎHow
much capacitance must be added to maximize the
power in the circuit (and thus bring it into resonance)?
‹ Want
XC = XL to minimize Z, so must decrease XC
‹
X C = 150Ω = 1/ ω d C
C = 17.7μF
‹
X C new = X L = 80Ω
Cnew = 33.2μF
‹
So we must add 15.5μF capacitance to maximize power
PHY2049: Chapter 31
39
Q Factor
ÎWe
can quantify low damping situations using Q factor
‹ Define
Q ≡ 2π
Q = 2π
ÎCan
using values at resonance
Energy in circuit
Energy lost per cycle
( 2π
1
2
LI m2
)
/ ω0 × 12 I m2 R
=
ω0 L
R
also be expressed (using resonance values) as
X L XC
Q=
=
R
R
PHY2049: Chapter 31
40
Power Example 2
ÎCircuit
parameters: C = 2.5μF, L = 4mH, εm = 10v
‹ ω0
= (1/LC)1/2 = 10000 rad/s
‹ Plot Power vs ωd / ω0 for different R values, using Q
R = 2Ω Q = 20
R = 5Ω Q = 8
Pave
Resonance
ω d = ω0
R = 10Ω Q = 4
R = 20Ω Q = 2
PHY2049: Chapter 31
41
Q Factor (cont)
2
Pave = I rms
R=
ÎFor
2
R
ε rms
2
−
+
L
1/
C
R
ω
ω
(
)
2
Q > 3 or so, can easily show that Q ≈
ω0
FWHM
FWHM = Full Width at Half Maximum
PHY2049: Chapter 31
42
Radio Tuner
Set RLC tuner to 103.7 (ugh!)
Other radio stations.
RLC response is less
Circuit response Q = 500.
Maximized for f = 103.7
PHY2049: Chapter 31
43
Quiz
ÎA
generator produces current at a frequency of 60 Hz with
peak voltage and current amplitudes of 100V and 10A,
respectively. What is the average power produced?
‹ (1)
‹ (2)
‹ (3)
‹ (4)
‹ (5)
1000 W
707 W
1414 W
500 W
250 W
Pave = 12 ε peak I peak = ε rms I rms
PHY2049: Chapter 31
44
Quiz
ÎThe
figure shows the current and emf of a series RLC
circuit. To increase the rate at which power is delivered to
the resistive load, which option should be taken?
‹ (1)
Increase R
‹ (2) Decrease L
‹ (3) Increase L
‹ (4) Increase C
Current lags applied emf (peak occurs later, φ > 0), thus circuit is
inductive. Max power is at φ = 0, so need to either (1) reduce XL by
decreasing L or (2) cancel XL by increasing XC (decrease C).
PHY2049: Chapter 31
45
RLC Circuit Example
If you wanted to increase the power
delivered to this RLC circuit, which
modification(s) would work?
(a) increase R
(d) decrease R
(b) increase C
(e) decrease C
(c) increase L
(f) decrease L
Again, current lags emf.
See previous page for details.
PHY2049: Chapter 31
46
Example (Prob. 31-48)
frequency EMF source with εm=6V connected to a
resistor and inductor. R=80Ω and L=40mH.
ÎVariable
‹ At
what frequency fd does VR = VL?
ω d L = R ⇒ ω d = 2000 rad/s
f d = ω d / 2π = 318Hz
‹ At
εm
VL
VR
Im
that frequency, what is phase angle φ?
tan φ = 1 ⇒ φ = 45°
‹ What
is current amplitude?
I m = ε m / 802 + 802 = 6 /113 = 0.053A
PHY2049: Chapter 31
47