1. RLC Circuits
PHY2049: Chapter 31 1

2. ω = ω = Work out equation for LC circuit (loop rule) C dt C − − L = 0
LC Oscillations
Work out equation for LC circuit (loop rule)
q di
C dt C
− − L
= 0 L
Rewrite using i = dq/dt
= 0 ⇒ ω2q = 0
dt C dt2 1 LC
d q
q d q
ω = L
ω (angular frequency) has dimensions of 1/t
Identical to equation of mass on spring 2 2 k m
d x
d x
ω =
+kx = 0 ⇒
+ω x = 0 2 m dt dt
PHY2049: Chapter 31 3

3. ω = Solution is same as mass on spring ⇒ oscillations
LC Oscillations (2)
Solution is same as mass on spring ⇒ oscillations
q = qmax cos(ωt +θ) k m
ω =
qmax is the maximum charge on capacitor
θ is an unknown phase (depends on initial conditions)
Calculate current: i = dq/dt
i = −ωqmax sin(ωt +θ) = −imax sin(ωt +θ)
Thus both charge and current oscillate
Angular frequency ω, frequency f = ω/2π
Period: T = 2π/ω 4

4. Plot Charge and Current vs t
q(t)
i(t)
PHY2049: Chapter 31 5

5. (i2)+ (q2) = 0 Total energy in circuit is conserved. Let’s see why
Energy Oscillations
Total energy in circuit is conserved. Let’s see why
di q
dt C
= 0 L
Equation of LC circuit
di q dq
dt C dt
L i = 0
Multiply by i = dq/dt
(i2)+
1 d
2C dt
(q2) = 0 2
= 2x
dt dt Ld
2 dt
dx dx
Use
⎛ q2 ⎞ q2 d 2
⎜ ⎜ ⎟ ⎟ = 0 Li
1 Li = const 1 C
UL + UC = const
dt ⎝ ⎠ C
PHY2049: Chapter 31 6

6. ωt +θ Energies can be written as (using ω2 = 1/LC)
Oscillation of Energies
Energies can be written as (using ω2 = 1/LC) q2 2C 2 2C
qmax
cos2(ωt +θ)
UC = = 2 2C
UL = 1 Li2 = 1 Lω2qmax sin2( ) =
qmax 2
sin2(ωt +θ)
ωt +θ 2 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
PHY2049: Chapter 31 7

7. Plot Energies vs t
UL(t)
PHY2049: Chapter 31
UC(t)
Sum 8

8. ( Parameters Calculate ω, f and T ω =1/ LC =1/ Calculate qmax and imax
LC Circuit Example
Parameters
C = 20μF
L = 200 mH
Capacitor initially charged to 40V, no current initially
Calculate ω, f and T
ω = 500 rad/s (
2×10−5)(0.2) = 500
ω =1/ LC =1/
f = ω/2π = 79.6 Hz
T = 1/f = sec
Calculate qmax and imax
qmax = CV = 800 μC = 8 × 10-4 C
imax = ωqmax = 500 × 8 × 10-4 = 0.4 A
Calculate maximum energies
UC = q2max/2C = 0.016J UL = Li2max/2 = 0.016J
PHY2049: Chapter 31 9

9. Note how voltages sum to zero, as they must!
LC Circuit Example (2)
Charge and current
q = cos(500t) i = −0.4sin(500t)
Energies
UC = 0.016cos2(500t)
UL = 0.016sin2(500t)
Voltages
VC = q/C = 40cos(500t)
VL = Ldi/dt = −Lωimax cos(500t) = −40cos(500t)
Note how voltages sum to zero, as they must!
PHY2049: Chapter 31 10

10. Work out equation using loop rule
RLC Circuit
Work out equation using loop rule
di q
dt C
+ Ri = 0 L
Rewrite using i = dq/dt 2
d q
R dq q
L dt LC
= 0 dt
Solution slightly more complicated than LC case
−tR/2L
cos(ω′t +θ) ω′ = 1/LC −(R/2L) 2
q = qmaxe
This is a damped oscillator (similar to mechanical case)
Amplitude of oscillations falls exponentially
PHY2049: Chapter 31 11

11. e i(t) Charge and Current vs t in RLC Circuit q(t) −tR/2L
PHY2049: Chapter 31 12

12. (0.012)(1.6×10−6) = 7220 Circuit parameters Calculate ω, ω’, f and T
RLC Circuit Example
Circuit parameters
L = 12mL, C = 1.6μF, R = 1.5Ω
Calculate ω, ω’, f and T
ω = 7220 rad/s
(0.012)(1.6×10−6) = 7220
ω =1/
ω’ = 7220 rad/s
f = ω/2π = 1150 Hz
ω′= −(1.5/0.024) 2 ω
T = 1/f = sec
Time for qmax to fall to ½ its initial value e−tR/2L =1/2
t = (2L/R) * ln2 = s = 11.1 ms
# periods = / ≈ 13
PHY2049: Chapter 31 13

13. RLC Circuit (Energy) (UL C) = −i R di q dt C + Ri + = 0
Basic RLC equation L
L i + Ri = 0
di q dq
dt C dt
Multiply by i = dq/dt
d q2 ⎞ ⎛
Collect terms
(similar to LC circuit)
Li2 + 2
⎜ ⎜ 2 ⎟
= −i R 2
dt ⎝ ⎠ ⎟ C
Total energy in circuit
decreases at rate of i2R
(dissipation of energy) d dt
Utot ∼ e−tR/L
(UL C) = −i R +U
PHY2049: Chapter 31 14

14. Energy in RLC Circuit
UC(t)
UL(t)
Sum
−tR/L e
PHY2049: Chapter 31 15

15. time to fully discharge the capacitors during the oscillations?
Quiz
Below 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 C C C A B
C has smallest capacitance, therefore highest
frequency, therefore shortest period
ω =1/ LC
PHY2049: Chapter 31 16

16. Enormous impact of AC circuits
Power delivery
Radio transmitters and receivers
Tuners
Filters
Transformers
Basic components R L C
Driving emf
Now we will study the basic principles
PHY2049: Chapter 31 17

17. RLC + “driving” EMF with angular frequency ωd
AC Circuits and Forced Oscillations
RLC + “driving” EMF with angular frequency ωd
ε =εmsinωdt
di q
dt C
+ Ri =εm dt
sinω L
General solution for current is sum of two terms
“Transient”: Falls
exponentially & disappears
Ignore
i ∼ e−tR/2L cosω′t
PHY2049: Chapter 31
“Steady state”:
Constant amplitude 18

18. ωd Assume steady state solution of form i = Imsin(ωdt −φ)
Im is current amplitude
φ is phase by which current “lags” the driving EMF
Must determine Im and φ
Plug in solution: differentiate & integrate sin(ωt-φ)
i = Imsin(ωdt −φ)
Substitute di dt
=ωdIm cos(ωdt −φ)
di q
dt C
+ Ri =εmsinωt L Im ωd
cos(ωdt −φ)
q =−
ImωdLcos(ωdτ −φ)+ ImRsin(ωdt −φ)− Im
ωdC
cos(ωdt −φ) =εmsinωdt
PHY2049: Chapter 31 19

19. Steady State Solution for AC Current (3)
(ωdL−1/ωdC)cosφ − Rsinφ = Same equations
Im(ωdL−1/ωdC)sinφ + ImRcosφ =εm
Solve for φ and Im in terms of
tanφ = ≡
R R
ωdL−1/ωdC XL − XC εm Z
Im =
R, XL, XC and Z have dimensions of resistance
XL =ωdL
XC =1/ωdC
Inductive “reactance”
Capacitive “reactance”
Z = R +(XL C) 2 2
− X
Total “impedance”
Let’s try to understand this solution using “phasors”
PHY2049: Chapter 31 21

20. ωdC XC = ("X R"= R ) Think of it as a frequency-dependent resistance
What is Reactance?
Think of it as a frequency-dependent resistance
ωd → 0, XC → ∞
- Capacitor looks like a break
ωd → ∞, XC → 0
- Capacitor looks like a wire (“short”) 1
ωdC
XC =
ωd → 0, XL → 0
- Inductor looks like a wire (“short”)
ωd → ∞, XL → ∞
- Inductor looks like a break
Independent of ωd
X L =ωdL
("X R"= R )
PHY2049: Chapter 31 26

21. AC Source and RLC Circuit (2)
Right triangle with sides VR, VL-VC and εm
VR mR
VR VL = ImXL
VC = ImXC
VL −VC = I εm φ
tanφ =
εm =VR +(VL −VC)
VL−VC VR
Solve for current: i = Imsin(ωdt −φ)
(Magnitude = Im, lags emf by phase φ)
Im =εm /Z
XL − XC ωdL−1/ωdC
R R
Z = R2 +( L C) = R2 +( d dC) 2
ω L−1/ω 2
X − X
PHY2049: Chapter 31 29

22. AC Source and RLC Circuit (3)
XL − XC R
tanφ =
Z = R2 +(XL − XC) 2
Only XL − XC is relevant, reactances cancel each other
When XL = XC, then φ = 0
Current in phase with emf, “Resonant circuit”:ωd
Z = R (minimum impedance, maximum current)
When XL < XC, then φ < 0
Current leads emf, “Capacitive circuit”: ωd < ω0
When XL > XC, then φ > 0
Current lags emf, “Inductive circuit”: ωd > ω0
PHY2049: Chapter 31
=ω0 = 1/LC 30

23. ε t an RLC circuit. We can conclude the following I RLC Example 1
Below 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 ε I t
PHY2049: Chapter 31 31

24. εm εm εm RLC circuit dominated by inductance? (3) Circuit 3
Quiz
Which 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 εm εm 3 1 2
PHY2049: Chapter 31 32

25. εm εm εm RLC circuit dominated by capacitance? (1) Circuit 1
Quiz
Which 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 εm εm 3 1 2
PHY2049: Chapter 31 33

26. Im Circuit parameters: C = 2.5μF, L = 4mH, εm = 10v Resonance
RLC Example 3
Circuit 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
ωd =ω0
PHY2049: Chapter 31 35