OUTLINE
1. RLC circuit and damped oscillation
(continued)
2. AC circuits and forced oscillations
Phasor method
Reactance
Impedance
Charge and Current vs t in RLC Circuit
q (t )
i (t )
e
−tR / 2 L
RLC Circuit (Energy)
di
q
L + Ri + = 0
dt
C
Loop-rule equation for RLC
circuit
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 B + U E ) = −i 2 R
dt
Total energy in circuit
decreases at rate of i2R
(dissipation of energy)
Energy in RLC Circuit
U B (t )
U E (t )
Sum
e
−tR / L
AC Circuits and Forced Oscillations
Classification of emf devices
• Constant emf: “direct-current” sources (dc sources)
Battery, solar cell, fuel cell, · · ·
• Sinusoidal emf: alternating-current sources (ac sources)
Generator
ac
Ciruit 1 (RC circuit)
Take a snapshot.
• Current i is the same everywhere,
because all the circuit elements
are in series.
• It appears as if i flows through the capacitor.
• Loop rule:
+q
-q
i
i
ℰ = ℰmsin(ωd t)
• Loop law: ℰ – i R – q /C = 0.
ac
(1)
+q
-q
• Method 1: solve for q, after turning
Eq. 1 into a differential equation by
using i = dq/dt.
Method 2: solve for i, which is more
useful than q.
Our choice.
i
i
• i must be: i = I sin(ωd t + φ). Finding amplitude I and phase φ is our
goal.
Since i = dq/dt,
q = –I /ωd cos(ωd t + φ) + const,
but const must be 0, since the sinusoidal emf can produce only
sinusoidally oscillating q. Substitute into Eq. 1:
(2)
ℰm sin(ωd t) – I R sin(ωd t + φ) – [– I / ωdC cos(ωd t + φ) ] = 0.
• Method 1´: use trigonometry and brute-force algebra to solve this.
Method 2´: phasor method.
Our choice.
Phasor Method
• What is a phasor? For a physical
quantity a that oscillates with amplitude
A at (angular) frequency ω, the phasor
is a vector of length A that rotates
around the origin at (angular) frequency
ω. The y component of the phasor is
the physical quantity a, which oscillates
as the phasor rotates counterclockwise
at a constant rate.
a = A sin(ωt)
A
ωt
Let’s solve Eq. 2 by using phasors.
• Step 1: all cos must be converted to sin. Note
sin(θ – π/2) = sin θ cos(π/2) – cos θ sin(π/2) = – cos θ .
Using this, write Eq. 2 as
ℰm sin(ωd t) – I R sin(ωd t + φ) – I / ωdC sin(ωd t + φ – π/2) = 0.
vR = VR sin(ωd t + φ)
vC = VC sin(ωd t + φ – π/2)
Here, the amplitudes VR = I R
VC = I / ωdC.
• Step 2: draw the phasors for the emf
and the current.
I
ℰ
ωt
ωt + φ
(3)
• Step 3: draw the phasors for the
potential differences vR and vC. Note:
phasor VR is in phase with phasor I
for the current; so the angle between
phasors ℰ and VR is φ. Phasor VC lags
behind by π/2.
I
ℰ
VR
• Step 4: note that the vector sum of
the two phasors as vectors, VR + VC ,
represent vR + vC, because the y
component of VR + VC is vR + vC . Eq. 3,
which states that ℰ = vR + vC at any
moment, is equivalent to the phasor
equation
ℰm = VR + VC .
Since VR and VC are orthogonal:
ℰm = (VR2 + VC2)1/2 = [(I R)2 + (I/ ωd C)2] 1/2
I = ℰm / [R2 + (1/ ωd C)2 ] 1/2
tan φ = VC /VR = (I /ωd C) / I R = 1 / ωd R C
VC
ωt + φ