Chapter 28 – AC Circuits
LC Circuit
•  Suppose switch closed at t = 0 with capacitor initially
charged to Q. What is q(t)?
�
�
dI
� = −L
� · dl
∆Vi = −Vc
E
=−
dt
-q
i
q
dq
VC =
+q
I=−
C
dt
I
d2 q
q
like harmonic
L 2 +
= 0 Looks
oscillator!
dt
C
d2 x
m 2 + kx = 0 (mass on spring)
dt
q(t) = Q cos(ωt) ω =
�
1
LC
(resonant frequency)
•  Initially, energy is stored by capacitor in the form of an electric field.
1
1 Q2
2
U = CV =
2
2 C
•  When capacitor is drained (q=0), what form is the
�energy?
1
•  Notice that when q=0, |I| = Imax = Qω = Q
LC
2
1Q
1 2
U=
= LImax
2 C
2
All energy in inductor!
•  What form is energy in inductor?
1 2
LI = uB × volume
2
2
B
•  Magnetic energy density:
uB =
2µ0
•  Magnetic! for solenoid:
•  Always true (general result is based on a much more formal
derivation)
•  Energy in an LC circuit oscillates between electric and
magnetic
•  similar to mechanical harmonic oscillator (kinetic and potential)
Damped LC oscillations
•  With resistance in the circuit, LC oscillations damp out.
I
-q
I
�
+q
dI
�
E · d� = VC + IR = −L
dt
dq
I=
dt
d2 q
dq
1
L 2 +R + q =0
dt
C
•  underdamped case: dt
q(t) = q0 exp(−R/(2L)t) cos
��
1
R2
−
t
LC
4L2
�
•  Similar to a damped mass on spring
Driven AC Circuits
•  Driven resistor:
Vemf (t) − IR = 0
Vemf
I=
R
•  Current in phase with driving voltage source.
Driven Capacitor
Vemf − VC = 0
I
q
Vc =
= Vp sin ωt
C
dq
I=
= CVp ω cos ωt = Vp (Cω) sin(ωt + π/2)
dt
Imax
Vp
=
XC
1
=⇒ XC =
ωC
Capacitive reactance
+
-
•  A light bulb is attached in series to a capacitor and an
AC voltage source with an adjustable frequency.
Keeping the peak voltage constant, the light bulb will be
brightest for
1.
2.
3.
4.
low frequencies
high frequencies
same for all frequencies
Bulb doesn’t shine for any frequency
ξ = Vp sin(ωt)
3
Driven RC Circuit
(t)
q
ξ(t) − IR −
=0
C
R
dq
q
R+
= Vp sin(ωt)
dt
C
C
dq
Vp
I=
=�
cos(ωt + δ)
dt
1 2
δ = tan−1 (−ωRC)
( ωC
) + R2
Driven Inductor
•  Faraday’s Law of induction
dI
−Vp sin(ωt) = −L
dt
Vp
I=
L
�
sin(ωt) dt
Vp
I(t) = −
cos(ωt)
ωL
Imax
Vp
=
XL
XL = ωL
Inductive reactance
�
dI
� = −L
� · dl
E
dt
Driven LR Circuit
dI
−V (t) + IR = −L
dt
if V (t) = V0 cos(ωt)
V0
I(t) = �
cos(ωt − φ)
R2 + (Lω)2
φ = tan
−1
wL
R
Jumping ring demo
•  why does ring always repel from solenoid (with AC
circuit)?
http://www.youtube.com/watch?v=Pl7KyVIJ1iE
Jumping Ring Analysis
I(t)
•  Suppose current in coils goes as I(t) = I0 sin(ωt)
•  The emf in loop will go as
Vemf ∝ −dI/dt ∝ − cos(ωt)
•  The induced current will therefore go as
Iring ∝ − cos(ωt − φ)
φ = tan
−1
wL
R
•  Repulsive force when Iring and I are in opposite directions
Jumping Ring Cont’d
•  If there were no phase difference between induced
current and emf, induced current would be in same
direction as I(t) half the time, and the ring would not
jump!
I(t)
Vemf
Jumping Ring Cont’d
•  Now add phase shift between induced emf and induced
curent, given by
−1 wL
φ = tan
R
•  For large inductance/and or high frequencies, phase shift
is near 90°. Here is a phase shift of 70°:
Vemf
Now currents
are in opposite
directions more
often than in
same direction!
I(t)
Iind
speaker crossover
Driven LRC Circuit
•  An RLC circuit driven by
an AC voltage source
exhibits frequencydependent behavior.
V (t) = Vp sin(ωd t)
-q +q
q
dI
−V (t) + IR +
= −L
C
dt
dq
I=
dt
d2 q
dq
q
L 2 +R +
= Vp sin(ωd t)
dt
dt
C
Ip = �
Vp
R2
+ (ωL −
Vp
1 2
ωC )
Vp
Ip = �
=
2
2
Z
R + (XL − XC )
I(t) = Ip sin(ωd t − φ)
XL − XC
tan φ =
R
Amplitude and phase in the driven RLC
circuit
•  Peak current has a
maximum at resonance:
•  resonant frequency
ω =1
LC .
(XL = XC =
•  Ip = Vp/Z
�
•  At resonance, voltage and
current are in phase
L
)
C
Transformers and power supplies
•  A transformer is a pair of
coils linked by mutual
inductance.
•  An AC current in the primary induces
a current in the secondary.
•  The secondary voltage differs from
the primary voltage by the ratio of
the number of turns.
•  Both step-up and step-down
transformers are possible.
dΦB
ξ1 = −N1
(ΦB is flux per winding)
dt
dΦB
ξ2 = −N2
dt
ξ2
N2
V2
=
=
ξ1
N1
V1
|V1 |max = |ξ1 |max
|V2 |max
N2
= |V1 |max
N1
Question 34.18 Transformers I
1) 30 V
What is the voltage
2) 60 V
3) 120 V
across the
4) 240 V
lightbulb?
5) 480 V
120 V
Question 34.20 Transformers III
A 6 V battery is connected to one
1) greater than 6 V
side of a transformer. The output
2) 6 V
voltage across coil B is:
3) less than 6 V
4) zero
A
6V
B
•  Transformers are used to produce low voltages for
electronic equipment.
•  Then they’re combined with diodes that convert AC to
DC and capacitors to smooth the DC voltage.
Input Power Vs Output Power
•  How does the input power compare to the output power?
P1 = P2
•  Ignoring self inductance,
V1
N1
I2 = I1
= I1
V2
N2
P1 = V 1 I 1
Clicker Question
If the resistance R is decreased what happens to I1?
a) Increases b) Decreases c) Remains the same
Transmission of Electrical Power
•  Electric power is most efficiently transmitted at high voltages.
•  This reduces I2R energy losses in the power lines.
•  But most end uses require lower voltages.
•  Transformers accomplish voltage changes throughout the power grid.
Slide 26-11
Two-Phase Power
to Your Home
center tap is
connected to ground
rms = root mean squared:
Slide 26-12