Chapter 33 Alternating Current
Circuits
Most of the electrical energy is produced by electrical generators in the form of
sinusoidal alternating current.
Why do we use the sinusoidal electric potential but neither the triangular nor the
square waves?
33.1 AC Sources
Φ m = BA , ε = −
dΦ m
dB
dA
d cosθ
=−
A cosθ + B cosθ + BA
dt
dt
dt
dt
What’s the best and the cheapest way to make the flux change?
Φ m = NBA cosθ
if the angular velocity is ω
--> Φ m = NBA cos(ωt + δ )
ε =−
dΦ m
= ωNBA sin (ωt + δ ) = ε peak sin (ωt + δ )
dt
33.2 Resistors in an AC Circuit
VR = ε = ε peak sin (ωt + δ )
Let δ =
π
2
--> sin (ωt + δ ) = cos(ωt )
VR = VR , peak cos(ωt )
Applying Ohm’s law (What is the Ohm’s law? Why do we use this? --> Linear
relation between I and V)
1
VR = VR , peak cos(ωt ) = IR --> I =
VR , peak
R
cos(ωt ) = I peak cos(ωt )
Root Mean Square Values
(I
I rms =
(I
2
instantane ous
2
instantane ous
) = (I
av
)
av
2
peak
, I instantane ous = I peak cos(ωt )
)
cos 2 (ωt ) av =
1 2
I peak
2
1
I peak ≈ 0.707 I peak
2
I rms =
Example: Find the average current and the rms current
for the sawtooth waveform. In the region 0<t<T, the
current is given by I = I 0
t
.
T
T
1
t
I
I av = ∫ I 0 dt = 0
T0 T
2
(I )
2
av
T
=
1 2 t2
I 02
I
I
dt
=
--> I rms = 0
0
2
∫
3
T0 T
3
33.3 Inductors in an AC Circuit
VL = − L
dI
dt
Apply Kirchhoff’s loop rule:
ε −L
dI
dI
= 0 --> ε peak cos(ωt ) − L = 0
dt
dt
2
ε peak cos(ωt ) = L
ε cos(ωt )
dI
--> peak
dt = dI
dt
L
ε peak
π

sin (ωt ) = I peak sin (ωt ) = I peak cos ωt − 
2
ωL

ε
V
V
--> I peak = peak = L , peak = L , peak --> VL , peak = I peak X L compare with V = IR
ωL
ωL
XL
X L = ωL --> resistance like --> impedance --> inductive reactance
The resistance R is not changed with the frequency ω but the impedance may
--> I =
change with the frequrncy.
I
V
V
I rms = peak , Vrms = peak --> I rms = rms
XL
2
2
Instantaneous Power Delivered:
1
P = VI = (V peak cos(ωt ))(I peak sin (ωt )) = V peak I peak sin (2ωt )
2
T
1 1
The average of dissipated energy is:
V peak I peak sin (2ωt )dt = 0 . --> No energy is
T ∫0 2
dissipated in an inductor.
Example: Inductive Reactance
The potential drop across a 40-mH inductor is sinusoidal with a peak potential drop of
120 V. Find the inductive reactance and the peak current when the frequency is 60 Hz.
(
)
X = ωL = (2π )(60 ) 40 × 10 −3 = 15.1Ω , I peak =
V peak
X
33.4 Capacitos in an AC Circuit
V=
Q
Q
, ε − =0
C
C
3
--> ε peak cos(ωt ) =
--> I =
Q
C
dQ d
= (ε peak cos(ωt )C ) = −ωε peak C sin (ωt ) = − I peak sin (ωt )
dt dt
π

I = − I peak sin (ωt ) = I peak cos ωt + 
2

ε peak ωC = I peak --> ε peak =
1
1
I peak = I peak X --> X =
ωC
ωC
Example: Capacitive Reactance
A 20-µF capacitor is placed across an ac generator that applies a potential drop with
an amplitude (peak value) of 100 V. Find the capacitive reactance and the current
amplitude when the frequency is 60 Hz.
X=
V
1
1
=
= 133Ω , I peak = peak
−6
ωC 2π (60) 20 × 10
X
(
)
33.5 The RLC Series Circuit
Phasors
The current in a steady-state ac circuit varies with time as I = I peak cos(ωt )
--> The voltage drop across a resistance is VR = IR = I peak R cos(ωt ) .
--> The voltage drop across an inductor is VL = L
dI
= − I peakωL sin (ωt ) .
dt
π
π


VL = − I peakωL cos − ωt  = I peak ωL cos ωt + 
2
2


4
--> The voltage drop across a capacitor is VC =
VC =
1
1
Idt =
I sin (ωt )
∫
ωC peak
C
π
π
 I

cos − ωt  = peak cos ωt − 
ωC
2
2
 ωC

I peak
Series RLC Circuit
Vapp = V peak cos(ωt )
Diff. Eq.: V peak cos(ωt ) − L
V peak cos(ωt ) − L
m
dI
Q
− IR − = 0 -->
dt
C
d 2Q
dQ 1
−R
− Q = 0 (forced oscillation
2
dt
dt C
d 2x
dx
+ b + kx = F0 cos(ωt ) )
2
dt
dt
Forced Oscillations --> Force The Oscillator Vibrate with The Frequency ω
Let Q = A sin (ωt − δ )
--> V peak cos(ωt ) + Lω 2 A sin (ωt − δ ) − RAω cos(ωt − δ ) −
1
A sin (ωt − δ ) = 0
C
V peak cos(ωt ) + Lω 2 A(sin (ωt )cos δ − cos(ωt )sin δ )
− RAω (cos(ωt )cos δ + sin (ωt )sin δ ) −
1
A(sin (ωt )cos δ − cos(ωt )sin δ ) = 0
C
cos(ωt ) : V peak − Lω 2 A sin δ − RAω cos δ +
sin (ωt ) : Lω 2 cos δ − Rω sin δ −
V peak R 2 + ( X L − X C )
1
cos δ = 0 --> tan δ =
C
2
--> A =
(
ω R + (X L − X C )
2
2
)
1
A sin δ = 0
C
=
1
ωC = X L − X C
R
R
Lω −
V peak
ω R2 + (X L − X C )
2
5
Q=
V peak
sin (ωt − δ ) , I =
ω R2 + (X L − X C )
2
I peak =
V peak
R2 + (X L − X C )
2
=
V peak
R2 + (X L − X C )
2
cos(ωt − δ )
V peak
Z
Total reactance: X L − X C
Impedance: Z = R 2 + ( X L − X C )
2
Phasor:
Vapp = V peak cos(ωt ) & I = I peak cos(ωt − δ )
--> VR = I peak cos(ωt − δ )R
dI
π
π


= −VL , peak sin (ωt − δ ) = −VL , peak cos − (ωt − δ ) = VL , peak cos (ωt − δ ) + 
dt
2
2


r
r r
= VR + VL + VC
VL = L
r
Vapp
r
r r
2
Vapp , peak = VR + VL + VC = VR2, peak + (VL , peak − VC , peak )
VR , peak = I peak R , VL , peak = I peak X L , VC , peak = I peak X C
--> Vapp = I peak R 2 + ( X L − X C ) = I peak Z (Z: impedance)
2
r r
VL + VC
X − XC
tan δ =
= L
r
R
VR
Example: A resistor R and capacitor C are in series with a generator. The generator
applies a potential drop across the RC combination given by Vapp = 2Vapp , rms cos(ωt ) .
Find the rms potential drop across the capacitor as a functional of frequency ω .
I = I 0 cos(ωt + δ )
VR = I 0R cos(ωt + δ )
6
VC =
1
I cos(ωt + δ − π / 2 )
ωC 0
Vapp = I 0 R 2 + I 02
2
2
1
(ωC )2
Vout = I 0 / ωC =
Vapp
I0 =
2Vapp , rms
R 2 + 1 / (ωC )
2
= 2Vout , rms
1 + (ωC ) R 2
2
=
2Vapp , rms
R 2 + 1 / (ωC )
Vout , rms =
2
Vapp , rms
1 + ω 2C 2 R 2
Parallel RLC Circuit
ε 
ε   ε
I =   +
−

 R   ωL 1 / ωC 
2
2
33.6 Power in an AC Circuit
The instantaneous power dissipation:
2
P = I 2 R = I peak
cos 2 (ωt )R
The average power dissipation:
( )
(
2
Pav = I 2 R av = I peak
cos 2 (ωt )R
)
av
(
)
2
= I peak
R cos 2 (ωt ) av
What’s the time average? What’s the spatial average?
λ
T
(P )time _ av = 1 ∫ Pdt
T
(P )spatial _ av = 1 ∫ Pdx
λ
0
 2π
1 + cos 2
1
1
 2π 
 T
cos 2 
t dt = ∫
∫
T0
T0
2
T 
T
T
(
)
2
Pav = I peak
R cos 2 (ωt ) av =
--> Pav =
0

t
 dt = 1 T = 1
T 2 2
1 2
I peak R
2
1 2
2
I peak R = I rms
R
2
7
The average power delivered by the generator is:
(
)
1
Pav = (εI )av = (ε peak cos(ωt )I peak cos(ωt ))av = ε peak I peak cos 2 (ωt ) av = ε peak I peak
2
I rms =
I peak
2
, ε rms =
ε peak
2
--> Pav = ε rms I rms
For the RLC circuit:
(
P = I peakVpeak cos(ωt ) cos(ωt − φ ) = I peakV peak cos 2 (ωt ) cos φ + cos(ωt ) sin (ωt ) sin φ
Pavg = I rmsVrms cos φ = I rmsVrms
)
R
2
= I rms
R
Z
33.7 Resonance in a Series RLC Circuit
Z = R + (X L − X C )
2
2
2
2
--> Z 2 = R 2 +
1 

= R +  ωL −

ωC 

--> Z = R + ( X L − X C )
2
2
(
L2  2
1 
L2 2
2
ω
−
=
R
+
ω − ω02

2 
2
ω 
ω
LC 
2
2
)
2
2
The power supplied to the resistor R is Pav = I rms
R.
2
Pav = I
2
rms
2
Vapp
 Vapp , rms 
, rms
 R =
R = 
R
2
L
2
2
2 2
 Z 
R + 2 ω − ω0
ω
(
)
The dissipated power has a maximum value at the resonance frequency
1
ω = ω0 =
.
LC
Full width at half maximum: ∆ω = R / L
Quality Factor: Q =
ω0 ω 0 L
=
∆ω
R
33.8 The Transformer and Power
8
Transmission
Assume that the magnetic flux through a single turn of coil is φm .
Loop 1: V1 − N1
dφm
dφm V1
= 0 -->
=
dt
dt
N1
Loop 2: (The coils behave as a voltage source.)
dφ
N
V2 = N 2 m = 2 V1
dt
N1
Energy conserved: the power input from the generator is equal to the power output,
P1 = P2 --> V1I1 = V2 I 2 .
If the input voltage is 110 V and the output voltage is 12 V, which copper wire of the
coils is thick? (the input coils or the output coils)
33.9 Rectifiers and Filters
9
Low Pass Filter – High Pass Filter
10