Short Version :
28. Alternating Current Circuits
28.1. Alternating Current
Reminder: All waves can be analyzed in terms of sinusoidal waves (Fourier analysis Chap 14).
Sinusoidal wave (Chap 13) :
1
2
Vp sin 
2
1
2
sin
x
dx

0
2
Vrms 
2
 cos x dx 
0
Vp
I rms 
2
Angular frequency :
=/6
2
  2 f
V  Vp sin  t  V 
 = phase
1
2
Ip
2
[] = rad/s
I  I p sin  t  I 
Example 28.1. Characterizing Household Voltage
Standard household wiring supplies 110 V rms at 60 Hz.
Express this mathematically, assuming the voltage is rising through 0 at t = 0.
 156 V
V p  2 Vrms
  2 f
 2  60 Hz   377 s 1
V  Vp sin  t   
V  t  0  0

 0

0  V p sin 
V  V p sin  t  156 sin  377 t  V
28.2. Current Elements in AC Circuits

Resistors

Capacitors

Inductors

Phasor Diagrams

Capacitors & Inductors: A Comparison
Displacing Functions
g
f
g is moved to the right (forward) by  to give f.
f  x   g  x  
x
x
sin
cos
Displacement:
Phase:
sin is cos moved forward by /2.
sin lags cos by /2.


sin  x   cos  x  
2



sin x  cos x  sin  x  
2



sin
x
dx


cos
x

sin
x




2

Derivative: moves sinusoidal functions backward by /2.
phase is increased by /2.
Integral: moves sinusoidal functions forward by /2.
phase is decreased by /2.
Resistors
V  t   VR  0
V  t  Vp

sin  t
I
R
R
I
+

V p sin  t  I R  0
+
VR

I  I p sin  t
I rms 
Vrms
R
Ip 
Vp
R
I & V in phase
Capacitors
When V(t) > 0 :
V  t   VC  0
I
+

+
VC

V p sin  t 
q  C V t 
I
 C V p sin  t
d q C dV
dt
dt
 C V p  cos  t


I  C V p  sin   t  
2

I peaks ¼ cycle before V
I p  C Vp  
XC 
q
0
C
1
C
I leads V by 90
Vp
XC
Capacitive reactance
 XC   
X C   as   0
DC: open ckt.
X C  0 as   
HF: short ckt.
Inductors
When V(t) > 0 :
V  t   EL  0
I
+


L
+
V p sin  t  L
1
I
L
 V t  d t 
I
I peaks ¼ cycle after V
d I Vp
 sin  t
dt
L
dI
0
dt
Vp
L
 sin  t


sin   t  
L 
2
Vp
Ip 
Vp
L
XL   L

d t 
Vp
L
cos  t
I trails V by 90
Vp
XL
Inductive reactance
XL  
X L  0 as   0
DC: short ckt.
X L   as   
HF: open ckt.
Table 28.1. Amplitude & Phase in Circuit Elements
Circuit Element
Peak Current vs Voltage
Ip 
Resistor
Capacitor
Inductor
Ip 
Vp
Ip 
Vp
XC
XL
Vp
V & I in phase
R


Phase Relation
Vp
1/  C
Vp
L
V lags I 90
V leads I 90
Example 28.2.
Equal Currents?
A capacitor is connected across a 60-Hz, 120-V rms power line,
and an rms current of 200 mA flows.
(a)
Find the capacitance.
(b)
What inductance, connected across the same powerline,
would result in the same current?
(c)
How would the phases of the inductor & capacitor currents compare?
I
 rms
C
(a)
Vrms
(b)
Vrms   L I rms
(c)
C
L
I rms
Vrms 
200  103 A

120 V   2 60 Hz 
 4.42  106 F  4.42  F
120 V
Vrms

I rms 
 200  103 A  2 60 Hz 
Capacitor: IC leads V by 90.
Inductor: V leads IL by 90.
 1.59 H
 IC leads IL by 180.
Phasor Diagrams
Phasor = Arrow (vector) in complex plane. Length = mag. Angle = phase.
V  I X
V leads I by 0.
( same phase )
V  V e i  V e it
V leads I by 90.
V leads I by 90.
( V lags I by 90 )
Capacitors Revisited
I
Vp e i t
q CV
+
VC

I
 C Vp ei  t
d q  C dV
dt
dt
I  i  CV  CV e
i
 C Vp i  ei  t

2
I leads V by 90
Taking the real part as physical
V  V p cos  t

I  CV p  sin  t  CV p  cos   t  
2

Taking the imaginary part as physical
V  V p sin  t

I  CV p  cos  t  CV p  sin   t  
2

V I Z
Z 
1
i
C
Impedance
Inductors Revisited
I
Vp e
V  t   EL  0

L
i t
I
Vp
+
I
L
e
V
i L
L
it
dt

d I
 Vp e i  t
dt
Vp
i L
V i 2

e
L
e it
I lags V by 90
Taking the real part as physical
V  V p cos  t
V
I  p sin  t
L



cos   t  
L
2

Vp
Taking the imaginary part as physical
V  V p sin  t
I 
Vp
L
cos  t



sin   t  
L
2

Vp
ZL   L i
Capacitors & Inductors: A Comparison
C  L translator:
EB
q  B
VI
ZY
Table 28.2. Capacitors & Inductors
Defining relation
Defining relation; differential form
Opposes change in
Energy storage
Capacitor
q
C
V
I C
dV
dt
V
UE 
1
CV2
2
Inductor
L
B
V L
I
dI
dt
I
UB 
1
L I2
2
Behavior in low freq limit
Open circuit
Short circuit
Behavior in high freq limit
Short circuit
Open circuit
Reactance
Admittance / Impedance
Phase
1 / XC   C
XL   L
YC  i  C  1 / ZC
ZL  i  L
I leads by 90
V leads by 90
Application: Loudspeaker Systems
Loudspeaker
C passes High freq
 1

Q


R
V   IC R
iC
 IC
C


V L
d IL
 ILR
dt
 i  L  R  I L
Loudspeaker system with
high & low frequency filters.
L passes low freq
I

V
+
28.3. LC Circuits
Analyzing the LC Circuit
I

V
+
U  UB  UE 
dU
0
dt
q
V
C
LI
1
1
L I2  CV2
2
2
dI
dV
CV
dt
dt
dq
I
dt
dV 1 d q

dt C dt
d q d2 q
q 1dq
0L

C
d t d t2
C C dt
d2 q
q
L

0
d t2
C
q  q p cos  t

1
LC

q
dI
L
0
C
dt
d I d2 q

d t d t2
Resistance in LC Circuits – Damping
I

VC
+
+ VR 
U

L
1
1
L I L2  C VC2
2
2
dI
d VC
dU
 C VC
 I 2R  L I
dt
dt
dt
+
q
VC 
C

LI
dVC I

dt
C
dI
I
 q  I 2R  0
dt
C
I
L
dq
dt
dI q
  IR  0
dt C
d2q
dq q
L 2 R  0
dt
dt C
q  t   q p e R t / 2 L cos  t
(see next page)
Resistance in LC Circuits – Damping
I

VC
+
+ VR 


L
q
dI
I RL
0
C
dt
I
dq
dt
+

d2q
dq q
L 2 R  0
dt
dt C
q  t   q p e R t / 2 L cos  t
(see next page)
Solutions to Damped Oscillator
d2q
dq q
L 2 R  0
dt
dt C
Ansatz:
1

2
at
L
a

R
a

q
e
0
0


C

q  q0 e a t
a
L a2  R a 
1
0
C
1 
 R 
2L 
R2  4
q  e  R t / 2 L  c ei  t  c e  i  t 
L
R


i

C
2L
1

2L
L
4  R2 
C
 e R t / 2 L  A cos t  B sin t 
 q0 e  R t / 2 L cos  t

4L  R 2C
R
   
 2L 
2
0
2
1  R 


LC  2 L 
0 
2
1
LC
28.4. Driven RLC Circuits & Resonance
I
+ VR 
V I RL
dI q
 0
dt C

+
L

+
 VC +
V  I R  i d L I 
I
i d C
0


1 
V  I  R  i  d L 


C
d



Driven damped oscillator :
Long time: oscillates with frequency d.
Resonance if d = 0.
 02 
Z  R  i d L 1  2 
 d 
Resonance in the RLC Circuit

 02  
V  I  R  i d L  1  2  
 d  

VL  i d L I
02
VC  VL 2
d
VL  VC  V  I R
VC & VL are 180 out of phase.
IC p 
VC p
1 / d C
 IL p 
VL p
d L

VL p  VL p
if
1
d C
 d L i.e., d2  02
Frequency Response of the RLC Circuit
Series circuit  same I phasor for all.
VR in phase with I.
VC lags I by 90.
VL leads I by 90.


1
Q
dI
L
V  I R 
 i  L
C
dt
iC



1 
Z  R  i  L 


C


I R
High Q
Low Q
Vp  I p Z
Z 

1 
R2    L 
 C 

2
See Prob 71 for definition of Q.
tan  
VL p  VC p
VR p
X  XC
 L

R
At resonance,  = 0.
 L
R
1
C
L
02 

1 2 

R   
Example 28.4. Designing a Loud Speaker System
Current flows to the midrange speaker in a loudspeaker system through a
2.2-mH inductor in series with a capacitor.
(a) What should the capacitance be so that a given voltage produces the
greatest current at 1 kHz ?
(b)
If the same voltage produces half this current at 618 Hz,
what is the speaker’s resistance ?
(c)
If the peak output voltage of the amplifier is 20 V,
what will be the peak capacitor voltage be at 1 kHz ?
(a) Greatest I is at resonance:
C
1
L
2
0

 2.2  10
02 
1
LC
1
3
H  2  10 Hz 
3
2
 11.5  106 F
 11.5  F
(b) If the same voltage produces half this current at 618 Hz, what is the speaker’s resistance ?
Vp  I p Z
Ip
2
Z 

1 
R2    L 
 C 

2
At resonance:
Z R
2
Z  Ip R

1 
R2    L 
 2R

C


1 
1
1 
1 
3
  2  618 Hz   2.2  10 H  

R
 L   C  
6

3
 2  618 Hz  11.5  10 F  
3

 8.0 
(c) If the peak output voltage of the amplifier is 20 V,
what will be the peak capacitor voltage be at 1 kHz ?
Peak voltage is at resonance (1 kHz).
VC p  I p X C 
Vp
Vp  I p R
20 V
1
1

8.0   2  103 Hz  11.5  106 F 
R C
 35V
Capacitor:
I leads V by 90 ,  P  = 0
28.5. Power in AC Circuits
  I p sin  t    
PIV
P 
Resistor:
I & V in phase ,  P  > 0
I
p
sin  t    
p
p
sin  t 
sin  t 
 I p V p sin  t    sin  t
 I p Vp
 I p Vp
I & V out of phase ,  P  
V
V
P 
 sin  t cos   cos  t sin   sin  t
 sin  t
2
cos   cos  t sin  t sin 

1
I p V p cos   I rms Vrms cos 
2
Power factor
Dissipative power = I2 R
 large power factor reduces I & hence heat loss.
Conceptual Example 28.1. Managing Power Factor
You’re chief engineer of a power company.
Should you strive for a high or a low power factor on your lines?
P  I rms Vrms cos 
Power factor
Generator : fixed Vrms .
To maintain fixed <P>, Irms cos  = const.
Smaller power factor
Ans.:

higher Irms .

higher power loss.
keep power factor close to 1.
Making the Connection
Transmission losses on a well-managed electric grid average about
8% of the total power delivered.
How does this figure change if the power factor drops from 1 to 0.71?
P  I rms Vrms cos 
To deliver the same power
Transmission losses:
I new 
2
PL  I rms
R
I
0.71

 1.4 I
PL , new  1.4  PL
2
 2 PL
( doubles to 16% )
28.6. Transformers & Power Supplies
Transformer: pair of coils wound
on the same (iron) core.
Vsec ondary
V primary

N sec ondary
N primary
Works only for AC.
Direct-Current Power Supplies
Diode passes + half of each cycle
Diode
Diode cuts off  half of each cycle
RC (low freq) filter