Announcements
• Assignment 0 due now.
– solutions posted later today
• Assignment 1 posted,
– due Thursday Sept 22nd
• Question from last lecture:
– Does VTH=INRTH
– Yes!
Lecture 5 Overview
• Alternating Current
• AC Components.
• AC circuit analysis
Alternating Current
pure DC
V
• pure direct current = DC
• Direction of charge flow
(current) always the same and
constant.
pulsating DC
V
• pulsating DC
• Direction of charge flow
always the same but variable
• AC = Alternating Current
pulsating DC
V
V
• Direction of Charge flow
alternates
-V
AC
Why use AC? The "War of
the Currents"
• Late 1880's: Westinghouse backed AC, developed by Tesla, Edison backed DC (despite
Tesla's advice). Edison killed an elephant (with AC) to prove his point.
• http://www.youtube.com/watch?v=RkBU3aYsf0Q
• Turning point when Westinghouse won the contract for the Chicago Worlds fair
• Westinghouse was right
• PL=I2RL: Lowest transmission loss uses High Voltages and Low Currents
• With DC, difficult to transform high voltage to more practical low voltage efficiently
• AC transformers are simple and extremely efficient - see later.
• Nowadays, distribute electricity at up to 765 kV
AC circuits: Sinusoidal waves
• Fundamental wave form
• Fourier Theorem: Can construct
any other wave form (e.g. square
wave) by adding sinusoids of
different frequencies
• x(t)=Acos(ωt+)
• f=1/T (cycles/s)
• ω=2πf (rad/s)
•  =2π(Δt/T) rad/s
•  =360(Δt/T) deg/s
RMS quantities in AC circuits
• What's the best way to describe the
strength of a varying AC signal?
• Average = 0; Peak=+/• Sometimes use peak-to-peak
• Usually use Root-mean-square (RMS)
• (DVM measures this)
I rms 
Ip
2
, Vrms 
Vp
2
, Pave  I rmsVrms
i-V relationships in AC circuits:
Resistors
Source vs(t)=Asinωt
vR(t)= vs(t)=Asinωt
iR (t ) 
vR (t ) A
 sin t
R
R
vR(t) and iR(t) are in phase
Complex Number Review
Phasor representation
2
2
i-V relationships in AC circuits:
Resistors
Source vs(t)=Asinωt
vR(t)= vs(t)=Asinωt
iR (t ) 
vR (t ) A
 sin t
R
R
vR(t) and iR(t) are in phase
Complex representation: vS(t)=Asinωt=Acos(ωt-90)=real part of [VS(jω)]
where VS(jω)= A[cos(ωt-90)-jsin(ωt-90 )]=Aej (ωt-90)
Phasor representation: VS(jω) =A(ωt-90)
IS(jω)=(A/R) (ωt-90)
Impedance=complex number of Resistance Z=VS(jω)/IS(jω)=R
Generalized Ohm's Law: VS(jω)=ZIS(jω)
http://arapaho.nsuok.edu/%7Ebradfiel/p1215/fendt/phe/accircuit.htm
Capacitors
What is a capacitor?
Definition of Capacitance: C=q/V
Capacitance measured in Farads
(usually pico - micro)
Energy stored in a Capacitor = ½CV2
(Energy is stored as an electric field)
In Parallel:
V=V1=V2=V3
q=q1+q2+q3
q q1  q2  q3
Ceq  
 C1  C2  C3
V
V
i.e. like resistors in series
Capacitors
In Series:
V=V1+V2+V3
q=q1=q2=q3
1 V V1  V2  V3 1
1
1
 
 

Ceq q
q
C1 C2 C3
i.e. like resistors in parallel
No current flows through a
capacitor
In AC circuits charge buildup/discharge mimics a current flow.
A Capacitor in a DC circuit acts like a
break (open circuit)
Capacitors in AC circuits
Capacitive Load
vC  A sin t
qC  CvC
dqC
 CA cos(t )
dt
VC ( j )  A(t  90)
iC 
I C ( j )  CA(t  0)
ZC 
VC ( j ) 1

  90

C
I C ( j )
cos( 90)  j sin( 90)   j
 j  j. j
1



C jC jC "capacitive reactance"
• Voltage and current not in phase:
• Current leads voltage by 90 degrees (Physical - current must conduct charge to capacitor
plates in order to raise the voltage)
• Impedance of Capacitor decreases with increasing frequency
http://arapaho.nsuok.edu/%7Ebradfiel/p1215/fendt/phe/accircuit.htm
Inductors
What is an inductor?
Definition of Inductance: vL(t)=-LdI/dt
Measured in Henrys (usually milli- micro-)
Energy stored in an inductor: WL= ½ LiL2(t)
(Energy is stored as a magnetic field)
• Current through coil produces
magnetic flux
• Changing current results in
changing magnetic flux
• Changing magnetic flux induces a
voltage (Faraday's Law v(t)=-dΦ/dt)
Inductors
Inductances in series add:
Inductances in parallel combine
like resistors in parallel (almost
never done because of magnetic
coupling)
An inductor in a DC circuit behaves like a short (a wire).
Inductors in AC circuits
Inductive Load
vS  A sin t
vL   L
diL
dt
A sin t  L
(back emf )
diL
dt
from KVL
A
A
iL   sin tdt  
cos t
L
L
A
A
iL 
sin( t  90) 
cos(t  180)
L
L
VL ( j )  A(t  90)
A
(t  180)
L
V ( j )
ZL  L
 L90
I L ( j )
cos(90)  j sin( 90)  j
Z L  jL
I L ( j ) 
• Voltage and current not in phase:
• Current lags voltage by 90 degrees
• Impedance of Inductor increases with increasing frequency
http://arapaho.nsuok.edu/%7Ebradfiel/p1215/fendt/phe/accircuit.htm
AC circuit analysis
• Effective impedance: example
• Procedure to solve a problem
–
–
–
–
Identify the sinusoid and note the frequency
Convert the source(s) to complex/phasor form
Represent each circuit element by it's AC impedance
Solve the resulting phasor circuit using standard circuit solving
tools (KVL,KCL,Mesh etc.)
– Convert the complex/phasor form answer to its time domain
equivalent
Example
( Z R1  Z C ) I1 ( j )
 Z C I 2 ( j )  VS ( j )
 Z C I1 ( j )  ( Z C  Z L  Z R 2 ) I 2 ( j )  0
VS ( j )
 ZC
0
ZC  Z L  Z R2
( Z C  Z L  Z R 2 )VS ( j )
I 1 ( j ) 

2
Z R1  Z C
 ZC
( Z R1  Z C )( Z C  Z L  Z R 2 )  Z C
 ZC
ZC  Z L  Z R2
1
1
66.7


 66.7 j ()
6
jC j1500 10
j
Z L  jL  j1500  0.5  750 j ()
ZC 
(75  683 j )150
I1 ( j ) 
(100  66.7 j )(75  683 j )  4450
(75  683 j )150
I1 ( j ) 
(100  66.7 j )(75  683 j )  4450
Top:
Bottom:
(75  683 j )150  68783.7  150
  tan 1
b
683
 tan 1
 83.7
a
75
A  a 2  b 2  687
(100  66.7 j )(75  683 j )  4450
 7500  45600  5000 j  683 j  4450  57550  63300 j
 8550047.8
68783.7  150
I 1 ( j ) 
8550047.8
 0.1235.9
 0.120.63 radians
i1 (t )  0.12 cos(1500t  0.63) Amps