Transcript Exam 3 review

Review
Part 3 of Course
Passive Circuit Elements
+
+
i
v
R
-
v  iR
1
i v
R
v
i
+
i
+
L
-
di
vL
dt
1
i   vdt
L
C
v
-
-
1
v   idt
C
dv
iC
dt
Energy stored in the capacitor
The instantaneous power delivered to the capacitor is
dv
p (t )  vi  Cv
dt
The energy stored in the capacitor is thus
t
dv
w   p (t )dt  C  v dt  C  vdv


dt
t
1 2
w  Cv (t ) joules
2
Energy stored in the capacitor
Assuming the capacitor was uncharged at t = -, and knowing that
q  Cv
2
1 2
q (t )
w  Cv (t ) 
2
2C
represents the energy stored in the electric field established
between the two plates of the capacitor. This energy can be
retrieved. And, in fact, the word capacitor is derived from this
element’s ability (or capacity) to store energy.
Parallel Capacitors
+
i
v
i1
C1
i2
iN
+
i
CN
C2
-
v
i
Ceq
-
dv
i1  C1
dt
dv
i2  C2
dt
dv
iN  C N
dt
dv
dv
i  i1  i2    iN   C1  C2    C N   Ceq
dt
dt
N
Ceq   Ck
k 1
Thus, the equivalent capacitance of N capacitors in parallel is the
sum of the individual capacitances. Capacitors in parallel act like
resistors in series.
Series Capacitors
C1
C2
CN
+
+
DC
v 1-
+ v2
-
+
vN -
DC
v
Ceq
-
i
1
v1 
idt

C1
v
i
1
v2 
idt

C2
vN 
1
idt

CN
 1
1
1 
1
v  v1  v2    vN   
  
idt
  idt 

CN 
Ceq
 C1 C2
N
1
1

Ceq k 1 Ck
The equivalent capacitance of N series connected capacitors is the
reciprocal of the sum of the reciprocals of the individual capacitors.
Capacitors in series act like resistors in parallel.
Energy stored in an inductor
The instantaneous power delivered to an inductor is
di
p (t )  vi  Li
dt
The energy stored in the magnetic field is thus
t
di
wL (t )   p(t )dt  L  i dt  L  idi
 dt

t
1 2
wL (t )  Li (t ) joules
2
Series Inductors
L1
L2
LN
+
+
DC
v 1-
+ v2
-
+
vN -
DC
v
v
i
Leq
-
i
di
v1  L1
dt
di
v2  L2
dt
di
vN  LN
dt
di
di
v  v1  v2    vN   L1  L2    LN   Leq
dt
dt
N
Leq   Lk
k 1
The equivalent inductance of series connected inductors is the sum
of the individual inductances. Thus, inductances in series combine
in the same way as resistors in series.
Parallel Inductors
+
i
v
i2
i1
L1
iN
L2
LN
i
v
i
Leq
-
-
1
i1   vdt
L1
+
1
i2 
vdt

L2
1
iN 
vdt

LN
1 1
1 
1
i  i1  i2    iN      
vdt

vdt


LN 
Leq
 L1 L2
N
1
1

Leq k 1 Lk
The equivalent inductance of parallel connected inductors is the
reciprocal of the sum of the reciprocals of the individual
inductances. Thus inductances in parallel combine like resistors in
parallel.
Complex Numbers
A  x  jy
A  A cos j  jA sin j
j  tan
1
A
y
x
j
e jj  cos j  j sin j
e
j j  
j measured positive
counter-clockwise
A  Ae jj
Note:
real
x  A cos j
Euler's equation:

A
y  A sin j
A  x2  y 2
Complex
Plane
imag
 e e  cos j     j sin j   
jj
j
Comparing Sinusoids
-sin t
cos t      cos t
Im

-cos t
Re
cos t
2

Note: positive angles are counter-clockwise
Im

45
135
45



cos  t    sin t
2

sin t
cos t  45
sin t      sin t


sin  t     cos t
sin t
Re
cos t
cos t leads sin t by 90
cos t lags - sin t by 90



sin t  45  cos t  135
 leads cost by 45

and leads sint by 135
AC Circuits and Phasors
A phasor is a complex number that represents the amplitude
and phase of a sinusoid.
X M e jj  X M j  X
XM
j
j t  
Recall that when the assumed form of the current i(t )  I M e
was substituted into the differential equations, the e j t cancelled out.
We were then left with just the phasors
Impedance
Impedance Z 
i(t)
VM cos t
+
AC
-
V phasor voltage

I phasor current
R
+ VR
I
- +
VL
-
L
Units = ohms
V
R  j L
V
Z   R  j L  Z  z
I
Z  R  jX
R  resistance
Z  R2   2 L2
X  reactance
1  L
 z  tan
R
Note that impedance is a complex number containing a real, or
resistive component, and an imaginary, or reactive, component.
Admittance
1 I
phasor current
Admittance Y = = 
Z V phasor voltage
i(t)
VM cos t
+
Units = siemens
R
+ VR
AC
-
conductance G 
V
I
R  j L
- +
VL
-
R
R 2   2 L2
L
I
1
R  j L
Y= 
 G  jB  2
V R  j L
R   2 L2
susceptance
B
 L
R 2   2 L2
Im
V
V
I
I
Re
I in phase
with V
V
Im
I
V
Re
I lags V
V
I
Im
I
V
I
I leads V
Re
I
Z3Z 4
Zin  Z1  Z 2 
Z3  Z 4
DC
Z1
V
Z2
I1
Z3
I2
Z4
Zin
We see that if we replace Z by R the impedances add like resistances.
Impedances in series add like resistors in series
Impedances in parallel add like resistors in parallel
Voltage Division
Z1
+ V1 DC
V
I
Z2
+
V2
-
V
I
Z1  Z 2
But
V1  Z1I
V2  Z2I
Therefore
Z1
V1 
V
Z1  Z 2
Z2
V2 
V
Z1  Z 2
Instantaneous Power
v(t )  VM cos t  v 
i(t )  I M cos t  i 
p(t )  v(t )i(t )  VM I M cos t  v  cos t  i 
VM I M
cos v  i   cos  2t   v  i  
p(t ) 
2
Note twice the frequency
Average Power
T  2 
1
P
T

t0 T
t0
1
p(t )dt 
T

t0  T
t0
VM I M cos t  v  cos t  i  dt
1 t0 T VM I M
cos  v  i   cos  2t   v  i   dt
P 
T t0
2
P  1 VM I M cos v  i 
2
Purely resistive circuit
 v  i  0
P  1 VM I M
2
Purely reactive circuit
v  i  90
P  1 VM I M cos  90   0
2
Effective or RMS Values
We define the effective or rms value of a periodic current
(voltage) source to be the dc current (voltage) that delivers the
same average power to a resistor.
1
PI R
T
2
eff
I eff

t0  T
t0
i 2 (t )Rdt
1 t0 T 2

i (t )dt

T t0
I eff  I rms
Veff2
1 t0 T v 2 (t )
P
 
dt
t
R T 0
R
1 t0 T 2
Veff 
v (t )dt

T t0
root-mean-square
Veff  Vrms
Effective or RMS Values
Vrms
Using
Vrms
Vrms
1 t0 T 2

v (t )dt

T t0
v(t )  VM cos t  v 
cos   1  1 cos 2
2
2
2

 VM 
 2

 VM 
 2


2 
0
2 
0
T  2 
and
 1  1 cos  2t  2v dt 
2
 2
 
1 
dt 
2 
1
2

 VM 
 2
2 
t
 
20



1
2
1
2
VM

2
Ideal Transformer - Voltage
AC
i1
i2
+
+
v1
N1
N2
-
d
v1 (t )  N1
dt
v2
Load
-

This changing flux through
coil 2 induces a voltage, v2
across coil 2
d
v1 N1 dt
N1


v2 N 2 d N 2
dt
The input AC voltage, v1,
produces a flux
1

v1 (t )dt

N1
d
v2 (t )  N 2
dt
N2
v2 
v1
N1
Ideal Transformer - Current
AC
i1
i2
+
+
v1
N1
N2
-
Magnetomotive force, mmf
v2
Load
F  Ni
-

The total mmf applied to core is
F  N1i1  N 2i2  R 
For ideal transformer, the reluctance R is zero.
N1i1  N 2i2
N1
i2 
i1
N2
Ideal Transformer - Impedance
AC
i1
i2
+
+
v1
N1
v2
N2
-
-
Load
V2
ZL 
I2
N1
V1 
V2
N2
Input impedance
V1
Zi 
I1
Load impedance
2
 N1 
Zi  
 ZL
 N2 
ZL
Zi  2
n
Turns ratio
N2
I1 
I2
N1
N2
n
N1
Ideal Transformer - Power
AC
i1
i2
+
+
v1
N1
N2
-
Load
P  vi
-
Power delivered to primary
Power delivered to load
P2  v2i2
P1  v1i1
N2
v2 
v1
N1
v2
N1
i2 
i1
N2
P2  v2i2  v1i1  P1
Power delivered to an ideal transformer by the source
is transferred to the load.
Force on current in a magnetic field
Force on moving charge q -- Lorentz force
F  q (v  B )
Current density, j, is the amount of charge passing per unit area
per unit time. N = number of charges, q, per unit volume moving
with mean velocity, v.
L
S
j
F   N V  q(v  B)
F  ( j  B)V
F  (i  B)L
vt
V  SL
dQ NqSvt
i

 j S
dt
t
j  Nqv
Force per unit length on a wire is
iB
Rotating Machine
B
Force out
commutator
i
+
i
brushes
X
Force in
Back emf
B
B
d

dt
Force out
+-
i
a
b
r

l
i
X
X
Force in
area  A  lw  l 2r cos
w  2r cos
flux  BA  Β2rl cos


emf   E ds    flux     B 2rl cos  

t
t

emf  eab  B 2rl sin 
 2Brl sin 
t
B
Back emf
Force out
commutator
i
+
-
eab  2Brl sin 
i
brushes
X
Force in
eab

Armature with four coil loops
eab
S
X
X
X
X
N

Motor Circuit
Ia

Ra
Vt
Ea
Ea  K a
Power and Torque
Pd  Ea I a   d 

 d  Ka I a
Vt  Ea  I a Ra
I a  Vt  Ea  / Ra
I a  Vt  Ka  / Ra
d 
Vt K a

Ra
 K a 
Ra
2

AC Nodal Analysis
For steady-state AC circuits we can
use same the method of writing nodal
equations by inspection that we
learned for resistive circuits. To do
so, we must replace resistances with
impedances.
We solved Problem 4.31 in Class:
V1
j2
V2
40 A
20 A

-j1
Change impedances to admittances
V1
j2
V2
-j/2 S

S
20 A
40 A
-j1
j1 S
Nodal Analysis for Circuits Containing
Voltage Sources That Can’t be
Transformed to Current Sources
1.
2.
3.
Assume temporarily that the current across each voltage
source is known and write the nodal equations in the same
way we did for circuits with only independent voltage
sources.
Express the voltage across each independent voltage
source in terms of the node voltages and replace known
node voltages in the equations.
Rewrite the equations with all unknown node voltages and
source currents on the l.h.s. of the equality and all known
currents and voltages on the r.h.s of the equality.
We solved #4.33 in text in Class:
V1
-j4
Note: V2 = 10
V2
j/4 S
645 A

1/2 S
I0
+
j2
-j/2 S
100 V
AC
I2
assume I2
 0.5  j 0.25 0   V1   4.243  j 6.743 

 I   


j
0.25
1
j
2.5

 2  

AC Mesh Analysis
For steady-state AC circuits we can
use same the method of writing mesh
equations by inspection that we
learned for resistive circuits. To do
so, we must replace conductances
with admittances.
We solved Problem 4.38 in text in Class:
Find I1 and I2:

j1
+
I1
AC
120 V

-
-j1
+
I2
60 A
AC
-
 2  j1 1  j1  I1   12 

I    
 1  j1 1  j 0   2   6 
What happens if we have independent
current sources that can’t be
transformed in the circuit?
1. Assume temporarily that the voltage across each
current source is known and write the mesh
equations in the same way we did for circuits
with only independent voltage sources.
2. Express the current of each independent current
source in terms of the mesh currents and replace
known mesh currents in the equations.
3. Rewrite the equations with all unknown mesh
currents and voltages on the left hand side of the
equality and all known voltages and currents on
the r.h.s of the equality.
We solved Problem 4.40 in text in Class:
Find I0:

Assume you know V2
-j1
+
j2
+
AC
60 V
-
I2
I1

I0
 3  j1

 2  j 2
20 A V2
-
Note I2 = -2
0   I1   2  j 4 
 V   

1   2   4  j4 
Matlab Solution: Know how to program
 I1  1.414  81.87 

 
V
 2   7.37612.53 
AC Thevenin's Theorem
AC Thevenin's Theorem
Thevenin’s theorem states that the two circuits given below are
equivalent as seen from the load ZL that is the same in both cases.
I
a
Zth
Linear
Circuit
ZL
b
AC
I
a
ZL
VTh
b
Zin
Zin
VTh = Thevenin’s voltage = Vab with ZL disconnected (= ) = the
open-circuit voltage = VOC
AC Thevenin's Theorem
I
a
Zth
Linear
Circuit
ZL
b
AC
I
a
ZL
VTh
b
Zin
Zin
ZTh = Thevenin’s impedance = the input impedance with all
independent sources turned off (voltage sources replaced by short
circuits and current sources replaced by open circuits). This is the
impedance seen at the terminals ab when all independent sources
are turned off.
We solved Problem 4.40 using Thevenin's Thm. in Class:

-j1
+

j2
+
AC
60 V
-j1
20 A
-
+
VOC
-
AC
60 V
-

20 A
I0

+
8  j2
VOC  6  2(1  j )  8  j 2
-j1
j2

AC
-j1

I0
I0 
8  j2
8  j 2 8.246  14.04


1  j1  j 2  2 3  j1
3.16218.43
I 0  2.608  32.47
ZTH  1  j1
AC Superposition
Superposition Principle
The superposition principle states
that the voltage across (or the current
through) an element in a linear circuit
is the algebraic sum of the voltages
across (or currents through) that
element due to each independent
source acting alone.
Steps in Applying the Superposition Principle
1. Turn off all independent sources except
one. Find the output (voltage or current)
due to the active source.
2. Repeat step 1 for each of the other
independent sources.
3. Find the total output by adding
algebraically all of the results found in
steps 1 & 2 above.
Example Done in Class:

30sin 5t
+
+
v0(t)
-
-
AC
0.2F
2 cos10t
1H
Note that the voltage source and the current source have two
different frequencies. Thus, if we want to use phasors, the
only way we've solved sinusoidal steady-state problems, we
MUST use superposition to solve this problem. We
considered each source acting alone, and then found v0(t) by
superposition.
Remember that

sin t  cos t  90


Example
30sin 5t
+
+
v0(t)
-
-
AC
0.2F
2 cos10t
1H
By superposition:
v0(t) = Response due to voltage source + Response due to voltage source
v0 (t )  v01 (t )  v02 (t )
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v0 (t )  4.631sin 5t  81.12  1.05cos 10t  86.24
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