EXAM
2
What
To
ENERGY
Know
CONSUMED
I
Wt
BY CAPACITOR
BELOW
cut
4 Steps CAPACITOR
INDUCTOR
CONVERT EQUATION FROM TIME DOMAIN TO PHASOR DOMAIN
WHAT IS MEAN SQUARE ROOT VOLTAGE
CALCULATE AVERAGE POWER OF A CIRCUIT
GENERIC METHOD
EQUIVALENT IMPEDANCE
CALCULATE
IMPEDANCE NETWORK
OF
SERIES
Eq
PARALLEL
SERIES
Capacitance
Equivalent
CEq PARALLEL
11
IG
Eat
I
EQ
I
C3
i
I
i ti ti
Cz
C
CEq
3
OR
C'XC XC
Ci t Cz
CEq
ALCULATE
VOLTAGE
F
3,5
IOOV
I
Turned
OF
Ix
Oort
Vx
WE
3,9
DIVIDER
ONLY
2
CAPACITORS
gov
Turned
97
on
F
I
O lov I
3OF
VOLTAGE
WORK
ON
off
3,9
41000
3505 1251
F
TECHNIQUE
I 900
350 F
V
USE
DIVIDER
VOLTAGE
NEED SUPERPOSITION
9ft
3,5
90
TO
on
Ux
CAPACITOR
97
3OF
00 V
OF A
CEE301
1 957
125 F
I2SF
73.7W
100 u
off
95
CEE301 1 3505
3805
I
Igor
0
X 900
180
3OF
2
ASF
95
380
In
3805
I 90W
Vx
Vx Vx 2
Vx
91.7
73.76
180
U
STATED WHAT TO KNOW
ENERGY CONSUMED
EXAMPLE
1
CALCULATE THE
N THE
SO MF
Solution
Step 1
BY
ENERGY
STORED
CAPACITOR
125nF Some
Solve
2
ENERGY
For
10
EXAMPLE
250W I
Equation
W
E C v2
capacitor
178.5712
0.797
6
J
797 my
129mF
2
3300 I
40mF
CAPACITOR
simplify
circuit
129mF
330W I
Step
2
40mF
Determine
Voltage
SOME
178.57 V
Energy
CALCULATE THE STORED
350mF
ENERGY IN THE
Step 1
I
in
Stored
50
W
RMF
Vi
Determine
Voltage Divider
V
125mF X 2500
step
A CAPACITOR
divider
I
575mF
V1
to
determine
voltage
350mF
225mF
575mF
120mF
Step 3
For
Solve
350
W
10
MAYBE
l
step
2
step
3
Find
Step
4
Solve
Find
Velo
Find
Ucla
EXAMPLE
1
VC o
constant
time
at
Velt
Step 1
56.9872
6
for
for
Solve
I
E
Vc
Step
3
Find
k
I
320
yokn
Rth
Rt
I
8mF
circuit
Ou
voltage
MEER
I
b
yoke
UCO
v
divider
UX RER XV
Vx
c
ca XV
Voltage divider
40kt
I
g
ME at
as
Mtr
320W
CAPAC
Resistance
open
Solve for
2
steady state
to
v
Step
INDUCTOR
starting
o
Velo
yoke
INDUCTOR
VC t
and
60ms
568mi
CAPACITOR
CAPACITOR
the
Vc o
t
J
BUT ON
t
The venin
Ikr
320W
0.568
METHOD
Step
ICU
W
capacitor
EXPAND
GENERIC
Find
Energy Equation
stored in
Energy
56.980
X 3300
Velo
T
Rt
T
24K 40k
24kt 40k
Rt XC
Yoke
4Okrtzuk
X 3200
200 u
ISKA
15,000
0
8 10
0.12 Sec
or
120ms
Step
Solve
4
Volt
Vocal
Volt
200
VCC60ms
solve
e
t Roms
6
200 e
1
t
and
Step
Solve for
2400 I 40k
Ico
3
an
Find Rt H
moi
I
240
girth
in
Step
4
Solve Eq
14.4mA
10K 6.67k
8K
x
40K
40kt 8k
Ict
t
12 mA
16km
81448,1
0.72
T
16k
r
6
Ge
45ns
67.5ms
th
co
ico
lect ices
e
t 45ns
24 12 e
ie t
12
14
iL 67.5ns 12 1246751 45
ROOT
6.67k
81k
yo
T
Rt
in
24mA
8K
icco 14.4
Step
inductor
circuit
340,4
k
40kg in
8km
ices
it
240W I
WE IE't
Short
young in
2
78.6940
for Inco
we
240W I
20
the
20054 120
200
INDUCTOR
constant
Step 1
200
200
III 416
time
60ms
Ucla Vocal e the
O
EXAMPLE
solve
Eq Volt
SQUARE
t 45
678mA
P
Vct
Um
Cos
REFIT
wt
It Et
III
rag
power
AC circuit
Plt
Vct
S Plt
dt
PLAUG
V
got
Street
ULt
W
go
T
0
PAVE V2
MS
Plt
2
R
Periodic
sinusodial
cost
100
If
2
a
I
I
100W
to
TRANSIENT
R
V'Rms
at
at
t
4
Cos O 900
IZ
Y
0.707 Um
1
Rms
t
R
YI
EXAMPLE
I
I
I
y
RMS
t
P
E
RMS
T
wave
if
we 2
Sin 04
DC Circuit
W ENERGY
Sinusodial
O
t
loopy
10.05
2
If
f SO Hz
20ms
Um
i
sinusodial
Um Amplitude
p
70.7 V
70.72
So
VET
100W
10000 COS
50
100Mt
200 cos
loot
what
paget
I
I
2
a
o
o
4
6
8
T
g at
TIME
DOMAIN
Form
Polar
1
Rec Form
a
b
V2 t
Vit
72
at
bj
a
Wtt 450
Find
UT
1002450
VT
1502 300
to
convert
t
V2 t
Rec
V2 150Cost 307
VI
Convert
Vm
Vict
IN POLAR
PHASOR
VT 100 Cos 45 t 1005in 45
3
t
Rec
150 sin wt 600
150 Cos wt 300
VT
DOMAIN
polar
Reca
a
1
É
O
Polar
1.60
PHASOR
y imaginary
tt
100 Cos
t
Um
Vm since
EXAMPLE
1
sec
8 2
y
o
Vm Cos O
851
V
4
Urms
Vrmsiytficzidttft.int
VRMs
2
is
t 1505in
70.7
back
130
to
7 1200.712 1 4.312
tan
4 3200.7
70.7
70.7
30
70.71
Polar
200.75
1.20
130
75
75 j
j
j
200.7
4.3 j
Polar
4
200.75
Convert to
time domain
Vlt
Um COS
Uct
200.75 Cos wt
PHASOR
PHASOR
DIVISION
V LO
V22 02
EXAMPLES
Z
0
wt
1.20
MULTIPLICATION
V2CO2
UV2 O 02
V LO X
Fl
1.20
L
Y
MUL
Q 02
Z
DIV
Zz
300
2
Zi
30 135
25 3
75
25
Zz
23 L 300 1350
1050
8.332 1650
EQUIVALENT IMPEDANCE
Impedance
of
Impedance
of
Impedance
of
EXAMPLE
a
a
inductor
an
Resistor
1
f
D
son
as
b
W
2nF
NETWORK
jwh
z
z
capacitor
do mfs't
ab
321350
72
Zc
SO Ht
2T SOHz
R
j
Find
Zab
100 it
ZR 150
PAKALLEL
ZL I
72.544 30.030
rag
62812
M
1001T 0.121
tan
37 7
6 36.312 72.5
30.030
33
I
j
ZE
Zab 22
t ZR
ZC
37.7
loot
25 10 6
j 0.007854
52 0.007857
37.75
74
127.3 j
t
1
to
t
62 8
Zab
62.8
36.3 j
0
7854
EXAMPLE
k
do
2
a
z
bo
Zoll Zizi
Zab
d
Find
2
Ze
jxc
ZRz
3525
z
Zee 7122
Zeller
0
XC
125
Zab
67.5
in
40
e
zh
21
in
are
50
series
255
I
jets'otasi
XetSOXC 3125
3125
2011214
XC
Zab
25xc XCHES i
50 02
XC 50
h
125 O
125
St
SOX
so
3,2
t
ÉÉÉÉÉo V