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TEST 2 SOLUTION
ELECTRIC CIRCUIT
Find Rab
60uf
40uf26uf100uf
50uf
100uf
25uf
150uf
25uf
200uf
50uf
Rab
a
b
Find Rab
24uf26uf100uf
50uf
100uf
25uf
150uf
25uf
200uf
50uf
Rab
a
b
Find Rab
50uf100uf
50uf
100uf
25uf
150uf
25uf
200uf
50uf
Rab
a
b
Find Rab
25uf100uf
100uf
25uf
150uf
25uf
200uf
50uf
Rab
a
b
Find Rab
25uf25uf
100uf
150uf
25uf
200uf
50uf
Rab
a
b
100uf
Find Rab
150uf
100uf
150uf
25uf
200uf
50uf
Rab
a
b
Find Rab
75uf
100uf
25uf
200uf
50uf
Rab
a
b
Find Rab
75uf
100uf
25uf
200uf
50uf
Rab
a
b
Find Rab
200uf
200uf
50uf
Rab
a
b
Find Rab
33.33ufRab
a
b
Find Rab
60uf
40uf26uf100uf
50uf
100uf
25uf
150uf
25uf
200uf
50uf
Rab
a
b
c
d e f
200ufa bc d25uf 50uf
e25uf 150uf
100uf
f50uf26uf
60uf 40uf
100uf
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
RLa b
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Remove the RL First thing to notice is that this circuit contained a dependent source, thus ‘normal’ method can’t be used. (because we need source to calculate the dependent source)
Therefore, turn off (shorted) the independent voltage source and take out (open) the independent current source.
We will put a 1A current source at terminal a-b. Then use linearity,
0
0
ivRth
Obtain the Thevenin equivalent circuit for terminal a-b
RTH: Obtain V1, V2 and V3.
+-
4 Ω 3 Ω
2 Ω
6 Ω
1 A
ix30ix
V1 V2 V3
5 Ω
a b
At V1,
042
1 211
VVV
Straightaway we get,
412 VVix
024 211 VVV
034 21 VV
Obtain the Thevenin equivalent circuit for terminal a-b
RTH: Obtain V1, V2 and V3.
+-
4 Ω 3 Ω
2 Ω
6 Ω
1 A
ix30ix
V1 V2 V3
5 Ω
a bAt V2,
0634
23212
VVVVV
0122
1244
1233 23212
VVVVV
0439 312 VVV
Obtain the Thevenin equivalent circuit for terminal a-b
RTH: Obtain V1, V2 and V3.
+-
4 Ω 3 Ω
2 Ω
6 Ω
1 A
ix30ix
V1 V2 V3
5 Ω
a bAt V3,
0530
31 323
xiVVV
0905815 23 xiVV
04
905815 2123
VVVV
04545101630 2123 VVVV
30354516 213 VVV
Obtain the Thevenin equivalent circuit for terminal a-b
Therefore RTH
VV 16.11
VV 51.02
VV 27.03 +-
4 Ω 3 Ω
2 Ω
6 Ω
1 A
ix30ix
V1 V2 V3
5 Ω
a b
89.01
13
0
0 VVivRth
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain VTH: Find V1, V2 and V3
Straightaway we get,
412 VVix
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain VTH: Find V1, V2 and V3
02
01284
4 121
VVV
02128
44 121
VVV
0256216 121 VVV
2403 21 VV
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain VTH: Find V1, V2 and V3
0364
32212
VVVVV
012
44122
1233 32212
VVVVV
0639 212 VVV
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain VTH: Find V1, V2 and V3
0530
34 323
xiVVV
09035560 323 xiVVV
0905860 23 xiVV
04
905860 2123
VVVV
120163545 321 VVV
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain VTH: Find V1, V2 and V3
VV 1101
VV 902
VV 1203
VVVVVVab 1012011013
Obtain the Thevenin equivalent circuit for terminal a-b
+-
+-
4 Ω 3 Ω
2 Ω
6 Ω128 V
4 A
ix30ix
V1 V2 V3
5 Ω
a b
Obtain RTH: Find V1, V2 and V3
VV 1101
VV 902
VV 1203
VVVVVVab 1012011013
Obtain the Thevenin equivalent circuit for terminal a-b
+-
0.89 Ω
10V
a
b
Finally
Calculate the maximum power transfer
+-
0.89 Ω
10V
a
b
RL
Finally
WRVP
TH
TH 09.28)89.0(4
)10(4
max22
The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)
10Ω
2Ω
2H0.1Ω20 A
40Ω
t=0
iL+
-v0
The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)
10Ω
2Ω
0.1Ω20 A
40ΩiL+
-v0
t<0
Because of the inductor shorted, iL(0)=20A
The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)
t>0
10Ω
2Ω
2H 40ΩiL+
-v0
1082)40||10(2eqR
2.0102
eqRL
The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)
t>0 2.0
Aeiti t
L 0)(
Ae t520
The switch has been closed for a long time before it is opened at t=0. Find iL(t) and v0(t)
t>0Aeti t
L520)(
10Ω
2Ω
2H 40ΩiL+
-v0
Using current divider
)(1040
10)(40 titi L
Aeti t540 20)2.0()(
Aeti t540 4)(
Vetitv tL
50 160)(40)( Therefore:
The voltage pulse across a 0.5uF capacitor described by following equation
sstVetstVst
v(t))(t
14104
00
1
Solutions
• Current, i from equation
• Power, p from equation
s1 A,24105.0s 1s 0 A,24105.0s 0 A,00105.0
116
6
6
teett
itt
dtdvCi
s W,
s s W,s W,
18241082400
1211 teeetttt
pttt
dtdvCvvip
Solutions (cont.)
• Energy, w from equation
s1 J,4165.021
s 1s 0 J,4165.021
s 0 J,0
1212
22
tee
ttt
t
w
tt
Solutions (cont.)
• Energy is being stored in the capacitor whenever the power is positive. Hence the energy is being stored in the interval 0 – 1 s.
• Energy is being delivered by the capacitor whenever the power is negative. Hence the energy is being delivered for all t > 1 s.
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