INC 112 Basic Circuit Analysis Lecturer Assistant Prof. Dr. Poj Tangamchit.
INC 112 Basic Circuit Analysis
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Transcript of INC 112 Basic Circuit Analysis
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INC 112 Basic Circuit Analysis
Week 10
RLC Circuits
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RLC Circuits
Similar to RL and RC circuits, RLC circuits has two partsof responses, namely: natural response and forced response.
Force Response:Similar to RL and RC, a step input causes a step output.
Natural Response:Different and more difficult than RL, RC.
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Source-free RLC Circuits
We study the natural response by studying source-free RLC circuits.
LR+
v(t)-
iR(t)C
iC(t)iL(t)
Parallel RLC Circuit
0)(1)(1)(
0)(
)(1)(
0
2
2
tvLdt
tdv
Rdt
tvdC
dt
tdvCdttv
LR
tv
iii CLR
Second-orderDifferential equation
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This second-order differential equation can be solved by assuming solutions
The solution should be in form of
0)(1)(1)(
2
2
tvLdt
tdv
Rdt
tvdC
stAetv )(
If the solution is good, then substitute it into the equation will be true.
011
0)11
(
011
2
2
2
Ls
RCs
Ls
RCsAe
AeL
AseR
eCAs
st
ststst
which meanss=??
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0112 L
sR
Cs
LCRCRCs
LCRCRCs
1
2
1
2
1
1
2
1
2
1
2
2
2
1
Use quadratic formula, we got
tseAtv 11)( Both and tseAtv 2
2)( are solution to the equation
Therefore, the complete solution iststs eAeAtv 21
21)(
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LCRCRCs
1
2
1
2
12
2,1
From
Define resonant frequencyLC
10
Damping factorRC2
1
Therefore,20
22,1 s
in which we divide into 3 cases according to the term inside the bracket
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Solution to Second-order Differential Equations
1. α > ω0 (inside square root is a positive value) Overdamped case
2. α = ω0 (inside square root is zero) Critical damped case
3. α < ω0 (inside square root is a negative value) Underdamped case
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1. Overdamped case , α > ω0
7H6Ω+
v(t)-
iR(t)1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
5.32
1
RC 6
10
LC
α > ω0 ,therefore, this is an overdamped case
20
22,1 s s1 = -1, s2 = -6
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Therefore, the solution is in form oftt eAeAtv 6
21)(
Then, we will use initial conditions to find A1, A2
From vc(0) = 0 we substitute t=0
210
20
10)0( AAeAeAv
From KCLAt t=0
4206
0642
1)10(
0
0)(
)10()0(
0
21
0
621
0
AA
eAeAR
dt
tdvC
R
v
iii
t
tt
t
CLR
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Solve the equation and we got A1 = 84 and A2 = -84
and the solution is
)(848484)( 66 tttt eeeetv
t
v(t)
)(84)( 6tt eetv
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2. Critical damped case , α = ω0
7H8.573Ω+
v(t)-
iR(t)
1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
45.21
2
10
LCRC
α = ω0 , this is an critical damped case s1 = s2 = -2.45
The complete solution of this case is in form of
stst eAteAtv 21)(
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Then, we will use initial conditions to find A1, A2
From vc(0) = 0 we substitute t=0
20
20
1 )0(0)0( AeAeAv
Find A1from KCLat t=0
0)(42
110
0)45.2(42
1)10(
0
0)(
)10()0(
0
1
0
45.21
45.21
0
A
eAetAR
dt
tdvC
R
v
iii
t
tt
t
CLR
Therefore A2 =0 and the solution is reduced totteAtv 45.2
1)(
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Solve the equation and we got A1 = 420 and the solution is
t
v(t)
ttetv 45.2420)(
ttetv 45.2420)(
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3. underdamped case , α < ω0
from20
22,1 s
The term inside the bracket will be negative and s will be a complex number
define 220 d
Then djs 2,1
and
)()(
)(
21
)(2
)(1
tjtjt
tjtj
dd
dd
eAeAetv
eAeAtv
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)()( 21tjtjt dd eAeAetv
Use Euler’s Identity sincos je j
)sincos()(
sin)(cos)(()(
)sincossincos()(
21
2121
2211
tBtBetv
tAAjtAAetv
tjAtAtjAtAetv
ddt
ddt
ddddt
)sincos()( 21 tBtBetv ddt
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3. Underdamped case , α < ω0
7H10.5Ω+
v(t)-
iR(t)
1/42f
iC(t)iL(t)
Initial conditionvc(0) = 0, iL(0) = -10A
Find v(t)
22
1
RC 6
10
LC
α < ω0 ,therefore, this is an underdamped case
2220 d
and v(t) is in form )2sin2cos()( 212 tBtBetv t
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Then, we will use initial conditions to find B1, B2
From vc(0) = 0 we substitute t=0
Find B2from KCLat t=0
0)2(42
110
02sin22cos242
1)10(
0
0)(
)10()0(
0
2
0
22
22
0
B
teBteBR
dt
tdvC
R
v
iii
t
tt
t
CLR
Therefore B1 =0 and the solution is reduced to
1210 )0sin0cos()0( BBBev
)2sin()( 22 tBetv t
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we got and then the solution is
t
v(t)
29722102 B
)2sin297)( 2 tetv t
)2sin297)( 2 tetv t
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Natural responseat different time
Mechanical systems are similar to electrical systems
A pendulum is an example of underdamped second-order systems in mechanic.
t
displacement(t)
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Series RLC
0)(1)()(
0)(1)(
)(
0
2
2
tiCdt
tdiR
dt
tidL
dttiCdt
tdiLRti
vvv CLR
LR +
vL(t)-
+ vR(t) -
C
i(t)
-vC(t)
+
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0)(1)()(
2
2
tiCdt
tdiR
dt
tidL
Series RLCParallel RLC
0)(1)(1)(
2
2
tvLdt
tdv
Rdt
tvdC
tsts eAeAtv 2121)(
LCRCRCs
1
2
1
2
12
2,1
20
22,1 s
RC2
1
LC
10
tsts eAeAti 2121)(
LCL
R
L
Rs
1
22
2
2,1
20
22,1 s
L
R
2
LC
10
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Procedure for Solving RLC Circuits
1. Decide that it is a series or parallel RLC circuit. Find α and ω0. Then, decide which case it is (overdamped, critical damped, underdamped).
2. Assume the solution in form of (natural response+ forced response)
fstst VeAteA 21
ftsts VeAeA 21
21
fddt VtBtBe )sincos( 21
Overdamped
Critical damped
Underdamped
3. Find A, B, Vf using initial conditions and stable conditions
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Example
S1 30
t = 0 sec
+
Vc
-
1/27f3H4A 5AiL(t)
Find vc(t)
52
L
R 31
0 LC
9,120
22,1 s
This is overdamped case, so the solution is in form
ftt VeAeA 9
21
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ftt
C VeAeAtv 921)(
Consider the circuit we found that the initial conditions will bevC(0) = 150 V and iL(0) = 5 A and the condition at stable point will bevC(∞) = 150 V and iL(∞) = 9 A
fVAA 21150Using vC(0) = 150 V , we got
Using vC(∞) = 150 V , we got fV 00150
Therefore, Vf = 150, A1+A2 = 0
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Consider the circuit, we find that iC(0) = 4A
21
02
01
921
9108
)9(27
14)0(
)9(27
1)()(
AA
eAeAi
eAeAdt
tdvCti
C
ttC
A1 = 13.5, A2 = -13.5
Therefore, 1505.135.13)( 9 ttC eetv