inc 112 basic circuit analysis
DESCRIPTION
INC 112 Basic Circuit Analysis. Week 8 RL Circuits. First-order Differential equation. Objective: Want to solve for i(t) (in term of function of t). RL Circuit. KVL. Voltage source go to zero. consider. Assume that i(t) = g(t) make this equation true. - PowerPoint PPT PresentationTRANSCRIPT
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INC 112 Basic Circuit Analysis
Week 8
RL Circuits
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RL Circuit
AC
+
-
u(t)
i(t)
L
R
KVL )()(
)( tudt
tdiLtRi
First-order Differential equation
Objective: Want to solve for i(t) (in term of function of t)
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)()(
)( tudt
tdiLtRi
Assume that i(t) = g(t) make this equation true.
consider
However, g(t) alone may be incomplete. The complete answer isi(t) = f(t) + g(t)
where f(t) is the answer of the equation
0)(
)( dt
tdiLtRi
Voltage source go to zero
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)()()( tgtfti
Transient ResponseForced Response
i(t) consists of two parts
Therefore, we will study source-free RL circuit first
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Source-free RL Circuit
LR+
-
-
+
i(t)
Inductor L has energy stored so thatthe initial current is I0
Compare this with a pendulumwith some height (potential energy) left.
height
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0)()(
0)(
)(
tiL
R
dt
tdidt
tdiLtRi
There are 2 ways to solve first-order differential equations
LR+
-
-
+
i(t)
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Method 1: Assume solutionwhere A and s is the parameters that we want to solve for
stAeti )(
Substitute in the equationstAeti )( 0)()(
tiL
R
dt
tdi
0)(
0
st
stst
AeL
Rs
AeL
RAse
ซงเทอมทจะเป น 0 ไดกคอเทอม (s+R/L) เทานน ดงนนจะได
คาตอบจะอยในรป
L
Rs
tL
R
Aeti
)(
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Initial condition 0)0( Ii
tL
R
Aeti
)(from Substitute t=0, i(t=0)=0
AI
AeI
0
00
We gottL
R
eIti
0)(
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Method 2: Direct integration
tti
I
dtL
R
ti
tdi
dtL
R
ti
tdi
tiL
R
dt
tdi
tiL
R
dt
tdi
0
)(
0)(
)(
)(
)(
)()(
0)()(
tL
R
tti
I
eIti
tL
RIti
tL
Rti
0
0
0
)(
)(
)0(ln)(ln
)(ln0
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t
i(t)
I0
Approach zero
tL
R
eIti
0)(
Natural Response only
Natural Responseof RL circuit
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Time Constant
Ratio L/R is called “time constant”, symbol τ
R
L
Time constant is defined as the amount of time usedfor changing from the maximum value (100%) to 36.8%.
t
tL
R
eIeIti
00)(
Unit: second
368.01 e
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t
i(t)
2AtL
R
eti
1)(
Natural Response + Forced Response
1A
Natural Response
Forced Response
Forced response = 1A comes from voltage source 1V
Approach 1A
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Switch
Close at t =0 Open at t =0
t=0 t=0
t=0t=0
t=0
3-way switch
t < 0
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Switch
Close at t =0 Open at t =0
t=0 t=0
t=0t=0
t=0
3-way switch
t > 0
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t=0 t=0
R R1V 1V
t
v(t)
1V
t
v(t)
1V
0V 0V
Step function (unit)
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t=0
R=1Ω
2VL
1V
Will divide the analysis into two parts: t<0 and t>0
When t<0, the current is stable at 2A. The inductor acts likea conductor, which has some energy stored.
When t>0, the current start changing. The inductor discharges energy.Using KVL, we can write an equation of current with constantpower supply = 1V with initial condition (current) = 2A
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For t>0
1))(ln(
)(
)(
)(
)(
)()(
)()(
cttRiVR
L
dttRiV
tLdi
dttRiV
tLdi
tRiVdt
tdiL
Vdt
tdiLtRi
tL
R
tL
R
tL
R
cL
RtL
R
eR
c
R
Vti
ecVtRi
cetRiV
eetRiV
cL
Rt
L
RtRiV
cttRiVR
L
2
2
2
1
1
)(
)(
)(
)(
))(ln(
))(ln(
1
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tL
R
eR
c
R
Vti
2)( We can find c2 from initial condition
i(0) = 2 A
Substitute t = 0, i(0) = 2
RVcR
c
R
V
2
12
2
2
Therefore, we havetL
R
eR
RV
R
Vti
2)(
tL
R
eti
1)( Natural Response
Forced Response
Substitute V=1, R=1
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RL Circuit Conclusion
• Force Response of a step input is a step
• Natural Response is in the form where k1 is a constant, whose value depends on the initial condition.
tL
R
ek
1
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How to Solve Problems?
• Start by finding the current of the inductor L first
• Assume the response that we want to find is in form of
t
ekk
21
• Find the time constant τ (may use Thevenin’s)
• Solve for k1, k2 using initial conditions and status at the stable point
• From the current of L, find other values that the problem ask
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Examplet=0
R=1Ω
2VL=1H
1V
i(t)
The switch is at this position for a long timebefore t=0 , Find i(t)
t
ekkti
21)(
Time constant τ = 1 sec
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t
ekkti
21)(
At t=0, i(0) = 2 A212 kk
At t = ∞, i(∞) = 1 A 01 1 k
Therefore, k1 = 1, k2 = 1
The answer is teti 1)(
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2A
1A
teti 1)(2)( ti
)(ti
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Example
R1
L
R2
R3
R4i(0)=5
+ -
i2(t)
L has an initial current of 5A at t=0Find i2(t)
The current L is in form of t
L ekkti
21)(
Time constant = R/L, find Req 421
213 R
RR
RRRReq
Time constanteqR
L
(Thevenin’s)
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Find k1, k2 using i(0) = 5, i(∞) = 0
At t=0, i(0) = 5 A 215 kk
At t = ∞, i(∞) = 0 A 00 1 k
t
L ekkti
21)(
Therefore, k1=0, k2 = 5 t
L eti
5)(
i2(t) comes from current divider of the inductor current
21
12 5)(
RR
Reti
t
Graph?
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)(2 ti
21
12 5)(
RR
Reti
t
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Example
t=0
1Ω1V
1H2Ω
2Ω+
v1(t)-
L stores no energy at t=0Find v1(t)
5.02
1
21)2||2(
eq
eq
R
L
R
Find iL(t) first
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Find k1, k2 using i(0) = 0, i(∞) = 0.25
At t=0, i(0) = 0 A 210 kk
At t = ∞, i(∞) = 0.25 A 025.0 1 k
t
L ekkti
21)(
Therefore, k1=0.25, k2 = -0.25t
L eti 225.025.0)(
v1(t) = iL(t) R
tetv 21 25.025.0)(
Graph?
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tetv 21 25.025.0)(
)(1 tv
0)(1 tv