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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 Presentation
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INC 112 Basic Circuit Analysis
Week 8
RL Circuits
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)
)()(
)( 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
)()()( tgtfti
Transient ResponseForced Response
i(t) consists of two parts
Therefore, we will study source-free RL circuit first
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
0)()(
0)(
)(
tiL
R
dt
tdidt
tdiLtRi
There are 2 ways to solve first-order differential equations
LR+
-
-
+
i(t)
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
)(
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)(
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
t
i(t)
I0
Approach zero
tL
R
eIti
0)(
Natural Response only
Natural Responseof RL circuit
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
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
Switch
Close at t =0 Open at t =0
t=0 t=0
t=0t=0
t=0
3-way switch
t < 0
Switch
Close at t =0 Open at t =0
t=0 t=0
t=0t=0
t=0
3-way switch
t > 0
t=0 t=0
R R1V 1V
t
v(t)
1V
t
v(t)
1V
0V 0V
Step function (unit)
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
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
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
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
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
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
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)(
2A
1A
teti 1)(2)( ti
)(ti
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)
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?
)(2 ti
21
12 5)(
RR
Reti
t
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
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?
tetv 21 25.025.0)(
)(1 tv
0)(1 tv
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