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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Chapter (6)
First Order Transient Circuits
This chapter deals with the transient response of first order circuit .
first order circuit are those circuits whose response can beexpressed by a first order differential equations.
Example of such circuits include the RL and RC circuits
RL circuit : circuit that contains an indicator and a resistor.RC circuit : circuit that contains a capacitor and resistor
For example: RC circuitConsider the following circuit .
+
- iVS
C
R
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S
S
Vi(t)R(t)v
0i(t)R(t)vV
c
C
=+
=++
Sc
c
c
Vdt
(t)dvCR(t)v
dt
(t)dvCi(t)since
=+
=
Scc V
CR1(t)v
CR1
dt
(t)dv
=+
b(t)va
dt
(t)dv
VCR
1
b,CR
1
alet
cc
S
=+
==This is the general form of a linear
first order differential equation
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This is a first order linear ordinary non-homogenous differential
equation describing the response of the capacitor voltage.
It is first order because the highest degree of derivative is one.
it is called linear because the differential equation is linear
function of (dvc /dt) and vc(t)
Example of non-linear differential equation :
b(t)vdt
(t)dv(t)v2
cc
c =+
it is called ordinary because it deals only with ordinary
derivatives ( not partial derivative )
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It is called non-homogenous because b 0.
Example of first order linear ordinary homogenous differential
equation is
0(t)va
dt
(t)dvc
c =+
Example :
R L circuit
+
- iL(t)
R
VSL
KVL around the loop
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L
V(t)i
L
R
dt
(t)di
0dt
(t)diL(t)iRV
SL
L
LLS
=+
=++
b(t)iadt
(t)di
LVb,
LRalet
LL
S
=+
==
This is a first order linear ordinary non-homogenous differential
equation
To find the response of the Vc(t) in RC circuit or iL(t) in the RL
circuit we need to solve these differentials
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Solution of the first order differential equation :
Consider the first order linear ordinary non-homogenousdifferential equation :
We want to find X(t) that satisfies (*)
(*)bx(t)adt
dx(t)
LL=+
Theorem : ( in differential equation)
If x (t) = xP (t) is any solution of equation (*) and x (t) = xc (t) is
any solution of the homogenous differential equation
(**)0(t)xadt
(t)dxc
cLL=+
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then
(t)x(t)xx(t) cP +=
Where :
xP(t) = particular solution ( forced solution)
xC(t) = complementary solution ( natural solution)
Hence we need to solve 2 differential equations
(*)b(t)xadt
(t)dx PP
LL=+
What is the function xP(t) that if its differential is summed to a*xP(t)
will give a constant (b) The solution xp(t) must be constant
xP(t) = k1
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Use xP(t) in the non-homogenous differential equation
a
bk(t)x
a
bk
bka)(kdt
d
1P
1
11
==
=
=+
Particular (forced) response
Consider the homogenous differential equation:
adt
(t)dx
(t)x
1
(*)0(t)xa
dt
(t)dx
c
c
cc
=
=+ LL
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( )[ ]
( )[ ] a(t)xlndt
d
dt
(t)dx
(t)x
1(t)xln
dt
dsince
c
c
c
c
=
=
[ ]
[ ]Cat
c
c
c
e(t)x
Cta(t)xln
dta(t)xln
+=+=
= Take the integral of both sides
at
2c
c
2
Catc
ek(t)x
eklet
ee(t)x
=
==
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t/
21 ekkx(t)+=
Hence : let = 1/a time constant
Time constant : a parameter that determines the rate of decrease of
x(t)
Lets find the solution of RC & RL circuits :
at21
cP
ekkx(t)
(t)x(t)xx(t)
+=
+=
Hence, a general form of the solution is:
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Example :
+
- ic(t)RVS
C
S1
CRt
21
at
21c
S
c
c
VCR
1
CRVs
a
bk
ekkekk(t)v
CR
V(t)v
CR
1
dt
(t)dv
===
+=+=
=+
Assume vc(0) = v0
To find k2 , we need the initial condition of vc(t)
For example , if we know vc(0) = V0
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S02
2S0
CRt
2Sc
VVk(1)kVV
ekV(0)v
=+=
+=
CRt
S0S
CR
t
21c
e)V(VV
ekk(t)v
+=
+=
As a special case , lets consider the natural response
Natural response :
Circuit response when no source is affecting the response
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Case 1 : Natural RC response
ic(t) R
C
0(t)vCR
1
dt
(t)dv
0Rdt
(t)dv
C(t)v
0R(t)i(t)v
cc
c
c
cc
=+
=+
=+Assume vc(0)=V0
First order linear ordinary homogenous differential equation :
CRt
2c ek(t)v
=
We can find k2 from initial conditions
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CRt
0c
02
2CR
t
20c
0c
eV(t)v
Vk
kekV(0)v
thenV(0)vAssume
=
====
=vc(t)
t
V0
RCt
21 ekkx(t)+=
Note : the general solution of the forced response is
RCt
0c
0S02
S1S
eV(t)V
V)V(Vkand
0Vk0Vsince
=
==
===
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Case 2 : Natural response of RL circuit
iL(t)
R
L
0(t)iL
R
dt
(t)di
0dt
(t)diL(t)iR
L
L
LL
=+
=+
Assume :iL(0) = i0
KVL :
This is a first order linear ordinary homogenous differential equation
tL
R
2L ek(t)i
=
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To find k2, we need initial condition iL(0)
tL
R
0L
02
2(0)L
R
20L
ei(t)i
ik
keki(0)i
=
==== iL(t)
t
i0
Note :
In the forced response , we have
RVk
ekk(t)i
S1
tL
R
21L
=
+=
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Since VS here is 0 then k1 = 0
tL
R
0L
S0S
02
ei(t)i
0)V(sinceiR
V
ik
=
==
=
Analysis Techniques :
1. The differential equation approach
Here , a differential equation that describe the behavior of the
circuit is used.This first order differential equation is expressed in tems of the
voltage across the capacitor or current through the inductor.
Then the solution of this differential equation is obtained
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Example :
Find iL(t) , t 0 ?
First, find the initial condition iL(0- )
At t = 0-
, the inductor behaves as a short circuit .
2 H20 A 1.0
40iL (t)
2
10
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Hence , we assume that the current through the inductor doesnt
change instantaneously
iL(0- ) = iL(0) = iL(0
+) = 20 A
20 A 1.040
iL (0-)
= 20 A
2
10
iL(0-
) = 20 A
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At t = 0 the switch is open
Where Req = (40 // 10) + 2 = 400/5 + 2 =10
L= 2 H iL (t) 10
Req
=+
-
2 H 40iL (0-)= 20 A
10
2
OR
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KVL around the loop :
0(t)i5dt
(t)di
0(t)i10dt
(t)di2
0(t)iRdt
(t)diL
0v(t)v
LL
LL
LeqL
eqL
=+
=+
=+
=+
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We know that
5t
2L
t2
at2L
ek(t)i5
1
a
15awhere
ekek(t)i
=
===
==
We can find k2 from the initial condition
iL(0 ) = 20 = k2 e0 = k2
iL(t) = 20 e-5t A
iL(t)
t
20
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Example :
Find vc(t) , t 0 ?
The switch has been closed for long time .
The capacitor behave as open circuit .
7.5 m A
k80k50
k20
F0.4
t = 0
+
-
vc (t)
7.5 m Ak80 k50
k20
vc (0-)
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
At t = 0 , the switch is open
KVL around the loop :
0(t)v50dt
(t)dv
F0.4C,0
dt
(t)dvCk50(t)v
0(t)ik50(t)v
cc
cc
cc
=+
==
+
=+
( )
( ) V200k150
k80Am7.5k50
k70k80
k80Am7.5k50)(0v(0)v)(0v ccc
=
=
+=== +
k50+
-
vc(t)F0.4
Vc(t) = 200 e-50t V.
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Example :
Find vc(t) , t 0 ?
For t < 0 , the capacitor behave as open circuit .
V40)(0v(0)v)(0v ccc ===+
At t = 0 , the switch is moved
k60
k601
k20
F0.25
+
-
vc (t)
+
-+
-
40 V
75 V
k8 k40t = 0
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k601
+
-
vc (t) +-
75 V
k40k8
k601+
-
vc (t)
k40k8
1.875 m A
Source transformation
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+
-vc (t) +
-
60 VF0.25
32 k
8 k
+
-
vc (t)
k23
k8
1.875 m A
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KVL around the loop :
6000-b,100a
6000(t)v100dt
(t)dv
0(t)v
dt
(t)dvCk4060
0(t)v(t)ik4060
cc
cc
cc
==
=+
=++
=++
601006000
abk
ekk(t)v
1
at
21c
===
+=
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Ve10060(t)v
100k40k60
40kk(0)v
t100
c
22
21c
+=
==+
=+=
vc(t)
t
30
- 60
To find k2 , we use initial condition:
40
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Case 2 : Step by step approach
1. Assume the solution is x(t) = k1 + k2 e-t/2. Assume that the circuit is in steady state before the switch moves
replace a capacitor by open circuit
replace a inductor by short circuitThen find vc(0
-) or iL(0
-)
3. The switch is now in the new location :
Replace the capacitor by a voltage source = vc(0
-
) Replace the inductor by a current source = iL(0
-)
And solve for x(0)
4. Assume t = , find x (t = )replace capacitor by open circuit and inductor by short circuit
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5. Find the time constant :
HOW ??
- Find the Thevenin equivalent resistance w.r.t the terminals of
the capacitor or inductor.
= RTH C or = L/RTH
6. Find the constants :
k1 = x()
k1+k2 = x(0) k2 = x(0)-x()
x(t) = x()+[x(0) - x()] e-t/
x(t) = final value + [ initial value final value ] et/
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Example :
[ ] t
000
t
210
e)(V(0)V)(V
ekk(t)v(1)
+=
+=
Find V0 (t) ?
F2
+
-
v0 (t)+-
12 V 2
+
-
8 V
2
2
1
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(1)20i2-i5
08)i-(i2i312-
21
211
LL=
=++
(2) Assume steady state , replace capacitor by open circuit .
Mesh
V812412(1)i)(0v
A0i,A4i
(2)8i4i2
0)i(i28i2
1c
21
21
122
=+=+=
===+
=++
LL
v0 (0-)12 V
2
+
-8 V
2
2
1
+
-
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3. The switch is moved now t = 0 ,
replace the capacitor by a voltage source = vc(0-) and solve
for V0(0)
V4
2
18
22
2)(0v(0)V c0 =
=
+
=
vc (0-)
= 10 V
12 V
2
21
+
- +
-
v0 (t)
+
-
=8V
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4. At t = replace capacitor by open circuit
V5
24
5
212
122
212)(V0 =
=
++=
12 V
2
21
+
-
v0 (t)
+
-)(vc
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5. Find the time constant
find the Thevenin equivalent resistance w.r.t x ,y
RTH = 1 // (2+2) = 1 // 4 = 4/5
12 V2
21
+
- x
y
( )5
8F2
5
4CR TH =
==
t
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[ ]
t8
5
0
t8
5
t
0000
e5
4
5
24
(t)v
e5244
524
e)(v(0)v)(v(t)V6.
=
+=
+=
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Example :
[ ] t
000
t
210
e)(v)(0v)(v
ekk(t)v:Step(1)+
+=
+=
+
-
v0
(t)
24 V
Vx4
4
F2+
-
2 Vx
t = 0
3 A
+-
Step (2) : assume steady-state , replace capacitor by open circuit
Find v0
(0-) ?
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V60)(0v
V60243624(12)3
24v3v24v2)(0v
V12(3)4v
0
xxx0
x
==+=+=
+=++=
==
+
-
v0 (0-)
24 V
Vx4
4
+
-
2 Vx
3 A
+-
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Step (3) : now switch is moved .Find v0(0)
(t)vx(t)hence
V60)(0v(0)v)(0v
c
000
==== +
24 V
Vx4
4
+
-
2 Vx
+-
( )0v
Step (4) : assume t = Find V0() ?
V24)(V
24
v24v2)(V
0Vx
0
xx0
=
=
++=
=
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Step (5) : Find the time constantfind the RTH w.r.t the terminals of capacitor
Voc = 24 V
24 V
Vx4
+
-
2 Vx+
-
+
-
voc4
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A4
2
8
2
VI
V8V
24V3
0V24V2
xsc
x
x
xx
===
=
=
=++
24 V
Vx4
+
-
2 Vx+-
4 I SC
24 V
Vx2
+
-
2 Vx+-
I SC
Now, find Isc
sec.12F)(2)(6CR
64
24
I
VR
TH
SC
OCTH
===
===
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[ ]
[ ]
Ve3624(t)V
e246024(t)V
e)(V(0)V)(V(t)V:Step(6)
12
t
0
12
t
0
t
0000
+=
+=
+=
t
210 ekk(t)i
+=
Example :
Find i0(t) , t > 0 ?
Step (1) :
12 V
k4
k2
F200
t = 0
+-
i0(t)
k2
k2
k2
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Step 2 : assume steady state ( for t < 0 ) replace capacitor by
open circuit .
V4)(0v(0)v)(0v
V4k6
2k)12()(0v
ccc
c
===
==
+
12 V
k2
+-
i0(t)
k2
k2
k2
vC(0-) -+
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Step 3 : now switch is moved , replace capacitor by voltage
source = vc(0) ,
Now find i0(0)
12 V
k2
+
-
i0(0+)
k2
k2
k2
+-
4 V
k4
i 2
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Am2.66
k3
k1Am8(0)i0 =
=
i0
k2k1
8 m A
i0(0+)
k2k2
6 m A2 m A
k2
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Step 4 : assume t = , find i0() . Steady state
Replace capacitor by open circuit
12 V
k2
+-
k2
k2
k2
)(i0
k4
12 V
+-
k2k2
)(i0
12
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k2k2
k2
k2k4
RTH
Am3k4
12)(i0 ==
Step 5 : find time constant .First find RTH at terminals of the capacitor
( )
( ) ( ) sec0.6F200k3CRk3
k2k2//k4//k4R
TH
TH
===
=
+=
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Step 6 : find the solution i0(t)
[ ]
Ame0.333(t)i
Ame3)(2.663
e)(i(0)i)(i(t)i
0.6t
0
0.6t
t
0000
=
+=
+=
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Example:
k21
F50
k4 k8k3
k21
V12k4
+-t = 0
i0 (t)
Find io(t) , using Step by step approach.
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Step 1 : assume i0
(t) = k1
+ k2
e t/
Step 2 : assume t < 0 ( steady state)
Replace capacitor by open circuit and find voc(0-)
As we have done before , vc(0
-
) = vc(0)= vc(0+
) = - 4 V.
Step 3 : now the switch is moved
replace the capacitor by voltage source of value -4 and find i0(0)
k4 k8k21
k4
i0 (0+)
i
k3
4-+-+-
k21
244
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( )
Am2
1
16
12
3
2
k4k12
k12
3
2(0)i
Am3
2
k6
4
k12//k4k3
4i(0)
0
=
=
+
=
=
=
+
=
Step 4 : assume t = , steady statereplace capacitor by open circuit and find i0()
i0(
) = 0
k12k3 k4
)(i0
St 5 fi d
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Step 5 : find [ ]
( ) ( ) sec0.350k6CR
k6k3k4//k12R
TH
TH
===
=+=
[ ]
0tAme
2
1(t)i
Ame2
1
0
e)(i(0)i)(i(t)i
0.3t
0
0.3t
t
0000
=
=
+=
Step 6 :k12k3 k4
RTH
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Example :
Step 1 : assume i0(t) = k1 + k2 e -t/
Step 2 : assume t < 0 ( steady state )
Replace inductor by short circuit and find
iL(0-) = i0(0
-)
k4 k2
t = 0
i0 (t)
Am10
k4
k2k5
Hm10
find i0(t) using step by step approach ?
k5
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Am5k2k2
k2Am10)(0i0 =
+=
k4 k2i0 (0)
Am10
k4
k2
Am10
k2
k2k5
Step 3 : no the s itch is mo ed t 0 replace ind ctor b a
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Step 3 : now the switch is moved t = 0 replace inductor by a
current source of value = 5 mA.
Since it is inductor i0(0-) = i0(0)
Step 4 : assume t = ( steady state )
Replace inductor by short circuit and find i0()
k4
k2i0 (t)
Am10
k4
k2k5
k2
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Am5k2k2
k2Am10)(i0 =
+=
Step 5 : find where = L / RTHLets find RTH at the terminals of the inductor
RTH = (4 k // 4 k) + 2 k
= 4 k = L / RTH = 10m/4k = 2.5 sec
i0 (t)
Am10
k4
k2k5
k4
k2k5
k4 RTH
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
[ ]
[ ]Am5(t)i
Aem5m5m5
e)(i(0)i)(i(t)i
0
s2.5t
t
0000
=
+=
+=
Step 6 : find i0(t)
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Find i0(t) using step by step approach ?
Step 1 : assume i0(t) = k1 + k2 et /
Step 2 : assume t < 0 ( steady state )
Replace inductor by short circuit and find iL (0-)
4
86
6
21
4
+
-
t = 0
24 V
1 H
i L(t)
i 0(t)
46 4
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
iL(0-) = 24 / 6 = 4 A
i0(0-) = 0 A
Step 3 : switch is movedReplace inductor by current source of value (4 A) and find i0(0)
4
86
6
21
4
+
-
24 V
i L(0-)
i 0(0-)
4
4
6
+
-
24 V
i L(0-)
i 0(0-)
44
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4
86i L(0
-)
i 0(0+)
21 4 A
4
86
4
i 0(0+)
21
+
-
48 V
4
16
i 0(0+)
48 V+
-i0(0) = - 48 / 20i0(0) = -2.4 A
St 4 t ( t d t t )
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Step 4 : assume t = ( steady state)
Replace inductor by short circuit and find i0()
i0() = 0 A
Step 5 : find , = L / RTHSo find RTH across the terminal of the inductor
4
86
4
21
RTH
( )[ ]{ } 4 812//46//84R =++=
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Step 6 : find i0(t)
[ ]
( )
0tfor,0(t)i
Ae2.4(t)i
0t,Ae02.40
Ae)(i(0)i)(i(t)i
0
t4.8
0
t4.8
t
0000
=
=
+=
+=
i0(t)
t
- 2.4
( )[ ]{ }
sec4.8
1
R
L
4.812//46//84R
TH
TH
==
=++=
Example :
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
Example :
Solve previous example using differential equation approach
For t < 0 , steady stateReplace the inductor by short circuit and find iL(0
-) and i0(0
-)
As before iL(0-) = 4 A = iL(0
+
)i0(0
-) = 0 A
For t > 0 ,
4
86
4
i 0(t)
i L(t)
L = 1 H21
4
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(t)i5
3(t)i
(t)i
20
12
k8k12
k12(t)i(t)i
L0
LL0
=
=
+=
4
4
i 0(t)
i L(t)
21
So we need to find iL(t) first
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UAE University Department of Electrical Engineering Dr.Hazem N.Nounou
KVL :
0(t)i4.8dt
(t)di
0(t)i4.8dt
(t)diL
0(4.8)(t)i(t)V
LL
LL
LL
=+
=+
=+
218
i 0(t)
i L(t)4.8i L(t)
L
0k( )i t4 8
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( )
0t,0(t)i
0t,e2.4(t)i
e
5
12(t)i
e453(t)i
53(t)i
0
t4.8
0
t4.8
0
t4.8L0
=
=
=
==
i0(t)
t
- 2.4
0t,e4(t)i
k4(0)i
0t,ek(t)i
t4.8
L
L
t4.8
L
=
==
=