12/10/2019
1
L20Chapter 8 Secondβorder circuits
A circuit with two energy storage elements:One inductor, one capacitorTwo capacitors, orTwo inductors Such a circuit will be described with
secondβorder differential equation:
2202
0 1
2 ;
(0) ; (0)
d v dvv b
dt dtdv
v V Vdt
Equivalently:
+
+
C v
L
iL
We first learn how to solve 2ndβorder diff equ.
π£ 2πΌπ£ π π£ π;π£ 0 π , π£ 0 π
L20Solving 2ndβorder differential equations
Consider a secondβorder differential equation
with initial condition: π£ 0 π0,
Need to find v(t) for all t.
In chapter 8, b=0 corresponds to source free case, bβ 0 for step response.
Let π£ β = b/02
Let the roots to s2+2s+02= 0 (not =b) be s1, s2,
2 21 2 0,s s
For example: 2 3 2 0 s s ( 1)( 2) 0 s s 1 21; 2s s 2 4 5 0 s s ( 2 )( 2 ) 0 s j s j 1 2, 2s s j
2 10 25 0 s s 2( 5) 0 s 1 2 5s s
The solution will be constructed using the roots.
Three cases: Case 1: > 0, two distinct real roots
Case 3: < 0, two distinct complex roots
Case 2: = 0, two identical real roots
π 1
π£ 2πΌπ£ π π£ π π£ 0 π
(Assume πΌ 0. Then as π‘ β β, π£ π‘ goes to a constant . )
1
2
12/10/2019
2
L20
Case 1: > 0, two distinct real roots for s2+2s+02 = 0
2 21 2 0,s s
The solution to (1) is not unique. For any real numbers A1, A2, the following function satisfies (1),
This is called the general solution. To verify,
20 2
0
b
b v(t) satisfies (1) for any A1, and A2
The solution will be uniquely determined by the initial conditions (IC)
0 0
π£ π‘ π π΄ π π π΄ π π£ π‘ π π΄ π π π΄ π
π£ π‘ π£ β π΄ π π΄ π Where π£ β π/π
π π΄ π π π΄ π 2πΌ π π΄ π π π΄ π π π£ β π΄ π π΄ π
π΄ π π 2πΌπ π π΄ π π 2πΌπ π π π£ β
π π£ β
π£ 2πΌπ£ π π£ π; 1π£ 0 π , π£ 0 π (IC)
π£ 2πΌπ£ π π£
L20
The general solution
There is only one pair of A1, A2 that satisfy the IC
To find A1, A2, use IC to form two equations.
Evaluate π£(t) and π£(t) at t=0:
To satisfy the IC, we obtain
You may use other methods to solve for A1 and A2.
Recall:
π£ π‘ π£ β π΄ π π΄ π Where π£ β π/π
π£ 0 π£ β π΄ π΄
π£ 0 π π΄ π π΄
π£ β π΄ π΄ ππ π΄ π π΄ π
1 1π π
π΄π΄
π π£ βπ
π΄π΄
1 1π π
π π£ βπ
π£ 2πΌπ£ π π£ π; 1π£ 0 π , π£ 0 π (IC)
π£ π‘ π π΄ π π π΄ π
3
4
12/10/2019
3
L20
Example: Solve
202 4; 3; 6b 02; 3;
0; Case 1, two distinct real roots for π 4π 3 0.
21 2, 2 2 3 2 1,s s 1 21, 3s s
Step 1: Find s1, s2, π£ β
Step 2: Find A1,A2, use initial condition
The general solution:3
1 2( ) 2 t tv t A e A e
Solve (*) and (**) to obtain A1=β2; A2=1
Finally, 3( ) 2 2 t tv t e e
2 21 2 0,s s
Note:π , π are roots to:π 4π 3 0π 3 π 1 0π 1, π 3
Note:π£ π‘ π΄ π 3π΄ π
π£ π‘ π£ β π΄ π π΄ π
π 2πΌπ π 0.
π£ β π/π =6/3 =2
π£ 2πΌπ£ π π£ π (To compare )
π£ 4π£ 3π£ 6, π£ 0 1; π£ 0 1
π£ 0 π΄ 3π΄ 1 ββ π£ 0 2 π΄ π΄ 1 β
L20
An Example:
The general solution:
31 2( ) 2 t tv t A e A e
π£ π‘ π£ β π΄ π π΄ π , π£ βππ
Since π , π are negative, limβ
π£ π‘ 2 π£ β
Another way to see π£ β π/π
π£ π‘ π΄ π π π΄ π π ,
π£ π‘ π΄ π π π΄ π π ,
limβ
π£ π‘ 0
limβ
π£ π‘ 0
From 1 , limβ
π£ π‘ 2πΌπ£ π‘ π π£ π‘ π π£ β π
π£ βππ
π£ 2πΌπ£ π π£ π; 1π£ 0 π , π£ 0 π (IC)
π£ 4π£ 3π£ 6, π£ 0 1; π£ 0 1
5
6
12/10/2019
4
L20Case 2: = 0
Two identical roots, s1=s2= β to s2+2s+02=0
The general solution:
Its derivative:
Use IC to find A1, A2.
At t = 0,
π 2πΌπ πΌ π πΌ
π£ π‘ π£ β π΄ π΄ π‘ π π£ β π/π
π£ 0 π£ β π΄ π π΄ π π£ βπ΄ π πΌπ΄
π£ π‘ π΄ π πΌ π΄ π΄ π‘ π
π£ 0 π΄ πΌπ΄ π
L20Case 3: < 0
Two complex roots to s2+2s+02=0
2 20Let d
The general solution:
Its derivative:
Use IC to find B1, B2.
π 1
π , π πΌ πΌ π πΌ 1 π πΌ πΌ π π
π , π πΌ π π
π£ π‘ π£ β π π΅ cosπ π‘ π΅ sinπ π‘ π£ β π/π
π£ 0 π£ β π΅ π π΅ π π£ β π΅ π πΌπ΅ /π
π£ π‘ πΌπ π΅ cosπ π‘ π΅ sinπ π‘ π π π΅ sinπ π‘ π π΅ cosπ π‘
π£ 0 πΌπ΅ π π΅ π
7
8
12/10/2019
5
L20
0 5 10 15 20-5
0
5
10
π£ β
2
d
Case 3: < 0
π£ π‘ π£ β π π΅ cosπ π‘ π΅ sinπ π‘
π£ β π΄ π
π£ β π΄ π
Example: Solve π£ 8π£ 25π£ 200, π£ 0 0, π£ 0 12
Solution:
Compare with π£ 2πΌπ£ π π£ π
πΌ ?π ? , π ? π€βππβ πππ π?
πΌ 4,π 5, π 200, πΌ π
Step 1: Form general solution using roots and π£ β .
π π πΌ 25 16 3
Complex roots: πΌ ππ 4 π3
π£ β ? π£ βπ
π
20025
8
Imaginary part of the roots:
π£ π‘ π£ β π π΅ cosπ π‘ π΅ sinπ π‘ , π£ β ? π ?
π£ π‘ 8 π π΅ cos 3π‘ π΅ sin 3π‘ , General solution
Step 2: Find π΅ ,π΅ , using initial conditions, π£ 0 0, π£ 0 12
π£ π‘ 4π π΅ cos 3π‘ π΅ sin 3π‘ π 3π΅ sin 3π‘ 3π΅ cos 3π‘
π£ 0 4π΅ 3π΅ 0
π£ 0 8 π΅ 12 π΅ 4;π΅163
Final solution: π£ π‘ 8 π 4cos3t sin 3π‘ π
case 3, two complex roots : πΌ ππ
π£ β 8π£ β 25π£ β 2000 0
9
10
12/10/2019
6
R20
Practice 2: Solve
Practice 3: Solve
Practice 1: Solve
π£ 5 π£ 6π£ 18, π£ 0 0; π£ 0 4
π£ 4 π£ 4π£ 8, π£ 0 1; π£ 0 1
π£ 4 π£ 13π£ 39, π£ 0 1; π£ 0 1
L21Secondβorder RLC circuits
2202
0 1
2 ;
(0) ; (0)
c cc
cc
d v dvv b
dt dtdv
v V Vdt
+
+
C v
L
iL
General circuit:
Example : Find π π‘ , π£ π‘ for t > 0.
cv
0.5H
+
6
30V
π
t = 0
66
+ 10V
As always, π π‘ , π£ π‘ are continuous function of time.Use the circuit before switch to find π 0 , π£ 0 .
For t>0, π£ π‘ satisfies We also need 0 .
Generally, π‘ is not continuous
at t=0.
How to find 0 ?
11
12
12/10/2019
7
L21
To solve a second order circuit, we need to1. Derive the differential equation:
Converting a circuit problem to a math problem
2. Find the initial condition
Β§ 8.2. Finding initial values We need to find
(0 ) (0 )(0 ), (0 ), ,C Lc L
dv div i
dt dt
Key points:
1. Capacitor voltage and inductor current donβt jump
(0 ) (0 ), (0 ) (0 )C C L Lv v i i
Use circuit before switch (t<0) to find these two.
(0 ) (0 )2. For ,C Ldv di
dt dt
Consider the circuit right after switch, i.e., at t = 0+
β’ Apply KVL to find π£πΏ(0+), then β’ Apply KCL to find ππΆ(0+), then
(0 ) 1 (0 ),C
C
dvi
dt C
(0 ) 1 (0 ),L
L
div
dt L
π£ 0 π , π£ 0 π π£ 2πΌπ£ π π£ π;
π£ πΏ , π πΆRecall:
L21Example: Find
+
4
2
12V
0.25H
0.1FCv
ππΏ
(0 ) (0 )(0 ), (0 ), ,C L
C L
dv div i
dt dt
(0 ), (0 ),C Lv i Use circuit before switch to find At t = 0
t=0
+
4
212V Cv
ππΏ
To find ,
Consider t = 0+
+
4
12V
0.25H
0.1FCv
ππΏ
Lv
ππΆ
12 4 2 4 0V
By continuity, π 0 2π΄, π£ 0 4π
π 2π΄; π 4π
π 0 2π΄; π£ 0 2 2 4π
π£ 0
π 0 π 0 2A
ππ 0ππ‘
1πΏπ£ 0 0
ππ£ 0ππ‘
1πΆπ 0
20.1
20 π/π ,
π£
13
14
12/10/2019
8
L21Chapter 8 circuits
1). Source free series RLC circuits2). Source free parallel RLC circuits3). Step response of series RLC circuits4). Step response of parallel RLC circuits5). Other second order circuits
We will focus on 1) and 3). We study 3) first. 1) is a special case of 3).
L21Β§ 8.5 Step response of series RLC circuits
A general circuit
+
+
C v
L
ππΏ
Before switch (t<0), L and C may not be in series.
Determine π£ 0 , π 0from circuit before switch by considering πΏ as short,πΆ as open.
After switch ( t>0 ), L and C in series
+
+
C v
+
L R0
ππ By Theveninβs theorem, the two terminal circuit can be replaced by a voltage source and a resistor
15
16
12/10/2019
9
L21By Theveninβs theorem, the two terminal circuit can be replaced by a voltage source and a resistor
Rearranging the elements does not changeloop current π and capacitor π£
ππ‘β ππ‘ββ
πππ π‘β π π‘ββ
π 0
0
1
(0 ) ,
(0 )
v V
dvV
dt
Given
C v
+
L R0
ππ
+
π π‘ββ
ππ‘ββ
π
C v
L+
π π‘β
ππ‘β
π
We call (ππ‘β,π π‘β)Theveninβs equivalentw.r.t LC
π£ π£
L21
0
1
(0 ) ,
(0 )
v V
dvV
dt
Given C v
L+
Rth
Vth
π
Choose π£ as key variable. We derive diff. equation for π£Express other variables in terms of π£:
;dv
i Cdt
By KVL,
R L thv v v V 2
2 ;th th
dv d vR C LC v V
dt dt
Normalize:2
2
1;th thd v R dv V
vdt L dt LC LC
(Vth, Rth) :Theveninβs equivalent w.r.t LC
The equivalent circuit:
π£ π π π πΆππ£ππ‘
; π£ πΏππππ‘ πΏπΆ
π π£ππ‘
π£ π£
17
18
12/10/2019
10
L21
C v
L+
Rth
Vth
0 1
(0 )(0 ) ,
dvv V V
dt
π
2
2 ;1 ththd v dvv
dt d
VR
L t LCLC
Step response of series RLC circuit:
22
02 2 ;d v dv
bvdt dt
As compared with general form
0
1, , ,
2th thR V
bL LCLC
The solution:
Case 1: 2 20 1 2 0; ,s s 1 2
1 2( ) ( ) s t s tv t v A e A e
Case 2: 0 1 2; s s 1 2( ) ( ) ( ) tv t v A A t e
Case 3: 2 2
0 0 1 2; , ,d ds s j
1 2( ) ( ) ( cos sin )td dv t v e B t B t
A1,A2,or B1,B2 will be determined by initial conditions
If Vth= 0, source free RLC, A special case
π£ βππ
π
L21
Case 2: 0 1 2; s s 1 2( ) ( ) ( ) tv t v A A t e
Case 3: 2 20 0 1 2; , ,d ds s j
1 2( ) ( ) ( cos sin )td dv t v e B t B t
Case 1: 2 2
0 1 2 0; ,s s 1 21 2( ) ( ) s t s tv t v A e A e
Find A1,A2 from
Find A1,A2 from
Find B1,B2 from π£ 0 π£ β π΅ππ£ 0ππ‘
πΌπ΅ π π΅
π£ 0 π£ β π΄ππ£ 0ππ‘
πΌπ΄ π΄
π£ 0 π£ β π΄ π΄ππ£ 0ππ‘
π π΄ π π΄
19
20
12/10/2019
11
L21Key points for step resp. of series RLC circuits:
Obtain π£ 0 , ππΏ 0 from circuit before switch
+
+
C v
L
ππΏ
For circuit after switch, obtain Theveninβsequivalent (Vth,Rth) w.r.t LC
Assign ππΆ by passive sign convention
(0 ) (0)C Li i Depending on how ππΏ is assigned. Then
2
2
1;th thd v R dv V
vdt L dt LC LC
0
1, , ( )
2th
th
Rv V
L LC (0 ) (0 )Cdv i
dt C
Then follow straightforward math computation on previous slide
Key parameters to obtain: β’ π£ 0 , ππΏ 0 from circuit before switchβ’ ππ‘β ,π π‘β from circuit after switch.
C v
L
+
RthVth
ππΆ ππΏ
ππΏ
L21Example 1: Find v(t) for t > 0.
0.25F v
1H+
5
24V
ππΆ
ππΏt = 0
1
Step 1: find v(0), ππΏ(0) from circuit before switch ππΏ
v
+
5
24V
1
ππΏ 0 =24/6=4A
π£ 0 = 4V
Step 2: Find Vth, Rth, w.r.t LC from circuit after switch
Vth=24V, Rth=5 v(β) = 24V
Since ππΆ ππΏ, .16π/π
v
+
5
24V
ππΆππΏ
0.25F
1H
21
22
12/10/2019
12
L21Step 3: Math computation
The general solution:
2 21 2 0, 2.5 6.25 4 2.5 1.5s s
41 2( ) 24 t tv t A e A e
1 21, 4s s
0
5 1 12.5, 2
2 2 0.25thR
L LC 0 , Case 1
Find A1, A2 from initial condition: v(0) = 4V; dv(0+)/dt=16
π£ 0 24 π΄1 π΄2 4
π΄ 4π΄ =16π΄1 64/3, π΄2 4/3
Final solution: 464 4( ) 24
3 3t tv t e e V
Note: v t v β π΄ π π΄ π
R21Practice 4: Find
3u(t)A
2
20V
0.5F
0.6H
Cv
iL
(0 ) (0 )(0 ), (0 ), ,C L
C L
dv div i
dt dt
+
4
23
24
12/10/2019
13
R21Practice 5: Find v(t) for t > 0.
0.02F v
0.5H
+
4
30V
iC
t = 0 6
4
R21Practice 6: The switch is at position a for a long time beforeswinging to position b at t = 0. Find iL(t), vC(t) for t > 0.
16mFCv
0.5H
+
10
60V
iL
t = 0 10
a b
25
26
12/10/2019
14
L22Example 2: Find i(t), v(t) for t > 0.
1/8Fv
0.5H
+
6
30V
it = 0
66
+ 10V
Step 1: find v(0), i(0) from circuit before switch
A simple circuit in Chapter 2.
(30 10) 20(0) 5
6 //(6 6) 4i A
v
+
6
30V
i
66
+ 10V
6v
By voltage division,
6
6(30 10) 10
6 6v V
By KVL, v(0)=20V
20π π£
L22
1/8F
v
0.5H
6
i
66
+ 10V
Step 2: After switch
Rth=(6+6)//6=4, Vth=10V = v(β)
iC
Since ππΆ π, ππΆ 0 5π΄
5/ 1/8 40π/π
0
4 1 14; 4
2 2 0.5 0.5/8thR
L LC
Step 3: math
Case 2.
General solution:
Find A1,A2 using v(0), :
1
1 2
(0) 10 20
(0 )4 40
v A
dvA A
dt
A1=10, A2=0
Final solution:
Find inductor current,
π π‘ = β (1/8)*10(β4)eβ4t = 5 eβ4tA
π 0 5π΄,ππ£ 0ππ‘
?
Rth=?, Vth=?
πΌ ?π ? Which case?
v(t) = 10+(A1+A2t)eβ4t
π£ π‘ = 10 + 10 eβ4tV
π π‘ π π‘ πΆππ£ππ‘
π 0 5π΄π£ 0 20π
27
28
12/10/2019
15
L22
+
+
Cv
L
For parallel RLC circuit
iL
By Nortonβs Theorem C
v
L
iL
RIN
iR iC
Choose iL as key variable. Derive diff equ. with KCL
2
2, ,L L L
R C
di L di d iv L i i LC
dt R dt dt
R C L Ni i i I 2
2L L
L N
L di d iLC i I
R dt dt
Normalize: 2
2
1 1L L NL
d i di Ii
dt RC dt LC LC 0
1 1, ,
2NIb
RC LCLC
Same math problem.
No parallel RLC circuits in homework or Final Exam.
L22Example 3: Find vC(t) for t > 0.
+
6
3u(t)A
6
16mF
0.5H 12
12V
Cv
0, t < 0( )
1, t > 0u t
Recall:
Thus for t<0, 3u(t)A=0A
The current source is turned off.
Replace it with open circuit for t<0
Step 1: Find π£ 0 , π 0 from circuit for t<0
Key parameters to obtain: β’ v (0), iL(0) from circuit before switch (t<0)β’ Vth, Rth w.r.t. LC from circuit after switch (t>0) .
29
30
12/10/2019
16
L22Step 1: Find π£ 0 , π 0 from circuit for t<0
For t < 0, 3u(t)A=0, current source open, C open , L short
π£ 0 π£12
6 1212 8π,
+
6
6
16mF
12
12V
Cv
π 0 Cv
Since π =0, 12V, 6Ξ© πππ 12Ξ© in the outer loop are in series
By voltage division:
π 0 0, ππ£ 0ππ‘
0
L22
+
6
3A
6
16mF
0.5H 12
12V
Cv
Step 2: For t>0, 3u(t)A=3A. Need to find π π£ β ,π w.r.t LC
31
32
12/10/2019
17
L22
+
6
3A
6
16mF
0.5H12
12V
Cv
Step 2: Fot t>0, 3u(t)A=3A. Need to find π π£ β ,π w.r.t LC
For π£ β+
6
3A
6
12
12V
( )Cv
Rv
6v
By KVL, π£ β π£ π£
π£ 3 6 18π, by ohms law
π1218
π΄23π΄,
π
π π
π3A
π£ 12π 1223
8π
π£ β 18 8 26π
For π , turn off 12V with short, turn off 3A with open
66
12π
R 6 6//12=10Ξ©
Step 3: Math
L22Step 2: Another way to look at the circuit for t>0
+
6
3A
6
16mF
0.5H12
12V
Cv
To see connection better, move 12 Ξ© to the left
π 6 12//6 10Ξ©Under DC condition,
π£ 8π, π£ 18π,π£ β π 8 18 26π
6+
6
3A
16mF
0.5H12
12V
Cv
π£
Rv
33
34
12/10/2019
18
π 10Ξ©, π£ β π 8 18 26π, π£ 0 8π, 0
Step 3: math From step 1, step 2, found
πΌπ 2πΏ
102 0.5
10; π1
πΏπΆ
10.5 0.016
125 ,
v t 26 e B cos 5t B sin 5t
The general solution:
To satisfy initial condition: π£ 0 26 π΅ 8π£ 0 10π΅ 5π΅ 0Solving equations to get π΅ 18;π΅ 36Step response: v t 26 e 18 cos 5π‘ 36 sin 5π‘ π
Also note: πΏ 0.5π»; πΆ 16ππΉ 0.016πΉ πΌ ? ,π ? π€βππβ πππ π?
πΌπ 2πΏ
; π1
πΏπΆ ;
πΌ π , πππ π 3, π 125 100 5
π£ 0 π£ β π΅ππ£ 0ππ‘
πΌπ΅ π π΅
R22Practice 7: Find iL(t), vC(t), vR(t) for t > 0.
20mF
1H
+
4
60V
iLt = 0
6
2A4
2
Cv
Rv
35
36
12/10/2019
19
R22Practice 8: Find vC(t) and vR(t) for t > 0.
250mF
2H
2u(βt)A
Cv
Rv
2
43A
4
Chapter 7, Chapter 8 Review
Final exam: 12/16/19 (Monday), 8β11am, BL210
One page oneβsided note allowed, Calculators allowedDonβt include the solution to any circuit problem in the note!
37
38
12/10/2019
20
Step response of general RC circuit
i
+
+
v
C
One or more switches changesstructure of circuit at t = 0.
th
tR C v(t) = v( ) + (v(0) v( )) e
L23
πΉππ π‘ 0, step response:
Three key parameters:
1. Initial condition v(0)=V0, Established from circuit before switch. Obtained by solving a DC circuit, t < 0 (capacitor = open circuit)
2. Final condition v(βEstablished from circuit after switch. Obtained by solving another DC circuit, t > 0 (capacitor = open circuit)
3. Equivalent resistance Rth, or Reqwith respect to capacitor for circuit after switch
Since v(β) = Vth, (Vth, Rth) can be together considered as the Theveninβsequivalent, with respect to the capacitor, from circuit after switch.
Step response of RL circuits
i
+
+
L
Step response β For t > 0,
Three key parameters:
1. i(0), initial condition, from the DC circuit for t < 0. (inductor =short)2. i(β)=IN, final value, from DC circuit for t > 0. (inductor =short)3. RN=Rth, equivalent resistance w.r.t inductor from circuit for t > 0
(IN, RN) together as Nortonβs equivalent, w.r.t inductor, for t > 0.
NRt
Li(t) = i( )+ (i(0) i( ))e
L23
39
40
12/10/2019
21
L23Key points for step resp. of series RLC circuits:
Obtain v(0), iL(0) from circuit before switch
+
+
C v
L
iL
For circuit after switch, obtain Theveninβsequivalent (Vth,Rth) w.r.t LC
Assign iC by passive sign convention
(0 ) (0)C Li i Depending on how iL is assigned. Then
2
2
1;th thd v R dv V
vdt L dt LC LC
0
1, , ( )
2th
th
Rv V
L LC (0 ) (0 )Cdv i
dt C
Then follow straightforward math computation on next slide
Key parameters to obtain: β’ v(0), iL(0) from circuit before switchβ’ Vth, Rth from circuit after switch.π is the total equivalent resistance in the loop
C v
L+
RthVth
iCiL
iL
L23
1 2
1 1 2 2
(0) ( )
(0 )
v v A A
dvs A s A
dt
Case 2: 0 1 2; s s
1 2( ) ( ) ( ) tv t v A A t e
Case 3: 2 20 0 1 2; , ,d ds s j
1 2( ) ( ) ( cos sin )td dv t v e B t B t
Case 1: 2 2
0 1 2 0; ,s s 1 21 2( ) ( ) s t s tv t v A e A e
Find A1,A2 from
1
1 2
(0) ( )
(0 )
v v A
dvA A
dt
Find A1,A2 from
1
1 2
(0) ( )
(0 )d
v v B
dvB B
dt
Find B1,B2 from
41
42
12/10/2019
22
Problem 1: The switch in the circuit has been open for a long time before closed at t = 0.Find ππΏ π‘ πππ π£π π‘ for t β₯ 0. (π 1 1.2231π΄, π£ 1 1.3388π
t =0
2 4H
6 1A10V
ππΏπ£π
Problem 2: The switch in the circuit has been closed for a long time before opening at t = 0. Find π£ π‘ ,π£ π‘ for t β₯ 0.
t =0
24
4
24V
10
0.05F π£
8
π£
43
44
12/10/2019
23
Problem 3: The switch in the circuit has been closed for a long time before opening at t = 0.Find v π‘ πππ π£πΆ π‘ for t β₯ 0. π£ 1 17.78π, π£ 1 5.91π
t =0
6 25mF
4 1.5A12V
π£πΆ
π£
Problem 4: The switch has been in position βbβ for a long time before moving to position βaβ at t = 0. Find ππΏ π‘ πππ π£π π‘ for t β₯ 0. (π 1 0.2917π΄, π£ 1 2.0102π
b a
t =0
1A 2A4
6
15 3HππΏ
π£π
45
46
12/10/2019
24
t =0
5mF
3
10
20
708
1H
Problem 5: The switch has been closed for a long time before open at t = 0. Find vc(t) and v1(t) for t > 0. π£ 0.1 74.0827π, π£ 0.1 72.8574π
π£π_
810A
π£1_
Problem 6: The switch is open for a long time before closed at t = 0. Find π£π π‘ for t > 0.π£ 1 4.1762π
3030
2032V
0t
4V
2H
cv
20mF
4
47
48
12/10/2019
25
Problem 1: The switch in the circuit has been open for a long time before closed at t = 0.Find ππΏ π‘ πππ π£π π‘ for t β₯ 0. (π 1 1.2231π΄, π£ 1 1.3388π
t =0
2 4H
6 1A10V
ππΏπ£π
2
6 1A10V
ππΏπ£π
Step 1) Find π 0 from circuit before switch
6Ξ© is short circuited. π£ 0. No current through 6Ξ©. π 1π΄ . 10π πππ 2Ξ© have no effect
π β 1π΄
Step 2): For t > 0 (After switch):
2
6 1A10V
ππΏ
π£π
Use source transformation:
At t = β
2 6
6V10Vπ
π 0168
2π΄π 6Ξ©, π w.r.t inductor?
π β ?
π π‘ π β π 0 π β π
π πΏ
64
1.5
π π‘ 1 π . π΄
π£ π‘ 6π 6 π π‘ 1 6π . π
2 4H
6
1A10V
ππΏπ£π
π
Also, π£ π‘ πΏ
Problem 2: The switch in the circuit has been closed for a long time before opening at t = 0. Find π£ π‘ ,π£ π‘ for t β₯ 0.
t =0
24
4
24V
10
0.05F π£
8
π£
Step 1: π£ 0 from circuit for t < 0:
24
4
24V
10
π£
8
π£
π£ 0,
π£
By voltage division:
π£24
24 824 18π
π£ 0 π£ 18π
Step 2: At π‘ β, π£ β ?
24
4
24V
10
π£
8
π£
π£
π£ β π£24
12 2424 16π
Step 3: π w.r.t. C = ?
24
4
108
π π 10 8 4 //24
= 18Ξ©
π£ π‘ 16 18 16 π . 16 2π . V
π£ π‘ π£ π‘ 10π π‘ π£ π‘ 10πΆππ£ππ‘
16 0.89π . V
π πΆ 18 0.05 0.9π
49
50
12/10/2019
26
Problem 3: The switch in the circuit has been closed for a long time before opening at t = 0.Find v π‘ πππ π£πΆ π‘ for t β₯ 0. π£ 1 17.78π, π£ 1 5.91π
t =0
6 25mF
4 1.5A12V
π£πΆ
Step 1 Find π£ 0 from circuit before switch
π£ 0 π£ 4 1.5 6π
1.5A
12V
6
π£πΆ
π£4
Step 2: For t 0, switch is open
π£ β ?
π£ β 12 π£ β 12 6 18π
π w.r.t. Capacitor?
π 10Ξ©
π£ π‘ π£ β π£ 0 π£ β π
1π πΆ
110 0.025
4
π£ π‘ 18 12π π
π£ π‘ 4 π 1.5 4 πΆππ£ππ‘
1.5
π£
ππΆ
1.5A
12V
6
π£πΆ π£
4
π£ π‘ 4 0.025 12 4 π 1.5 4.8π 6 π
Always find π£ π‘ first, then derive other variables from π£ π‘ using basic laws andproperty of capacitor
Problem 4: The switch has been in position βbβ for a long time before moving to position βaβ at t = 0. Find ππΏ π‘ πππ π£π π‘ for t β₯ 0. (π 1 0.2917π΄, π£ 1 2.0102π
b a
t =0
1A 2A4
6
15 3HππΏ
π£π
π£ π‘ 2.4 2.88π π
Step 1: Find π 0 from circuit before switch
b a
1A 2A4
6
15ππΏ
π£π
b a
t =0
1A 2A4
6
15 3HππΏ
π£π
b
1A4
6
ππΏ
π 04
4 61 0.4π΄
Step 2: For t 0
1A4
6
ππΏ
π β ? π ?
46
15 π
π β4
4 6 1 0.4π΄ π 15// 6 4 6Ξ©
π π‘ π β π 0 π β π ,
π π‘ 0.4 0.8π π΄
π πΏ
63
2
π£ π‘ 6 ππ£15
6 ππΏ
15ππππ‘
π£
51
52
12/10/2019
27
t =0
5mF
3
10
20
708
1H
Problem 5: The switch has been closed for a long time before open at t = 0. Find vc(t) and v1(t) for t > 0. π£ 0.1 74.0827π, π£ 0.1 72.8574π
π£π_
810A
π£1_
π£ π‘ 80 π 36 cos 10π‘ 4 sin 10π‘ ππ 0
8011 70//30 8
2π΄
Step 1: Find π 0 ,π£ 0 . Consider t 0:
3
10
20
70
8
1H
8
π
π£π_
80V
π
π70
1002 1.4π΄
π£ 0 8π 20π 16 28 44π
π π
π 0 π 0 2π΄
π£ β ?π ? πΌ ?π ? πβππβ πππ π
π£ β 80π,π 11 10//90 20Ξ©
Step 2: For t 0
5mF
3
20701H
π£π_
810A
π£1_
π π
10
πΌ 10, π 200, Case 3
π 200 100 10
General solution: π£ π‘ 80 π π΅ cos 10π‘ π΅ sin 10 π‘ ππ£ 0ππ‘
π 0πΆ
20.005
400π/π
44 80 π΅400 10π΅ 10π΅
π΅ 36,π΅ 4
π£ π‘ 8 10 πΆππ£ππ‘
80 π 16 cos 10π‘ 12.8 sin 10π‘ π
Problem 6: The switch is open for a long time before closed at t = 0. Find π£π π‘ for t > 0.π£ 1 4.1762π
3030
2032V
0t
4V
2H
cv
20mF
4
Step 1: Find π£ 0 , π 0 from circuit before switch
3030
2032V 4V
cv
4ππ£
π£ 0 π£ 4, π 0 π 0 0
Since π 0, 20Ξ©, 30Ξ©, 30Ξ© are in series.By voltage division,
π£30
30 20 3032 12π
π£ 0 12 4 8π
Step 2: For t 0
3030
20
32V 4V
2H
cv
4
π£ π£ β ? π ?
Since 32V and 30Ξ© are short circuited, no effect.
π£ β 4π, π 4 20//30 16Ξ©
πΌ ?π ? πβππβ πππ π?
πΌ 4, π 5, Case 3
π π πΌ 3
The general solution?
π£ π‘ 4 π π΅ cos 3π‘ π΅ sin 3π‘
Use initial conditions to find coefficients
0
8 4 π΅0 4π΅ 3π΅
π΅ 12, π΅ 16
π£ π‘ 4 π 12 cos 3π‘ 16 sin 3π‘ π
53
54