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L20Chapter 8 Secondโorder circuits
A circuit with two energy storage elements:One inductor, one capacitorTwo capacitors, orTwo inductors Such a circuit will be described with
secondโorder 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 2ndโorder diff equ.
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐;๐ฃ 0 ๐ , ๐ฃ 0 ๐
L20Solving 2ndโorder differential equations
Consider a secondโorder differential equation
with initial condition: ๐ฃ 0 ๐0,
Need to find v(t) for all t.
In chapter 8, b=0 corresponds to source free case, bโ 0 for step response.
Let ๐ฃ โ = 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
๐ 1
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐ ๐ฃ 0 ๐
(Assume ๐ผ 0. Then as ๐ก โ โ, ๐ฃ ๐ก goes to a constant . )
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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
๐ฃ ๐ก ๐ ๐ด ๐ ๐ ๐ด ๐ ๐ฃ ๐ก ๐ ๐ด ๐ ๐ ๐ด ๐
๐ฃ ๐ก ๐ฃ โ ๐ด ๐ ๐ด ๐ Where ๐ฃ โ ๐/๐
๐ ๐ด ๐ ๐ ๐ด ๐ 2๐ผ ๐ ๐ด ๐ ๐ ๐ด ๐ ๐ ๐ฃ โ ๐ด ๐ ๐ด ๐
๐ด ๐ ๐ 2๐ผ๐ ๐ ๐ด ๐ ๐ 2๐ผ๐ ๐ ๐ ๐ฃ โ
๐ ๐ฃ โ
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐; 1๐ฃ 0 ๐ , ๐ฃ 0 ๐ (IC)
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ
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 ๐ฃ(t) and ๐ฃ(t) at t=0:
To satisfy the IC, we obtain
You may use other methods to solve for A1 and A2.
Recall:
๐ฃ ๐ก ๐ฃ โ ๐ด ๐ ๐ด ๐ Where ๐ฃ โ ๐/๐
๐ฃ 0 ๐ฃ โ ๐ด ๐ด
๐ฃ 0 ๐ ๐ด ๐ ๐ด
๐ฃ โ ๐ด ๐ด ๐๐ ๐ด ๐ ๐ด ๐
1 1๐ ๐
๐ด๐ด
๐ ๐ฃ โ๐
๐ด๐ด
1 1๐ ๐
๐ ๐ฃ โ๐
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐; 1๐ฃ 0 ๐ , ๐ฃ 0 ๐ (IC)
๐ฃ ๐ก ๐ ๐ด ๐ ๐ ๐ด ๐
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L20
Example: Solve
202 4; 3; 6b 02; 3;
0; Case 1, two distinct real roots for ๐ 4๐ 3 0.
21 2, 2 2 3 2 1,s s 1 21, 3s s
Step 1: Find s1, s2, ๐ฃ โ
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=โ2; A2=1
Finally, 3( ) 2 2 t tv t e e
2 21 2 0,s s
Note:๐ , ๐ are roots to:๐ 4๐ 3 0๐ 3 ๐ 1 0๐ 1, ๐ 3
Note:๐ฃ ๐ก ๐ด ๐ 3๐ด ๐
๐ฃ ๐ก ๐ฃ โ ๐ด ๐ ๐ด ๐
๐ 2๐ผ๐ ๐ 0.
๐ฃ โ ๐/๐ =6/3 =2
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐ (To compare )
๐ฃ 4๐ฃ 3๐ฃ 6, ๐ฃ 0 1; ๐ฃ 0 1
๐ฃ 0 ๐ด 3๐ด 1 โโ ๐ฃ 0 2 ๐ด ๐ด 1 โ
L20
An Example:
The general solution:
31 2( ) 2 t tv t A e A e
๐ฃ ๐ก ๐ฃ โ ๐ด ๐ ๐ด ๐ , ๐ฃ โ๐๐
Since ๐ , ๐ are negative, limโ
๐ฃ ๐ก 2 ๐ฃ โ
Another way to see ๐ฃ โ ๐/๐
๐ฃ ๐ก ๐ด ๐ ๐ ๐ด ๐ ๐ ,
๐ฃ ๐ก ๐ด ๐ ๐ ๐ด ๐ ๐ ,
limโ
๐ฃ ๐ก 0
limโ
๐ฃ ๐ก 0
From 1 , limโ
๐ฃ ๐ก 2๐ผ๐ฃ ๐ก ๐ ๐ฃ ๐ก ๐ ๐ฃ โ ๐
๐ฃ โ๐๐
๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐; 1๐ฃ 0 ๐ , ๐ฃ 0 ๐ (IC)
๐ฃ 4๐ฃ 3๐ฃ 6, ๐ฃ 0 1; ๐ฃ 0 1
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L20Case 2: = 0
Two identical roots, s1=s2= โ to s2+2s+02=0
The general solution:
Its derivative:
Use IC to find A1, A2.
At t = 0,
๐ 2๐ผ๐ ๐ผ ๐ ๐ผ
๐ฃ ๐ก ๐ฃ โ ๐ด ๐ด ๐ก ๐ ๐ฃ โ ๐/๐
๐ฃ 0 ๐ฃ โ ๐ด ๐ ๐ด ๐ ๐ฃ โ๐ด ๐ ๐ผ๐ด
๐ฃ ๐ก ๐ด ๐ ๐ผ ๐ด ๐ด ๐ก ๐
๐ฃ 0 ๐ด ๐ผ๐ด ๐
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.
๐ 1
๐ , ๐ ๐ผ ๐ผ ๐ ๐ผ 1 ๐ ๐ผ ๐ผ ๐ ๐
๐ , ๐ ๐ผ ๐ ๐
๐ฃ ๐ก ๐ฃ โ ๐ ๐ต cos๐ ๐ก ๐ต sin๐ ๐ก ๐ฃ โ ๐/๐
๐ฃ 0 ๐ฃ โ ๐ต ๐ ๐ต ๐ ๐ฃ โ ๐ต ๐ ๐ผ๐ต /๐
๐ฃ ๐ก ๐ผ๐ ๐ต cos๐ ๐ก ๐ต sin๐ ๐ก ๐ ๐ ๐ต sin๐ ๐ก ๐ ๐ต cos๐ ๐ก
๐ฃ 0 ๐ผ๐ต ๐ ๐ต ๐
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L20
0 5 10 15 20-5
0
5
10
๐ฃ โ
2
d
Case 3: < 0
๐ฃ ๐ก ๐ฃ โ ๐ ๐ต cos๐ ๐ก ๐ต sin๐ ๐ก
๐ฃ โ ๐ด ๐
๐ฃ โ ๐ด ๐
Example: Solve ๐ฃ 8๐ฃ 25๐ฃ 200, ๐ฃ 0 0, ๐ฃ 0 12
Solution:
Compare with ๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐
๐ผ ?๐ ? , ๐ ? ๐คโ๐๐โ ๐๐๐ ๐?
๐ผ 4,๐ 5, ๐ 200, ๐ผ ๐
Step 1: Form general solution using roots and ๐ฃ โ .
๐ ๐ ๐ผ 25 16 3
Complex roots: ๐ผ ๐๐ 4 ๐3
๐ฃ โ ? ๐ฃ โ๐
๐
20025
8
Imaginary part of the roots:
๐ฃ ๐ก ๐ฃ โ ๐ ๐ต cos๐ ๐ก ๐ต sin๐ ๐ก , ๐ฃ โ ? ๐ ?
๐ฃ ๐ก 8 ๐ ๐ต cos 3๐ก ๐ต sin 3๐ก , General solution
Step 2: Find ๐ต ,๐ต , using initial conditions, ๐ฃ 0 0, ๐ฃ 0 12
๐ฃ ๐ก 4๐ ๐ต cos 3๐ก ๐ต sin 3๐ก ๐ 3๐ต sin 3๐ก 3๐ต cos 3๐ก
๐ฃ 0 4๐ต 3๐ต 0
๐ฃ 0 8 ๐ต 12 ๐ต 4;๐ต163
Final solution: ๐ฃ ๐ก 8 ๐ 4cos3t sin 3๐ก ๐
case 3, two complex roots : ๐ผ ๐๐
๐ฃ โ 8๐ฃ โ 25๐ฃ โ 2000 0
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R20
Practice 2: Solve
Practice 3: Solve
Practice 1: Solve
๐ฃ 5 ๐ฃ 6๐ฃ 18, ๐ฃ 0 0; ๐ฃ 0 4
๐ฃ 4 ๐ฃ 4๐ฃ 8, ๐ฃ 0 1; ๐ฃ 0 1
๐ฃ 4 ๐ฃ 13๐ฃ 39, ๐ฃ 0 1; ๐ฃ 0 1
L21Secondโorder 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 ๐ ๐ก , ๐ฃ ๐ก for t > 0.
cv
0.5H
+
6
30V
๐
t = 0
66
+ 10V
As always, ๐ ๐ก , ๐ฃ ๐ก are continuous function of time.Use the circuit before switch to find ๐ 0 , ๐ฃ 0 .
For t>0, ๐ฃ ๐ก satisfies We also need 0 .
Generally, ๐ก is not continuous
at t=0.
How to find 0 ?
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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
ยง 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 donโt 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+
โข Apply KVL to find ๐ฃ๐ฟ(0+), then โข Apply KCL to find ๐๐ถ(0+), then
(0 ) 1 (0 ),C
C
dvi
dt C
(0 ) 1 (0 ),L
L
div
dt L
๐ฃ 0 ๐ , ๐ฃ 0 ๐ ๐ฃ 2๐ผ๐ฃ ๐ ๐ฃ ๐;
๐ฃ ๐ฟ , ๐ ๐ถRecall:
L21Example: Find
+
4
2
12V
0.25H
0.1FCv
๐๐ฟ
(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
๐๐ฟ
To find ,
Consider t = 0+
+
4
12V
0.25H
0.1FCv
๐๐ฟ
Lv
๐๐ถ
12 4 2 4 0V
By continuity, ๐ 0 2๐ด, ๐ฃ 0 4๐
๐ 2๐ด; ๐ 4๐
๐ 0 2๐ด; ๐ฃ 0 2 2 4๐
๐ฃ 0
๐ 0 ๐ 0 2A
๐๐ 0๐๐ก
1๐ฟ๐ฃ 0 0
๐๐ฃ 0๐๐ก
1๐ถ๐ 0
20.1
20 ๐/๐ ,
๐ฃ
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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).
L21ยง 8.5 Step response of series RLC circuits
A general circuit
+
+
C v
L
๐๐ฟ
Before switch (t<0), L and C may not be in series.
Determine ๐ฃ 0 , ๐ 0from circuit before switch by considering ๐ฟ as short,๐ถ as open.
After switch ( t>0 ), L and C in series
+
+
C v
+
L R0
๐๐ By Theveninโs theorem, the two terminal circuit can be replaced by a voltage source and a resistor
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L21By Theveninโs theorem, the two terminal circuit can be replaced by a voltage source and a resistor
Rearranging the elements does not changeloop current ๐ and capacitor ๐ฃ
๐๐กโ ๐๐กโโ
๐๐๐ ๐กโ ๐ ๐กโโ
๐ 0
0
1
(0 ) ,
(0 )
v V
dvV
dt
Given
C v
+
L R0
๐๐
+
๐ ๐กโโ
๐๐กโโ
๐
C v
L+
๐ ๐กโ
๐๐กโ
๐
We call (๐๐กโ,๐ ๐กโ)Theveninโs equivalentw.r.t LC
๐ฃ ๐ฃ
L21
0
1
(0 ) ,
(0 )
v V
dvV
dt
Given C v
L+
Rth
Vth
๐
Choose ๐ฃ as key variable. We derive diff. equation for ๐ฃExpress other variables in terms of ๐ฃ:
;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) :Theveninโs equivalent w.r.t LC
The equivalent circuit:
๐ฃ ๐ ๐ ๐ ๐ถ๐๐ฃ๐๐ก
; ๐ฃ ๐ฟ๐๐๐๐ก ๐ฟ๐ถ
๐ ๐ฃ๐๐ก
๐ฃ ๐ฃ
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L21
C v
L+
Rth
Vth
0 1
(0 )(0 ) ,
dvv V V
dt
๐
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
๐ฃ โ๐๐
๐
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 ๐ฃ 0 ๐ฃ โ ๐ต๐๐ฃ 0๐๐ก
๐ผ๐ต ๐ ๐ต
๐ฃ 0 ๐ฃ โ ๐ด๐๐ฃ 0๐๐ก
๐ผ๐ด ๐ด
๐ฃ 0 ๐ฃ โ ๐ด ๐ด๐๐ฃ 0๐๐ก
๐ ๐ด ๐ ๐ด
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L21Key points for step resp. of series RLC circuits:
Obtain ๐ฃ 0 , ๐๐ฟ 0 from circuit before switch
+
+
C v
L
๐๐ฟ
For circuit after switch, obtain Theveninโsequivalent (Vth,Rth) w.r.t LC
Assign ๐๐ถ by passive sign convention
(0 ) (0)C Li i Depending on how ๐๐ฟ 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: โข ๐ฃ 0 , ๐๐ฟ 0 from circuit before switchโข ๐๐กโ ,๐ ๐กโ from circuit after switch.
C v
L
+
RthVth
๐๐ถ ๐๐ฟ
๐๐ฟ
L21Example 1: Find v(t) for t > 0.
0.25F v
1H+
5
24V
๐๐ถ
๐๐ฟt = 0
1
Step 1: find v(0), ๐๐ฟ(0) from circuit before switch ๐๐ฟ
v
+
5
24V
1
๐๐ฟ 0 =24/6=4A
๐ฃ 0 = 4V
Step 2: Find Vth, Rth, w.r.t LC from circuit after switch
Vth=24V, Rth=5 v(โ) = 24V
Since ๐๐ถ ๐๐ฟ, .16๐/๐
v
+
5
24V
๐๐ถ๐๐ฟ
0.25F
1H
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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
๐ฃ 0 24 ๐ด1 ๐ด2 4
๐ด 4๐ด =16๐ด1 64/3, ๐ด2 4/3
Final solution: 464 4( ) 24
3 3t tv t e e V
Note: v t v โ ๐ด ๐ ๐ด ๐
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
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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
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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
20๐ ๐ฃ
L22
1/8F
v
0.5H
6
i
66
+ 10V
Step 2: After switch
Rth=(6+6)//6=4, Vth=10V = v(โ)
iC
Since ๐๐ถ ๐, ๐๐ถ 0 5๐ด
5/ 1/8 40๐/๐
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,
๐ ๐ก = โ (1/8)*10(โ4)eโ4t = 5 eโ4tA
๐ 0 5๐ด,๐๐ฃ 0๐๐ก
?
Rth=?, Vth=?
๐ผ ?๐ ? Which case?
v(t) = 10+(A1+A2t)eโ4t
๐ฃ ๐ก = 10 + 10 eโ4tV
๐ ๐ก ๐ ๐ก ๐ถ๐๐ฃ๐๐ก
๐ 0 5๐ด๐ฃ 0 20๐
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L22
+
+
Cv
L
For parallel RLC circuit
iL
By Nortonโs 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 ๐ฃ 0 , ๐ 0 from circuit for t<0
Key parameters to obtain: โข v (0), iL(0) from circuit before switch (t<0)โข Vth, Rth w.r.t. LC from circuit after switch (t>0) .
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L22Step 1: Find ๐ฃ 0 , ๐ 0 from circuit for t<0
For t < 0, 3u(t)A=0, current source open, C open , L short
๐ฃ 0 ๐ฃ12
6 1212 8๐,
+
6
6
16mF
12
12V
Cv
๐ 0 Cv
Since ๐ =0, 12V, 6ฮฉ ๐๐๐ 12ฮฉ in the outer loop are in series
By voltage division:
๐ 0 0, ๐๐ฃ 0๐๐ก
0
L22
+
6
3A
6
16mF
0.5H 12
12V
Cv
Step 2: For t>0, 3u(t)A=3A. Need to find ๐ ๐ฃ โ ,๐ w.r.t LC
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L22
+
6
3A
6
16mF
0.5H12
12V
Cv
Step 2: Fot t>0, 3u(t)A=3A. Need to find ๐ ๐ฃ โ ,๐ w.r.t LC
For ๐ฃ โ+
6
3A
6
12
12V
( )Cv
Rv
6v
By KVL, ๐ฃ โ ๐ฃ ๐ฃ
๐ฃ 3 6 18๐, by ohms law
๐1218
๐ด23๐ด,
๐
๐ ๐
๐3A
๐ฃ 12๐ 1223
8๐
๐ฃ โ 18 8 26๐
For ๐ , turn off 12V with short, turn off 3A with open
66
12๐
R 6 6//12=10ฮฉ
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 ฮฉ to the left
๐ 6 12//6 10ฮฉUnder DC condition,
๐ฃ 8๐, ๐ฃ 18๐,๐ฃ โ ๐ 8 18 26๐
6+
6
3A
16mF
0.5H12
12V
Cv
๐ฃ
Rv
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12/10/2019
18
๐ 10ฮฉ, ๐ฃ โ ๐ 8 18 26๐, ๐ฃ 0 8๐, 0
Step 3: math From step 1, step 2, found
๐ผ๐ 2๐ฟ
102 0.5
10; ๐1
๐ฟ๐ถ
10.5 0.016
125 ,
v t 26 e B cos 5t B sin 5t
The general solution:
To satisfy initial condition: ๐ฃ 0 26 ๐ต 8๐ฃ 0 10๐ต 5๐ต 0Solving equations to get ๐ต 18;๐ต 36Step response: v t 26 e 18 cos 5๐ก 36 sin 5๐ก ๐
Also note: ๐ฟ 0.5๐ป; ๐ถ 16๐๐น 0.016๐น ๐ผ ? ,๐ ? ๐คโ๐๐โ ๐๐๐ ๐?
๐ผ๐ 2๐ฟ
; ๐1
๐ฟ๐ถ ;
๐ผ ๐ , ๐๐๐ ๐ 3, ๐ 125 100 5
๐ฃ 0 ๐ฃ โ ๐ต๐๐ฃ 0๐๐ก
๐ผ๐ต ๐ ๐ต
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(โt)A
Cv
Rv
2
43A
4
Chapter 7, Chapter 8 Review
Final exam: 12/16/19 (Monday), 8โ11am, BL210
One page oneโsided note allowed, Calculators allowedDonโt include the solution to any circuit problem in the note!
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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
๐น๐๐ ๐ก 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(โEstablished 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(โ) = Vth, (Vth, Rth) can be together considered as the Theveninโsequivalent, with respect to the capacitor, from circuit after switch.
Step response of RL circuits
i
+
+
L
Step response โ For t > 0,
Three key parameters:
1. i(0), initial condition, from the DC circuit for t < 0. (inductor =short)2. i(โ)=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 Nortonโs equivalent, w.r.t inductor, for t > 0.
NRt
Li(t) = i( )+ (i(0) i( ))e
L23
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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 Theveninโsequivalent (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: โข v(0), iL(0) from circuit before switchโข Vth, Rth from circuit after switch.๐ 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
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Problem 1: The switch in the circuit has been open for a long time before closed at t = 0.Find ๐๐ฟ ๐ก ๐๐๐ ๐ฃ๐ ๐ก for t โฅ 0. (๐ 1 1.2231๐ด, ๐ฃ 1 1.3388๐
t =0
2 4H
6 1A10V
๐๐ฟ๐ฃ๐
Problem 2: The switch in the circuit has been closed for a long time before opening at t = 0. Find ๐ฃ ๐ก ,๐ฃ ๐ก for t โฅ 0.
t =0
24
4
24V
10
0.05F ๐ฃ
8
๐ฃ
43
44
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23
Problem 3: The switch in the circuit has been closed for a long time before opening at t = 0.Find v ๐ก ๐๐๐ ๐ฃ๐ถ ๐ก for t โฅ 0. ๐ฃ 1 17.78๐, ๐ฃ 1 5.91๐
t =0
6 25mF
4 1.5A12V
๐ฃ๐ถ
๐ฃ
Problem 4: The switch has been in position โbโ for a long time before moving to position โaโ at t = 0. Find ๐๐ฟ ๐ก ๐๐๐ ๐ฃ๐ ๐ก for t โฅ 0. (๐ 1 0.2917๐ด, ๐ฃ 1 2.0102๐
b a
t =0
1A 2A4
6
15 3H๐๐ฟ
๐ฃ๐
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46
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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. ๐ฃ 0.1 74.0827๐, ๐ฃ 0.1 72.8574๐
๐ฃ๐_
810A
๐ฃ1_
Problem 6: The switch is open for a long time before closed at t = 0. Find ๐ฃ๐ ๐ก for t > 0.๐ฃ 1 4.1762๐
3030
2032V
0t
4V
2H
cv
20mF
4
47
48
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Problem 1: The switch in the circuit has been open for a long time before closed at t = 0.Find ๐๐ฟ ๐ก ๐๐๐ ๐ฃ๐ ๐ก for t โฅ 0. (๐ 1 1.2231๐ด, ๐ฃ 1 1.3388๐
t =0
2 4H
6 1A10V
๐๐ฟ๐ฃ๐
2
6 1A10V
๐๐ฟ๐ฃ๐
Step 1) Find ๐ 0 from circuit before switch
6ฮฉ is short circuited. ๐ฃ 0. No current through 6ฮฉ. ๐ 1๐ด . 10๐ ๐๐๐ 2ฮฉ have no effect
๐ โ 1๐ด
Step 2): For t > 0 (After switch):
2
6 1A10V
๐๐ฟ
๐ฃ๐
Use source transformation:
At t = โ
2 6
6V10V๐
๐ 0168
2๐ด๐ 6ฮฉ, ๐ w.r.t inductor?
๐ โ ?
๐ ๐ก ๐ โ ๐ 0 ๐ โ ๐
๐ ๐ฟ
64
1.5
๐ ๐ก 1 ๐ . ๐ด
๐ฃ ๐ก 6๐ 6 ๐ ๐ก 1 6๐ . ๐
2 4H
6
1A10V
๐๐ฟ๐ฃ๐
๐
Also, ๐ฃ ๐ก ๐ฟ
Problem 2: The switch in the circuit has been closed for a long time before opening at t = 0. Find ๐ฃ ๐ก ,๐ฃ ๐ก for t โฅ 0.
t =0
24
4
24V
10
0.05F ๐ฃ
8
๐ฃ
Step 1: ๐ฃ 0 from circuit for t < 0:
24
4
24V
10
๐ฃ
8
๐ฃ
๐ฃ 0,
๐ฃ
By voltage division:
๐ฃ24
24 824 18๐
๐ฃ 0 ๐ฃ 18๐
Step 2: At ๐ก โ, ๐ฃ โ ?
24
4
24V
10
๐ฃ
8
๐ฃ
๐ฃ
๐ฃ โ ๐ฃ24
12 2424 16๐
Step 3: ๐ w.r.t. C = ?
24
4
108
๐ ๐ 10 8 4 //24
= 18ฮฉ
๐ฃ ๐ก 16 18 16 ๐ . 16 2๐ . V
๐ฃ ๐ก ๐ฃ ๐ก 10๐ ๐ก ๐ฃ ๐ก 10๐ถ๐๐ฃ๐๐ก
16 0.89๐ . V
๐ ๐ถ 18 0.05 0.9๐
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Problem 3: The switch in the circuit has been closed for a long time before opening at t = 0.Find v ๐ก ๐๐๐ ๐ฃ๐ถ ๐ก for t โฅ 0. ๐ฃ 1 17.78๐, ๐ฃ 1 5.91๐
t =0
6 25mF
4 1.5A12V
๐ฃ๐ถ
Step 1 Find ๐ฃ 0 from circuit before switch
๐ฃ 0 ๐ฃ 4 1.5 6๐
1.5A
12V
6
๐ฃ๐ถ
๐ฃ4
Step 2: For t 0, switch is open
๐ฃ โ ?
๐ฃ โ 12 ๐ฃ โ 12 6 18๐
๐ w.r.t. Capacitor?
๐ 10ฮฉ
๐ฃ ๐ก ๐ฃ โ ๐ฃ 0 ๐ฃ โ ๐
1๐ ๐ถ
110 0.025
4
๐ฃ ๐ก 18 12๐ ๐
๐ฃ ๐ก 4 ๐ 1.5 4 ๐ถ๐๐ฃ๐๐ก
1.5
๐ฃ
๐๐ถ
1.5A
12V
6
๐ฃ๐ถ ๐ฃ
4
๐ฃ ๐ก 4 0.025 12 4 ๐ 1.5 4.8๐ 6 ๐
Always find ๐ฃ ๐ก first, then derive other variables from ๐ฃ ๐ก using basic laws andproperty of capacitor
Problem 4: The switch has been in position โbโ for a long time before moving to position โaโ at t = 0. Find ๐๐ฟ ๐ก ๐๐๐ ๐ฃ๐ ๐ก for t โฅ 0. (๐ 1 0.2917๐ด, ๐ฃ 1 2.0102๐
b a
t =0
1A 2A4
6
15 3H๐๐ฟ
๐ฃ๐
๐ฃ ๐ก 2.4 2.88๐ ๐
Step 1: Find ๐ 0 from circuit before switch
b a
1A 2A4
6
15๐๐ฟ
๐ฃ๐
b a
t =0
1A 2A4
6
15 3H๐๐ฟ
๐ฃ๐
b
1A4
6
๐๐ฟ
๐ 04
4 61 0.4๐ด
Step 2: For t 0
1A4
6
๐๐ฟ
๐ โ ? ๐ ?
46
15 ๐
๐ โ4
4 6 1 0.4๐ด ๐ 15// 6 4 6ฮฉ
๐ ๐ก ๐ โ ๐ 0 ๐ โ ๐ ,
๐ ๐ก 0.4 0.8๐ ๐ด
๐ ๐ฟ
63
2
๐ฃ ๐ก 6 ๐๐ฃ15
6 ๐๐ฟ
15๐๐๐๐ก
๐ฃ
51
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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. ๐ฃ 0.1 74.0827๐, ๐ฃ 0.1 72.8574๐
๐ฃ๐_
810A
๐ฃ1_
๐ฃ ๐ก 80 ๐ 36 cos 10๐ก 4 sin 10๐ก ๐๐ 0
8011 70//30 8
2๐ด
Step 1: Find ๐ 0 ,๐ฃ 0 . Consider t 0:
3
10
20
70
8
1H
8
๐
๐ฃ๐_
80V
๐
๐70
1002 1.4๐ด
๐ฃ 0 8๐ 20๐ 16 28 44๐
๐ ๐
๐ 0 ๐ 0 2๐ด
๐ฃ โ ?๐ ? ๐ผ ?๐ ? ๐โ๐๐โ ๐๐๐ ๐
๐ฃ โ 80๐,๐ 11 10//90 20ฮฉ
Step 2: For t 0
5mF
3
20701H
๐ฃ๐_
810A
๐ฃ1_
๐ ๐
10
๐ผ 10, ๐ 200, Case 3
๐ 200 100 10
General solution: ๐ฃ ๐ก 80 ๐ ๐ต cos 10๐ก ๐ต sin 10 ๐ก ๐๐ฃ 0๐๐ก
๐ 0๐ถ
20.005
400๐/๐
44 80 ๐ต400 10๐ต 10๐ต
๐ต 36,๐ต 4
๐ฃ ๐ก 8 10 ๐ถ๐๐ฃ๐๐ก
80 ๐ 16 cos 10๐ก 12.8 sin 10๐ก ๐
Problem 6: The switch is open for a long time before closed at t = 0. Find ๐ฃ๐ ๐ก for t > 0.๐ฃ 1 4.1762๐
3030
2032V
0t
4V
2H
cv
20mF
4
Step 1: Find ๐ฃ 0 , ๐ 0 from circuit before switch
3030
2032V 4V
cv
4๐๐ฃ
๐ฃ 0 ๐ฃ 4, ๐ 0 ๐ 0 0
Since ๐ 0, 20ฮฉ, 30ฮฉ, 30ฮฉ are in series.By voltage division,
๐ฃ30
30 20 3032 12๐
๐ฃ 0 12 4 8๐
Step 2: For t 0
3030
20
32V 4V
2H
cv
4
๐ฃ ๐ฃ โ ? ๐ ?
Since 32V and 30ฮฉ are short circuited, no effect.
๐ฃ โ 4๐, ๐ 4 20//30 16ฮฉ
๐ผ ?๐ ? ๐โ๐๐โ ๐๐๐ ๐?
๐ผ 4, ๐ 5, Case 3
๐ ๐ ๐ผ 3
The general solution?
๐ฃ ๐ก 4 ๐ ๐ต cos 3๐ก ๐ต sin 3๐ก
Use initial conditions to find coefficients
0
8 4 ๐ต0 4๐ต 3๐ต
๐ต 12, ๐ต 16
๐ฃ ๐ก 4 ๐ 12 cos 3๐ก 16 sin 3๐ก ๐
53
54