1. Chap 07 First-Order Circuits

2. Chap 07 First-Order Circuits
2

3. Outline
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Introduction
The Source-Free RC Circuit
The Source-Free RL Circuit
Singularity Functions
Step Response of an RC Circuit
Step Response of an RL Circuit
Chap 07 First-Order Circuits
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4. Introduction
A first-order circuit is characterized by a first-order
differential equation.
• RC and RL circuits: the equation to analyze RC or
RL circuit is a first-order differential equation. Thus,
RC, RL circuits are also known as first-order circuits.
• To analyze the resistor circuits, algebraic equations
are needed.
• However, because of the memory effect of C or L, the
time-differential equations are involved.
Chap 07 First-Order Circuits
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5. Introduction (cont.)
• Lots of natural phenomena can be modeled as RC or
RL circuits. Hence, the RC or RL circuit can be seen
as a circuit itself, or can be seen as a model to
describe the behavior of other physical elements.
• Sources
– Source-free circuits: initial condition
• The energy is initially stored in C or L, and the energy
causes current to flow in the circuit and is gradually
dissipated in resistors.
– Independent sources: dc, sinusoidal and exponential
sources…. (dc only in this chapter)
Chap 07 First-Order Circuits
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6. The Source-Free RC Circuit
• Source-free RC circuit: The dc source of an RC
circuit is suddenly disconnected. The energy stored in
C is released to R.
Chap 07 First-Order Circuits
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7. The Source-Free RC Circuit (cont.)
By KCL  iC  iR  0
By i - v relations  iC  C
dv
v
; iR 
dt
R
Hence,
dv v
dv
v
 0

0
dt R
dt RC
dv
1


dt
v
RC
Integrating both sides, we get
C
t
 ln A
RC
 v  t   Ae t RC
A first-order differential equation.
ln v  
If v(0)  V0 , v(t )  V0e

t
RC

t
 V0e  , t  0;   RC 
(V0 is the initial voltage across the capacitor at t=0.)
Chap 07 First-Order Circuits
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8. The Source-Free RC Circuit (cont.)
• If the initial voltage across the capacitor is Vt0 at t=t0 ,
t t
then
 0
v(t )  Vt0 e
RC
; t  t0
Natural Response: The natural response of a circuit
refers to the behavior (in terms of voltages and
currents) of the circuit itself, with no external
sources of excitation.
Time Constant of a circuit: It is the time required for
the response to decay by a factor of 1/e or 36.8
percent of its initial value.
Chap 07 First-Order Circuits
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9. Time Constant of a Circuit
• The time constant  of RC circuit is equal to RC.
• Time required for a circuit to reach its final state or
steady state is 5 : According to Table 7.1, v(t) takes
about 5 time to decay to be less than 1% of V0.
Hence, customarily, it takes 5 for the circuit to
reach its final or steady states.
transient state
steady state
5
Chap 07 First-Order Circuits
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10. Graphical Determination of Time
Constant
• The time constant can be depicted in a way that it is
the intersection of the tangent line at t=0 with the
time-axis.
v(t )  V0e
 t /
v
  e t /
V0
d v 
1 t /
    e
dt  V0  t 0

Chap 07 First-Order Circuits

t 0
1

10

11. Various Values of Time Constant
• Small time constant →Fast system response
 A circuit with a small time constant gives a fast
system response in that it reaches the steady state
quickly due to quick dissipation of energy stored.
Chap 07 First-Order Circuits
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12. Energy Dissipated in Resistor of
RC Circuits
• The capacitor with an initial voltage stores energy in
terms of electrical field. This stored energy will be
dissipated gradually in the resistor.
Initially Stored Energy in C
0
1 2
1
w(0)  Cv (t )  CV02
2
2

Chap 07 First-Order Circuits
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13. Energy Dissipated in Resistor of
RC Circuits (cont.)
Energy Dissipated in the Resistor
v (t ) V0  t /
 e
R
R
V0 2 2 t /
p(t )  viR 
e
R
2
t
tV
wR (t )   pd   0 e 2 / d
0
0 R
iR ( t ) 

V0
2
2R
t
e 2 /
0
1
 CV02 1  e 2 t /  ,   RC
2
1
wR (t ) t  wR ()  CV02
2
Chap 07 First-Order Circuits
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14. Summary of Source-Free RC Circuit
• The key to working with a source-free RC
circuit
– The initial voltage v(t0)=V0 across the capacitor.
– The time constant .
v(t )  V0e
Chap 07 First-Order Circuits
 ( t t0 )/
14

15. Summary of Source-Free RC
Circuit (cont.)
• Both the current and the voltage in elements of
a source-free RC circuit are in the form below
x(t )  Ke
 ( t t0 )/
– K is the voltage or current at t=t0
– The time constant τ is the product of capacitance
and resistance.
  RC
Chap 07 First-Order Circuits
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16. Summary of Source-Free RC Circuit
(cont.)
• As t approaches ∞, the work (energy)
dissipated in the resistor becomes CV02/2
which equals the initial energy stored in the
capacitor
• This result demonstrates that the capacitor
does not consume but only store energy.
Chap 07 First-Order Circuits
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17. Example 7.1
Q: In Fig. 7.5, let vC = 15 V. Find vC, vx, and ix
for t > 0.
Chap 07 First-Order Circuits
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18. Example 7.1 (cont.)
Sol
• The equivalent resistance at the capacitor terminal is
Req 
20  5
 4
20  5
• The time constant
  ReqC  4  0.1  0.4s
Thus
vc  v  v(0)e
 t /
 15e
 t /0.4
V,
12
vx 
vc  0.6  15e2.5t  9e2.5t V
12  8
vx
ix   0.75e2.5t A
12
Chap 07 First-Order Circuits
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19. Example 7.2
Q: The switch in the circuit has been closed for a long
time, and it is opened at t = 0. Find v(t) for t ≥ 0.
Calculate the initial energy stored in the capacitor.
Chap 07 First-Order Circuits
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20. Example 7.2 (cont.)
Sol
• Using voltage division.
vc (t ) 
9
 20  15V, t  0
93
at t = 0- is the same at t = 0
vc (0)  V0  15V
Req  1  9  10
  ReqC  10  20 103  0.2s
v(t )  Vc (0)et /  15et /0.2 V
• The initial energy
1 2
1
wc (0)  Cvc (0)   20  103  152  2.25J
2
2
Chap 07 First-Order Circuits
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21. Example 7.2 (cont.)
t  0
t 0

Chap 07 First-Order Circuits
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22. The Source-Free RL Circuit
• Source-free RL circuit: The dc source of an RL
circuit is suddenly disconnected. The energy stored in
L is released to R.
Chap 07 First-Order Circuits
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23. The Source-Free RL Circuit (cont.)
Assume i  0   I 0
Applying KVL gives
vL  vR  0
vL  L
di
; vR  iR
dt
di
di R
 Ri  0 or
 i0
dt
dt L
di
R
   dt
i
L
Integrating both sides, we get
R
ln it   ln I 0   t
L
 it   I 0 e  Rt L
L
Chap 07 First-Order Circuits
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24. Duality of RC and RL Circuits
dv
1

dt
v
RC
RC Circuit
i(t)
R
G=1/R
C
Chap 07 First-Order Circuits
RL Circuit
v(t)
Time Constant
di
1

dt
i
GL
L
RC
GL
24

25. Graphical Determination of Time
Constant
• The time constant can be depicted in a way that it is
the intersection of the tangent line at t=0 with the
time-axis.
i (t )  I 0 e
 t /
i (t )

 e  t /
I0
d i 
1  t /
    e
dt  I 0  t 0

Chap 07 First-Order Circuits

t 0
1

25

26. Energy Dissipated in Resistor of
RL Circuits
• The inductor with an initial voltage stores energy in
terms of magnetic field. This stored energy will be
dissipated gradually in the resistor.
Initially Stored Energy in L
0
1 2
1 2
w(0)  Li (t )  LI 0
2
2

Chap 07 First-Order Circuits
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27. Energy Dissipated in Resistor of
RL Circuits (cont.)
Energy Dissipated in the Resistor
L
i (t )  I 0 e ;  
R
vR (t )  iR  I 0 Re t /
 t /
p(t )  vR i  I 02 Re 2t /
t
t
0
0
wR (t )   pd   I 02 Re 2 / d
1
   RI 02 e2 /
2
t
0
wR (t ) t   wR () 
Chap 07 First-Order Circuits
1 2
 LI 0 1  e 2t / 
2
1 2
LI 0
2
27

28. Summary of Source-Free RL
Circuits
• The key to working with a source-free RL
circuit
– The initial current i(t0) = I0 through the inductor.
– The time constant  of the circuit.
i(t )  I 0e
Chap 07 First-Order Circuits
 ( t t0 )/
28

29. Example 7.3
Q: Assuming that i(0)=10A, calculate i(t) and ix(t)
in the circuit.
Chap 07 First-Order Circuits
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30. Example 7.3 (cont.)
Method 1 : Thevenin's Equivalent Circuit
--Find Req at inductor terminal
Applying KVL
1

2(i1  i2 )  1  0  i1  i2   2


 6i  2i  3i  0  i  5 i
1
1
2
1
 2
6

 i1  3 A, i0  i1  3 A
Hence
v0 1
Req  RTh   
i0 3
 The time constant is
L
3

 s
Req 2
Thus, the current through theinductor is
i (t )  i (0)et /  10e  (2/3)t A, t  0
Chap 07 First-Order Circuits
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31. Example 7.3 (cont.)
Method 2 :
1 di1
 2(i1  i2 )  0
2 dt
KVL at loop 1:
(a)
5
6i2  2i1  3i1  0  i2  i1
6
KVL at loop 2:
Substituting (b) into (a) 
(b)
di1 2
di
2
 i1  0  1   dt
dt 3
i1
3
Integrating both sides
t
i (t )
ln i i (0)
2
 
3 0
Hence, i (t )  i (0)e  (2/3) t  10e  (2/ 3)t A; t  0
vL
ix (t ) 
di
2
10
 0.5(10)(  )e (2/ 3)t   e (2/3) t V
dt
3
3
v
 1.667e  (2/ 3)t A; t  0
2
Chap 07 First-Order Circuits
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32. Example 7.4
Q: The switch in the circuit has been closed for a
long time. At t = 0, the switch is opened.
Calculate i(t) for t > 0.
Chap 07 First-Order Circuits
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33. Example 7.4 (cont.)
At t =0 4 12
4 ||12 
 3
4  12
Hence
40
i1 
 8A
23
Using current division
12
i1  6A, t  0
12  4
i (0)  i (0 )  6A
i (t ) 
Chap 07 First-Order Circuits
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34. Example 7.4 (cont.)
Combining the resistors, we have Req at terminal inductor
Req  (12  4) ||16  8Ω
The time constant is
L
2 1

  s
Req 8 4
8
Thus,
i(t )  i(0)e
Chap 07 First-Order Circuits
t /
4t
 6e A
34

35. Example 7.5
Q: Find io, vo, and i for all time, assuming that
the switch was open for a long time.
Chap 07 First-Order Circuits
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36. Example 7.5 (cont.)
(a) For t  0
10
i (t ) 
 2A ,
23
v0 (t )  3i (t )  6V ,
i (0)  2A
RTh  3 || 6  2
(b) For t  0
L

 1s
RTh
i (t )  i (0)et /  2e t A
Chap 07 First-Order Circuits
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37. Example 7.5 (cont.)
Since the inductor is in the parallel with the 6Ω and 3Ω
di
v0 (t )  vL   L  2(2e  t )  4e  t V, t  0
dt
vL
2 t
i0 (t )    e A, t  0
6
3
Thus
t0
0A,

i0 (t )   2  t
;
 3 e A, t  0

t0
2A,
i(t )    t
2e A, t  0
Chap 07 First-Order Circuits
6V, t  0
v0 (t )    t
4e V, t  0
37

38. Example 7.5 (cont.)
Chap 07 First-Order Circuits
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