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CHAPTER 5
DC TRANSIENT ANALYSIS
1
Objectives
 Investigate the behavior of currents and
voltages when energy is either released or
acquired by inductors and capacitors when
there is an abrupt change in dc current or
voltage source.
 To do an analysis of natural response and step
response of RL and RC circuit.
2
Lecture’s contents
•
•
•
•
5-1 NATURAL RESPONSE OF RL CIRCUIT
5-2 NATURAL RESPONSE OF RC CIRCUIT
5-3 STEP RESPONSE OF RL CIRCUIT
5-4 STEP RESPONSE OF RC CIRCUIT
3
First – Order Circuit
• A circuit that contains only sources, resistor and inductor
is called and RL circuit.
• A circuit that contains only sources, resistor and capacitor
is called an RC circuit.
• RL and RC circuits are called first – order circuits because
their voltages and currents are describe by first order
differential equations.
R
i
L
An RL circuit
vs
–
+
–
+
Vs
R
i
C
An RC circuit
4
Review (conceptual)
• Any first – order circuit can be reduced to a Thévenin
(or Norton) equivalent connected to either a single
equivalent inductor or capacitor.
RTh
RN
L
VTh
–
+
IN
C
 In steady state, an inductor behave like a short circuit.
 In steady state, a capacitor behaves like an open circuit.
5
• The natural response of an RL and RC circuit is its
behavior (i.e., current and voltage ) when stored
energy in the inductor or capacitor is released to the
resistive part of the network (containing no
independent sources)
• The steps response of an RL and RC circuits is its
behavior when a voltage or current source step is
applied to the circuit, or immediately after a switch
state is changed.
6
5-1 Natural Response of an RL
circuit
 Consider the following circuit, for which the switch is
closed for t<0, and then opened at t = 0:
t=0
Is
Ro
i
L
+
R
V
–
 The dc voltage V, has been supplying the RL circuit with
constant current for a long time
7
Solving the circuit
• For t ≤ 0, i(t) = Io
• For t ≥ 0, the circuit reduce to
i
Io
Ro
L
+
R
v
–
• At t = 0, the inductor has initial current Io, hence i(0) = Io
• The initial energy stored in the inductor is,
w(0) 
1
LI 02
2
8
Cont.
• Applying KVL to the circuit:
v(t )  Ri (t )  0
di (t )
L
 Ri (t )  0
dt
di (t )
L
  Ri (t )
dt
di (t )
R
  dt
i (t )
L
(1)
(2)
(3)
(4)
•From equation (4), let say;
du
R
  dv
u
L
(5)
9
Cont.
• Integrate both sides of equation (5);
du
R t
i (to ) u   L to dv
i (t )
(6)
• Therefore,
i (t )
R
ln
 t
i (0)
L
(7)
• hence, the current is
i(t )  i(0)e
( R / L )t
 I 0e
 ( R / L )t
10
Cont.
• From the Ohm’s law, the voltage across the resistor R is:
v(t )  i (t ) R  I 0 Re
 ( R / L )t
• And the power dissipated in the resistor is:
p  vR i (t )  I Re
2
0
2 ( R / L ) t
• Energy absorb by the resistor is:
1
w  LI 02 (1  e  2 ( R / L ) t )
2
11
Time Constant, τ for RL circuit
• Time constant, τ determines the rate
at which the current or voltage
approaches zero.
• The time constant of a circuit is the
time required for the response to
decay to a factor of 1/e or 36.8% of its
initial current
• Natural response of the RL circuit is
an exponential decay of the initial
current. The current response is
shown in Fig. 5-1
• Time constant for RL circuit is
L
 
R
• And the unit is in seconds.
Figure 5-1
12
• The expressions for current, voltage, power and energy
using time constant concept:
i (t )  I 0 e
 t /
v (t )  I 0 Re
p  I Re
2
0
t / 
 2t /
1
2
 2t /
w  LI 0 (1  e
)
2
13
Switching time
• For all transient cases, the following instants of switching times are
considered.
 t = 0- , this is the time of switching between -∞ to 0 or time before.
 t = 0+ , this is the time of switching at the instant just after time t = 0s
(taken as initial value)
 t = ∞ , this is the time of switching between t = 0+ to ∞ (taken as final
value for step response)
• The illustration of the different instance of switching times is:
-∞
t  0
∞
t  0
14
Example 1
• For the circuit below, find the expression of io(t) and Vo(t). The
switch was closed for a long time, and at t = 0, the switch was
opened.
2Ω
i0
t=0
20A
0.1Ω
+
i
2H 10Ω
40Ω
V
L
–
15
Solution :
• When t < 0, switch is closed and the inductor is short circuit.
2Ω
0.1Ω
20A
i (0-)
10Ω
40Ω
L
Therefore;
iL(0-) = 20A
• When t > 0, the switch is open and the circuit become;
2Ω
Hence; iL(0+) = iL(0-) = 20A
io(0+)
+
20A
iL(0+)
2H 10Ω
40Ω
vo(0+)
–
Current through the
inductor remains
the same
(continuous)
16
RT = (2+10//40) = 10Ω
L
2

 0.2 sec
So, time constant,  
RT 10
By using current division, the current in the 40Ω resistor is:
10
io  iL
 4 A
10  40
Using ohm’s Law, the Vo is,
Hence:
Vo (t )  4  40  160V
5t
io (t )  4e A
V0 (t )  160e
5t
17
Example 2
The switch in the circuit below has been closed for a long time.
At t = 0, the switch is opened. Calculate i(t) for t > 0.
4Ω
2Ω
i(t)
t=0
40V +

12Ω
16Ω
2H
18
Solution :
• When t < 0, the switch is closed and the inductor is short circuit to dc.
The 16Ω resistor is short circuit too.
4Ω
i 2Ω
1
+
40V 
12Ω
i(0-)
using current division, calculate i(0-)
12
i(0 ) 
i1  6 A
12  4
-
calculate i1; i 
1
40
 8A
2  4 // 12
Since the current through
an inductor cannot
change instantaneously
Hence; i(0) = i (0-) = 6A
19
• When t > 0, the switch is open and the voltage source is disconnect.
4Ω
i(t)
16Ω
12Ω
Thus,
i(0+) = i(0) = i (0-) = 6A
2H
RT = Req = (12 + 4)// 16 = 8Ω
Time constant,  
hence;
Because current
through the
inductor is
continuously
L 1
 sec
Req 4
i(t )  i(0)e-t/  6e 4t A
20
5-2 Natural Response of an RC
Circuit
• The natural response of RC circuit occurs when its dc source is
suddenly disconnected. The energy already stored in the
capacitor, C is released to the resistors, R.
• Consider the following circuit, for which the switch is closed for t
< 0, and then opened at t = 0:
+
Vo 
Ro
C
t=0
+
v
–
R
21
Solving the circuit
• For t ≤ 0, v(t) = Vo
• For t > 0, the circuit reduces to
i
Vo
+

+
Ro
C
v
R
–
• At t = 0, the initial voltage v(0) = Vo
• The initial value of the energy stored is
1
w(0)  CVo2
2
22
Cont.
• Applying KCL to the RC circuit:
ic  iR  0
dv(t ) v(t )
C

0
dt
R
dv(t ) v(t )

0
dt
RC
dv(t )
v(t )

dt
RC
dv(t )
1

dt
v(t )
RC
(1)
(2)
(3)
(4)
(5)
23
Cont.
• From equation (5), let say:
dx
1

dy
x
RC
(6)
• Integrate both sides of equation (6):

v (t )
Vo
1
1
du  
x
RC
t
 dy
(7)
0
• Therefore:
ln
v (t )
t

Vo
RC
(8)
24
Cont.
• Hence,
• The voltage is:
v(t )  v(0)e
• Using Ohm’s law, the current is:
 t / RC
 Vo e
 t / RC
v(t ) Vo t / RC
i (t ) 
 e
R
R
• The power dissipated in the resistor is:
Vo2  2t / RC
p(t )  viR 
e
R
• The energy absorb by the resistor is:
1
w  CVo2 (1  e  2t / RC )
2
25
Time Constant, τ for RC circuit
• The time constant for the RC circuit equal the product of the
resistance and capacitance,
• Time constant,
sec
  RC
• The natural response of RC circuit illustrated graphically in
Fig 5.2
Figure 5.2
26
• The expressions for voltage, current, power and energy using
time constant concept:
v (t )  Vo e
 t /
Vo t /
i (t ) 
e
R
2
Vo  2t /
p (t ) 
e
R
1
2
 2t /
w(t )  CVo (1  e
)
2
27
Example 3
The switch has been in position a for a long time. At time t = 0,
the switch moves to b. Find the expressions for the vc(t), ic(t) and
vo(t).
5kΩ
a
b
18kΩ
t=0
90V
+

10kΩ
0.1μF
60kΩ
12kΩ
+
Vo
–
28
Solution
At t < 0, the switch was at a. the capacitor behaves like an open
circuit as it is being supplied by a constant source.
5kΩ
10
+
vc (0 ) 
 90  60V
15
Vc(0-)

90V
+

10kΩ
–
At t > 0, the instant when the switch is at b.
18kΩ
+
60kΩ
12kΩ
Vc(0+)
–
0.1μF
+
Vo
–
the voltage across
capacitor remains
the same at this
particular instant
vc(0+) = vc(0-) = 60V
29
RT = (18 kΩ + 12 kΩ) // 60 kΩ = 20 kΩ
time constant, τ = RTC = 20kΩ x 0.1 μF = 2ms
Vc(t) = 60e-500t V
Using voltage divider rule, Vo (0  ) 
Hence,
12
 60  24V
30
Vo(t) = 24e-500t V
dv
i c (t)  C
 - 3e -500t
dt
30
Example 4
The switch in the circuit below 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.
3Ω
20V +

t=0
9Ω
1Ω
+
v
–
20mF
31
Solution
For t<0, switch is closed and capacitor is open circuit.
3Ω
1Ω
20V +

+
vc(0-)
–
9Ω
vc (0  ) 
9
(20)  15V
93
For t>0, the switch is open and the RC circuit is
1Ω
9Ω
+
vc(0+)
–
20mF
32
Because;
Req = 1+9 = 10Ω
So ;
Time constant, τ = ReqC = 0.2s
Because; v (0) = v (0+) = v (0-) = 15 V
c
c
c
Voltage across the capacitor; v(t) = 15e-5t V
Initial energy stored in the capacitor is;
wc(0) = 0.5Cvc2 =2.25J
33
Summary
No RL circuit
1
2
3
L

R
Inductor behaves like a
short circuit when being
supplied by dc source for
a long time
Inductor current is
continuous
iL(0+) = iL(0-)
RC circuit
  RC
Capacitor behaves like an
open circuit when being
supplied by dc source for a
long time
Voltage across capacitor is
continuous
vC(0+) = vC(0-)
34
5-3 Step Response of RL Circuit
• The step response is the response of the circuit due to a sudden
application of a dc voltage or current source.
• Consider the RL circuit below and the switch is closed at time t = 0.
i
Vs
+

R
+
t=0
L
v(t)
–
• After switch is closed, using KVL
di
Vs  Ri (t )  L
dt
(1)
35
Cont.
• Rearrange the equation;
Vs 
di (t )  Ri (t )  Vs  R 


i
(
t
)



dt
L
L 
R
(2)
 R  Vs 
di 
 i  dt
L 
R
(3)
R
di
dt 
L
i (t )  Vs R
(4)
i (t )
R t
du
  dv  
0
L 0
u  (Vs R)
•Therefore:

i (t )  (Vs R )
R
t  ln
L
I 0  (Vs R )
(5)
(6)
36
Cont.
• Hence, the current is; i (t ) 
• Or may be written as;
Vs 
Vs  ( R / L ) t
  I o  e
R 
R
iL (t )  i()  [i(0)  i()]e
t /
Where i(0) and i(∞) are the initial and final values of i, respectively.
• The voltage across the inductor is;
• Or;
v(t )  (Vs  I o R)e  ( R / L )t
di
vL (t )  L
dt
37
Example 5
The switch is closed for a long time at t = 0, the switch opens.
Find the expressions for iL(t) and vL(t).
t=0
10V
+

2Ω
3Ω
i
1/4H
38
Solution
When t < 0, the 3Ω resistor is short – circuit, and the inductor acts like
short circuit.
10V
2Ω
+

iL (0 )  10 / 2  5 A
iL(0-)
When t< 0, the switch is open and the both resister are in series.
So;
10V +

2Ω
3Ω
i(0) = i(0+) = i(0-) = 5A
iL(0+)
1/4H
Because inductor current
cannot change
instantaneously
39
RT  ( 2  3)  5
Time constant;  
L
1

s
RT
20
When t = ∞, the inductor acts as short circuit again.
10V +

2Ω
iL ()  Vs / RT  2 A
3Ω
iL(∞)
Thus: iL(t) = i(∞) +[i(0) – i(∞)]e-t/τ = 2 + 3e-20t A
And the voltage is: vL (t )  L
di
dt
= -15e-20t V
40
5-4 Step Response of RC Circuit
• Consider the RC circuit below. The switch is closed at time t = 0
+
t=0
Is
R
C
i
• From the circuit;
dvc vc
Is  C

dt R
vc(t)
–
(1)
• Division of Equation (1) by C gives;
I s dvc vc


C
dt RC
(2)
41
Cont.
• Same mathematical techniques with RL, the voltage is:
vc (t )  I s R  (Vo  I s R)e  t / RC
• Or can be written as:
v(t) = v(∞) + [v(0) – v(∞)]e-t/τ
Vo  t / RC

• And the current is: i (t )   I s  e
R

• Or can be written as:
dv
i (t )  C
dt
42
Example 6
The switch has been in position a for a long time. At t = 0, the
switch moves to b. Find Vc(t) for t > 0 and calculate its value at
t=1s and t=4s
3kΩ
24V
+

5kΩ
a
+
Vc
–
4kΩ
b
t=0
0.5mF
+

30V
43
Solution
When t<0, the switch is at position A. The capacitor acts like an open circuit.
3kΩ
Using voltage division:
24V
+

5kΩ
+
Vc (0-)
–
5
Vc (0 )  24   15V
8

Since voltage across
the capacitor remains
same.
When t <0, the switch is at position B.
4kΩ
0.5mF
V (0)  Vc (0  )  Vc (0  )  15V
+
 30V
And the time constant is:
  RC  2s
44
At t = ∞, the capacitor again behaves like an open circuit.
4kΩ
+
Vc(∞)
–
+
 30V
Hence;
Vc ()  30V
Since, v(t) = v(∞) + [v(0) – v(∞)]e-t/τ
So; vc(t) = 30-15e-0.5t V
And;
At t = 1s, Vc(t) = 20.9V
At t = 4s, Vc(t) = 28 V
45