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CHAPTER 5
DC TRANSIENT
ANALYSIS
1
SUB - TOPICS




NATURAL RESPONSE OF RL CIRCUIT
NATURAL RESPONSE OF RC CIRCUIT
STEP RESPONSE OF RL CIRCUIT
STEP RESPONSE OF RC CIRCUIT
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OBJECTIVES


To 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.
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FIRST – ORDER CIRCUITS



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.
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R
R
i
i
L
–
+
An RL circuit
vs
–
+
Vs
C
An RC circuit
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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.
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

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.
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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
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Solving for the circuit


For t ≤ 0, i(t) = Io
For t ≥ 0, the circuit reduce to
i
Io
Ro
L
+
R
v
–
Notation:

0  is used to denote the time just prior to switching.

0  is used to denote the time immediately after switching.
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Continue…

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)
10
Continue

From equation (4), let say;
du
R
  dv
(5)
u
L

Integrate both sides of equation (5);
i ( t ) du
R t
(6)
i (to ) u   L to dv
Where:
i(to) is the current corresponding to time to
i(t) ia the current corresponding to time t



11
Continue


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
12
Continue

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
13
Continue

Energy absorb by the resistor is:
1 2
2( R / L )t
w  LI 0 (1  e
)
2
14
Time Constant, τ


Time constant, τ determines the rate at which the
current or voltage approaches zero.
Time constant,
L
 
R
(sec)
15

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
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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)
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
The illustration of the different instance of switching
times is:
∞
-∞
t0

t0

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Example

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
L
40Ω
V
–
19
Solution :
Step 1:
Find τ for t > 0. Draw the equivalent circuit. The
switch is opened.
RT  (2  10 // 40)  10
So;
L
2

  0.2
RT 10
sec
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Step 2:
At t = 0- , time from -∞ to 0-, the switch was closed for
a long time.
2Ω
20A
0.1Ω
i (0-)
10Ω
40Ω
L
The inductor behave like a short circuit as it being
supplied for a long time by a dc current source. Current
20A thus flows through the short circuit until the switch
is opened.
Therefore;
iL(0-) = 20A
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Step 3:
At the instant when the switch is opened, the time t = 0+,
2Ω
io(0+)
+
20A
iL(0+)
2H 10Ω
40Ω
vo(0+)
–
The current through the inductor remains the same (continuous).
Thus,
which is the initial current.


iL (0 )  iL (0 )  20 A
Only at this particular instant the value of the current through the
inductor is the same.
Since, there is no other supply in the circuit after the switch is
opened, this is the natural response case.
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By using current division, the current in the 40Ω resistor
is:
10
io  iL
10  40
 4 A
So,
io (t )  4e
5t
A
Using Ohm’s Law, the Vo is:
Vo (t )  4  40  160
So,
V0 (t )  160e
5t
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NATURAL RESPONSE OF AN RC
CIRCUIT

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
Notation:
 0- is used to denote the time just prior to switching
 0+ is used to denote the time immediately after switching.
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Solving for the voltage (t ≥ 0)


For t ≤ 0, v(t) = Vo
For t > 0, the circuit reduces to
i
Vo
+

+
Ro
C
v
R
–
25
Continue

Applying KCL to the RC circuit:
iC  iR  0
(1)
dv(t ) v(t )
C

0
dt
R
(2)
dv(t ) v(t )

0
dt
RC
(3)
dv (t )
v (t )

dt
RC
(4)
dv(t )
1

dt
v(t )
RC
(5)
26
Continue

From equation (5), let say:
dx
1

dy
x
RC

Integrate both sides of equation (6):

v (t )
Vo

(6)
1
1
du  
x
RC
t
 dy
0
(7)
Therefore:
v (t )
t
ln

Vo
RC
(8)
27
Continue

Hence, the voltage is:
v(t )  v(0)e

 t / RC
 Vo e
 t / RC
Using Ohm’s law, the current is:
v(t ) Vo t / RC
i (t ) 
 e
R
R
28
Continue

The power dissipated in the resistor is:
2
o
V 2t / RC
p(t )  viR 
e
R

The energy absorb by the resistor is:
1
2
 2 t / RC
w  CVo (1  e
)
2
29
Continue

The time constant for the RC circuit equal the product
of the resistance and capacitance,

Time constant,
  RC
sec
30

The expressions for voltage, current, power and
energy using time constant concept:
v(t )  Vo e
 t /
Vo t /
i (t ) 
e
R
Vo2  2t /
p (t ) 
e
R
1
2
 2t /
w(t )  CVo (1  e
)
2
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
For the case of capacitor, two important observation
can be made,
1) capacitor behaves like an open circuit when being
supplied by dc source
(From, ic = Cdv/dt, when v is constant, dv/dt = 0.
When current in circuit is zero, the circuit is open
circuit.)
2) in capacitor, the voltage is continuous / stays the
same that is, Vc(0+) = Vc(0-)
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Example
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) and hence sketch them for t =
0 to t = 5τ.
5kΩ
a
b
18kΩ
t=0
90V
+

10kΩ
0.1μF
60kΩ
12kΩ
+
Vo
–
33
Solution
Step 1:
Find t for t > 5τ that is when the switch was at a. Draw
18kΩ
the equivalent circuit.
0.1μF
12kΩ
60kΩ
RT  (18k  12k) // 60k  20k
  RT C  20 10  0.110  2ms
3
6
34
Step 2:
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
90V
+

10kΩ
+
Vc(0-)
–
10
vc (0 )   90  60V
15

35
Step 3:
At t = 0+, the instant when the switch is at b.
18kΩ
+
60V
0.1μF
–
60kΩ
12kΩ
+
Vo
–
The voltage across capacitor remains the same at this
particular instant.
vc(0+) = vc(0-) = 60V
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Using voltage divider rule,
12
Vo (0 ) 
 60  24V
30

Hence;
vc (t )  60e
500t
vo (t )  24e
500t
V
ic (t )  0.03e
V
500t
A
37
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-)
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Step Response of RL Circuit

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)
39
Continue

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)
(5)
40
Continue

Therefore:
i (t )  (Vs R)
R
 t  ln
L
I 0  (Vs R)

(5)
Hence, the current is;
Vs 
Vs  ( R / L ) t
i (t )    I o  e
R 
R

The voltage;
v(t )  (Vs  I o R)e
( R / L )t
41
Example
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
1/4H
42
Solution
Step 1:
Find τ for t > 0. The switch was opened. Draw the
equivalent circuit. Short circuit the voltage source.
2Ω
3Ω
1/4H
RT  (2  3)  5
L
1


s
RT 20
43
Continue
Step 2:
At t = 0-, the switch was closed. Draw the equivalent
circuit with 3Ω shorted and the inductor behaves like a
short circuit.
10V
+

2Ω
iL(0-)

iL (0 )  10 / 2  5 A
44
Continue
Step 3:
At t = 0+, the instant switch was opened. The current in
inductor is continuous.


I 0  iL (0 )  iL (0 )  5 A
Step 4:
At t =∞, that is after a long time the switch has been left
opened. The inductor will once again be behaving like a
short circuit.
45
Continue
10V
Hence:
+

2Ω
3Ω
iL(∞)
iL ()  Vs / RT  2 A
Vs 
Vs  ( R / L )t
iL (t )    I o  e
R 
R
iL (t )  2  3e 20t A
46
Continue

And the voltage is:
vL (t )  (Vs  I o R)e
vL (t )  15e
( R / L )t
20t
V
47
Step Response of RL Circuit

The switch is closed at time t = 0
+
t=0
Is
R
C
i

vc(t)
–
From the circuit;
dvc vc
Is  C

dt
R
(1)
48
Continue

Division of Equation (1) by C gives;
I s dvc vc


C
dt RC

Same mathematical techniques with RL, the voltage is:
vc (t )  I s R  (Vo  I s R)e

(2)
 t / RC
And the current is:
Vo  t / RC

i (t )   I s  e
R

49
Example
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
50
Solution
Step 1:
To find τ for t > 0, the switch is at b and short circuit the
voltage source.
4kΩ
0.5mF
  RC  2s
51
Continue
Step 2:
The capacitor behaves like an open circuit as it is being
supplied by a constant dc source.
3kΩ
24V
+

+
Vc (0-)
–
5kΩ
From the circuit,
5
Vc (0 )  24   15V
8

52
Continue
Step 3:
At t = 0+, the instant when the switch is just moves to b.
Voltage across capacitor remains the same.


Vc (0 )  Vc (0 )  15V
Step 4:
At t = ∞, the capacitor again behaves like an open
circuit since it is being supplied by a constant source.
4kΩ
+
Vc(∞)
–
+

30V
Vc ()  30V
53
Continue
Step 5:
Hence,
Vc (t )  30  (15  30)e
0.5t
 30  15e
0.5t
V
At t = 1s, Vc(t) = 20.9V
At t = 4s, Vc(t) = 28 V
54
THE END
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