ELL 100 - Introduction to Electrical Engineering
LECTURE 9:
TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS
(NATURAL RESPONSE)
SOURCE-FREE RC CIRCUITS
EXAMPLE
Fluid-flow analogy:
water tank emptying
through a small pipe
Electrical circuit:
Capacitor discharging
through resistance
2
SOURCE-FREE RC CIRCUITS
EXAMPLE
Mechanical system
(velocity decays
through damping)
Electrical equivalent
(charge dissipated
through resistance)
3
SOURCE-FREE RC CIRCUITS
APPLICATIONS
High pass filter
Low pass filter
4
SOURCE-FREE RC CIRCUITS
APPLICATIONS
Timers
Camera Flash
Oscillators
555 Timer
circuits
5
SOURCE-FREE RC CIRCUITS
APPLICATIONS
Delay Circuits
Warning Blinkers
6
SOURCE FREE RC CIRCUITS
APPLICATIONS
Computer Circuits
Digital and Time delay circuits
7
SOURCE FREE RC CIRCUITS
APPLICATIONS
Pacemakers
Timing device in automobile
intermittent wiper system
8
SOURCE-FREE RL CIRCUIT
APPLICATIONS
Pulse Generators
Electronic filter
Tubelight choke
9
TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS
• A first-order circuit is characterized by a first-order
differential equation.
• Example :
• a circuit comprising a resistor and capacitor
(RC circuit)
• a circuit comprising a resistor and an inductor
(RL circuit)
Applying Kirchhoff’s laws to RC or RL circuit results in differential
equations involving voltage or current, which are first-order.
10
TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS
EXCITATION
There are two ways to excite the circuits.
• Initial conditions of the storage elements– Source-Free Circuits
(Energy stored in the capacitor, Energy stored in the inductor)
• Independent sources – Forced Excitation circuits
(DC sources, Sinusoidal sources, Exponential Sources)
11
TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS
NATURAL RESPONSE
• The natural response of a circuit refers to the behavior (in terms of
voltage or current) with no external sources of excitation.
• The circuit has a response only because of the energy initially stored in
the energy storage elements (i.e. capacitor or inductor).
12
SOURCE-FREE RC CIRCUIT
• A source-free RC circuit occurs when its DC
source is suddenly disconnected.
• The energy already stored in the capacitor(s)
is released to the resistor(s) & dissipated.
• RC source-free circuit is analyzed from its
initial voltage v(0) = V0 and time constant τ
13
SOURCE-FREE RC CIRCUIT
DERIVATION
• Assume the voltage v(t) across the capacitor.
• Since the capacitor is initially charged,
Assume that at time t = 0, the initial voltage is,
v(0)  V0
with the corresponding value of the energy stored as
1
w(0) 
CV0 2
2
14
SOURCE-FREE RC CIRCUIT
DERIVATION
Applying KCL at the top node of the circuit, yields
iC + i R = 0
By definition, iC = C dv∕dt and iR = v ∕ R. Thus,
dv v
C
 0
dt R
or
dv
v

0
dt RC
This is a first-order differential equation.
15
SOURCE-FREE RC CIRCUIT
DERIVATION
dv
v

 0 =>
dt RC
dv
1

dt
v
RC
t
Integrating both sides, we get ln v  
 ln A
RC
v
t

=> ln
A
RC
 t / RC
v
(
t
)

A
e
=>
But from the initial conditions, v(0) = A = V0.
Hence, v(t )  V0 et / RC (Exponentially Decaying)
16
SOURCE-FREE RC CIRCUIT
VOLTAGE RESPONSE
• As t increases, the voltage decreases exponentially towards zero.
The rapidity with which the voltage decreases is expressed in
terms of the time constant, denoted by τ.
17
SOURCE-FREE RC CIRCUIT
TIME CONSTANT
The time constant of a circuit is the time required for the response to
decay to a factor of 1/e or 36.8 percent of its initial value.
v(t )  V0 e
 t / RC
V0 e- /RC = V0e-1  0.368V0
  RC
v(t )  V0 et /
18
SOURCE-FREE RC CIRCUIT
TIME CONSTANT
t
τ
2τ
3τ
4τ
5τ
v(t)/V0
0.36788
0.13534
0.04979
0.01832
0.00674
Graphical determination of the time constant τ from the response curve.
19
SOURCE-FREE RC CIRCUIT
TIME CONSTANT
20
SOURCE-FREE RC CIRCUIT
POWER DISSIPATION
The power dissipated in the resistor is
p (t )  viR
V0 2 2t /

e
R
The energy absorbed by the resistor up to time t is
t
wR (t ) 

0

t
V0 2 2  /
p ( )d   
e
d
R
0
 V0 2
2R
e
2  /
1
-2t/τ
2  /
|  CV0 2 (1

e
(1 – e )),
2
t
0
  RC
1
t  , wR ( )  CV0 2
2
21
SOURCE-FREE RL CIRCUIT
• A circuit with series connection
of a resistor and inductor
• Current i(t) through the inductor is
considered as response of this system.
At t = 0, assume that the inductor has an initial
current I0, or i (0)  I 0
1
Initial energy stored in the inductor w(0)  LI 0 2
2
22
SOURCE-FREE RL CIRCUIT
RESPONSE OF THE CIRCUIT
Applying KVL around the loop, vL  vR  0
vL = L di/dt and vR = iR. Thus,
di
L  Ri  0
dt
di R
 i0
dt L

i (t )
=>

I0
t
di
R
   dt
i
L
0
i (t )
Rt

=> ln
I0
L

i (t )  I 0 e  Rt / L
23
SOURCE FREE RL CIRCUIT
RESPONSE OF THE CIRCUIT
• Current through inductor decays
exponentially i (t )  I 0e  Rt / L
• Time constant for the RL circuit is
L

R
i (t )  I 0 e t /
24
SOURCE FREE RL CIRCUIT
POWER DISSIPATION
- t /
v
(
t
)

iR

I
Re
Voltage across the resistor is R
0
2
-2 t /
p

v
i

I
Re
The power dissipated in the resistor is
R
0
The energy absorbed by the resistor is
t
wR (t ) 

p ( )d  
0
t
2
2  /
I
Re
d
 0
0
  I 0 Re
2
2  /
1
|  LI 0 2 (1  e 2  / ),
2
t
0
  L/R
1
t  , wR ()  LI 0 2
2
25
SOLVING NUMERICALS
Points to remember :
Elements
DC steady state
R
R
L
Short-circuit
(v = 0)
Open-circuit
(i = 0)
C
Continuous quantity
(from t=0- to t=0+)
Current i
Voltage v
26
SOURCE-FREE RC CIRCUIT
Q1. Consider the circuit below. Let vC (0)=15 V. Find vc , vx and ix for t > 0.
27
SOURCE-FREE RC CIRCUIT
Solution :
• We first convert the given circuit into a simple R-C circuit.
• Find the equivalent resistance or the Thevenin resistance at the
capacitor terminals.
28
SOURCE-FREE RC CIRCUIT
Req
20  5

 4
20  5
The time constant is   R e q C  4(0.1)  0.4s
29
SOURCE-FREE RC CIRCUIT
 t /
2.5 t
vC(t )  V0 ve(t )  15 e  tv/0.4
(tV
)  15 e V
we can use voltage division to get vx
12
2.5 t
2.5 t
vx 
vC  0.6(15e
)  9e
V
12  8
vx
2.5 t
ix 
 0.75e
A
12
30
SOURCE-FREE RC CIRCUIT
Q2. The switch in the circuit below is closed for a long time, and
then opened at t = 0. Find v(t) for t ≥ 0. Also calculate the energy
stored in the capacitor before opening of the switch.
31
SOURCE-FREE RC CIRCUIT
Solution:
For t < 0, the switch is closed and the capacitor is an open circuit in
steady state, as represented in Fig.(a).
Using voltage division
9
vC (t ) 
(20)=15 V , t  0
93
Since the voltage across a capacitor cannot change instantaneously, the
voltage across the capacitor at t = 0− is the same at t = 0+, or
vC (0)  V0  15V
32
SOURCE-FREE RC CIRCUIT
Solution:
For t > 0, the switch is open, and we have the RC circuit shown in Fig. (b),
R eq  1 + 9  10 
The time constant is
 = R eq C  10  20 10  0.2s
3
Thus, the voltage across the capacitor for t ≥ 0 is
v(t )  vC (0) e t/  15e t/0.2 V  15e5t V
1
1
The initial energy stored
2
w C (0)  Cv C (0)   20 103 152  2.25 J
2
2
in the capacitor is:
33
SOURCE-FREE RL CIRCUIT
Q3. Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit below
34
SOURCE-FREE RL CIRCUIT
Solution: There are two ways we can solve this problem
Method -1:
The equivalent resistance is the same as the Thevenin resistance at the
inductor terminals. Because of the dependent source, we insert a voltage
source with vo = 1 V at the inductor terminals a-b, as shown below
35
SOURCE-FREE RL CIRCUIT
Applying KVL to the two loops,
1
2(i1  i2 )  1  0  i1  i2  
2
5
6i2  2i1  3i1  0  i2  i1
6
(1)
(2)
Substituting Eq. (2) into Eq. (1) gives
=>
i1= -3A, i0= - i1 =3A
The
The
R e q  RTh
vo
1

 
io
3
1
L
2  3s



time constant is
1
R eq
2
3
current through the inductor is i (t )  i (0)e  t /  10e( 2/3) t A,
t 0
36
SOURCE-FREE RL CIRCUIT
Method-2:
Applying KVL to the circuit For loop 1,
1 di1
 2(i1  i2 )  0
2 dt
For loop 2,
6i2  2i1  3i1  0

(3)
5
i2  i1
6
Substituting above into Eq. (3) gives
i (t)
2 t
  t |0
=> ln
i (0)
3
di1 2
 i1  0
dt
3
 i (t )  i (0)e  ( 2/3) t  10e  ( 2/3) t A,
t 0
37
SOURCE-FREE RL CIRCUIT
The voltage across the inductor is
di
10  ( 2/3) t
 2   ( 2/3) t
vL
 0.5(10)    e

e
V
dt
3
 3
Since the inductor and the 2-Ω resistor are in parallel,
v
ix (t )   1.6667e  (2/3) t A,
2
t 0
38
SOURCE-FREE RL CIRCUIT
Q4. The switch in the circuit below is closed for a long time.
At t = 0, the switch is opened. Calculate i(t) for t > 0.
39
SOURCE-FREE RL CIRCUIT
Solution:
For t < 0, the switch is closed, and the inductor acts as a short circuit in
steady state. The 16-Ω resistor is short-circuited; the resulting circuit is
shown in Fig (a).
Req
4 12

 2  5Ω
4  12
Ω
40
i1 
8A
5
We obtain i(t) from i1 using current division,
12
i (t ) 
i1  6 A, t  0
12  4
40
SOURCE-FREE RL CIRCUIT
Since the current through an inductor cannot change instantaneously,
i (0)  i (0 )  6 A
For t > 0, the switch is open and the voltage source is disconnected.
We now have the source-free RL circuit in Fig.(b).
Combining the resistors, we have
R e q  (12  4) ||16  8 
The time constant is
L
2 1
 
  s
Req
8
4
i (t )  i (0)e  t /  6e 4t A
41
SOURCE-FREE RL CIRCUIT
Q5. In the circuit shown below, find io, vo, and i for all t > 0,
assuming that the switch was open for a long time and closed at t = 0.
42
SOURCE-FREE RL CIRCUIT
Solution :
It is better to first find the inductor current i and then obtain other
quantities from it.
For t < 0, the switch is open. Since the inductor acts like a short circuit to
DC, the 6-Ω resistor is short-circuited, so that we have the circuit shown
Hence, io = 0 and
10
i (t ) 
 2A t  0
23
vo (t )  3i(t)  6 V t  0
43
SOURCE-FREE RL CIRCUIT
Thus, i(0) = 2 A
For t > 0, the switch is closed, so that the voltage source is short-circuited
We now have a source-free RL circuit as shown. At the inductor
terminals, RTh  3 || 6  2
Thus the time constant is
L

 1s
RTh
Hence,
i (t )  i(0)e
 t /
 2e
 t /
t 0
44
SOURCE-FREE RL CIRCUIT
Because the inductor is in parallel with the 6-Ω and 3-Ω resistors,
di
vo (t )  vL   L  2(2et )  4etV ,
dt
vL 2  t
io (t )  
e A, t  0
6
3
Thus for all time,
t0
0 A

io (t )   2  t
 3 e , t  0
6V
vo (t )   t
4e ,
2 A
i (t )    t
2e ,
t0
t0
t 0
t0
t0
45
SOURCE-FREE RL CIRCUIT
Q6. The switch ‘S’ is
kept in position ‘1’ for
a long time and then
suddenly changed to
position ‘2’ at t = 0 as
shown
Compute the value of vL and iL
i. At the instant just prior to the switch changing (t = 0-)
ii. At the instant just after the switch changes (t = 0+)
Also find the rate of change of current through the inductor at t = 0+
46
SOURCE-FREE RL CIRCUIT
Solution :
At t = 0- the current through and the voltage across the inductor are
10
iL (0 ) 
10  5 A;
10  10
v L (0  )  0 V (inductor acts as short-circuit)
At t = 0+, iL (0 )  5 A;
v L (0 )  (10  10)  5  100 V

(KVL)
The rate of change of current through inductor at time t = 0+ is
dil (t )
dil (t )
100
L
 100 V 

 25 A / s
dt t 0
dt t 0
4
47
PRACTICE PROBLEMS
48
SOURCE FREE RC CIRCUIT
Q1. Calculate time constants of the following circuits.
(a)
Answer: (a) 6μs
(b)
(c)
(b) 1ms ( c) 0.25 sec
49
SOURCE FREE RC CIRCUIT
Q2. Switch ‘S’ shown in fig. is kept in position ‘1’ for a long time.
When the switch is thrown in position ‘2’, find at steady state condition
(i) the voltage across the each capacitor (ii) the charge across the each
capacitor (iii) the energy stored by the each capacitor
Answer: (i) V/2
(ii)CV/2
(iii)CV2/8
50
SOURCE FREE RC CIRCUIT
Q3. In the circuit shown in Fig.
v(t) = 56 e −200t V, t > 0
i(t) = 8 e −200t mA, t > 0
(a) Find the values of R and C.
(b) Calculate the time constant τ.
(c) Determine the time required for the voltage to decay half its initial
value at t = 0.
Answer: (a) 0.7143 μF, (b) 5 ms, (c) 3.466 ms
51
SOURCE FREE RC CIRCUIT
Q4. In the circuit shown in Fig. Determine the charge lost by the
capacitor from 25μs to 100 μs in coulombs. Consider, v(0) = 4 V, C=5μC,
R=5Ω.
Answer: 7 μC
52
SOURCE FREE RC CIRCUIT
Q5. Refer to the circuit in Fig. Let vC (0) = 60 V. Determine vC, vx, and io
for t ≥ 0.
Answer: 60e−0.25t V, 20e−0.25t V, −5e−0.25t A.
53
SOURCE FREE RC CIRCUIT
Q6. If the switch in Fig. opens at t = 0, find v(t) for t ≥ 0 and wC (0).
Answer: 8 e −2t V, 5.333 J.
54
SOURCE FREE RL CIRCUIT
Q7. Find i and vx in the circuit of Fig. Let i(0) = 7 A.
Answer: 7 e −2t A, −7 e −2t V, t > 0.
55
SOURCE FREE RL CIRCUIT
Q8. For the circuit in Fig., find i(t) for t > 0..
Answer: 2 e −2t A, t > 0.
56
SOURCE FREE RL CIRCUIT
Q9. For the circuit in Fig, find io for t > 0.
Answer: 1.2 e −3t A, t > 0.
57