First Order Circuits:
Simple RC and RL Circuits
Electric Circuits 2
Chapter Objectives
• RL and RC circuits are called first-order circuits. In
this chapter we will do the following:
– Develop a vocabulary that will help us talk about the
response of a first-order circuit.
– Analyze first-order circuits with inputs that are
constant after some particular time, to.
– Introduce the notion of a stable circuit and use it to
identify stable first-order circuits.
– Analyze first-order circuits that experience more than
one abrupt change.
– Introduce the step function and use it to determine
the step response of a first-order circuit.
– Analyze first-order circuits with inputs that are not
constant.
First-Order Circuits
• The order of a differential equation is the
order of the highest order derivative.
• The order of the differential equation is
usually equal to the number of capacitors plus
the number of inductors in the circuit.
• Circuits that contain only one inductor and no
capacitors or only one capacitor and no
inductors can be represented by a first-order
differential equation.
• These circuits are called first-order circuits.
Source – Free RC Circuit
• Consider the circuit,
KCL,
dv v
+ =0
C
dt R
or
dv
1
+
v=0
dt RC
Rearranging the terms and simplifying,
dv
1
=−
v
dt
RC
dv
1
=−
dt
v
RC
Source – Free RC Circuit
• From the given,
Taking the indefinite integral,
dv
1
∫ v = − RC ∫ dt
or
t
+K
ln v = −
RC
ln v(0) = ln V0 = K
v
t
ln v − ln V0 = ln = −
V0
RC
v
−t
= e RC
V0
Finally,
v(t ) = V0 e
−t
RC
Source – Free RC Circuit
• Graph of the voltage response in the simple
RC circuit,
v(t ) = V0 e
−t
RC
• The rate at which the voltage decays is determined solely by the
product of the resistance and the capacitance.
Source – Free RC Circuit
• Since the response is characterized by the
circuit elements and not by an external
voltage or current source, the response is
called the natural response of the circuit.
• Generally,
v(t ) = V0 e
− ( t −t 0 ) / RC
,
t ≤ t0
TIME CONSTANTS
Time Constants
• Considering the circuit,
• Where the voltage response is,
v = V0 e
− t / RC
Time Constants
• Graphs of v for different values for RC,
Time Constants
• The current in i the circuit, is given as
V0 −t / RC
i= e
R
• The current decreases in the same manner as
the voltage
Time Constants
• The time required for the natural response to
decay by a factor of 1/e is defined as the time
constant, τ
V0 e
− ( t +τ ) / RC
• Which yields,
V0 −t / RC
− ( t + RC ) / RC
= e
= V0 e
e
τ = RC
• Which has the units Ω-F = (V/A)(C/V)=(C/A) = s
Time Constants
• The voltage response in terms of the time
constant, τ
− t /τ
v = V0 e
• Knowledge of the time constant allows us to
predict the general form of the response.
• Another definition is that it is the time required
for the natural response to become zero if it
decreases at a constant rate equal to the rate of
decay.
• A first-order circuit is stable if the time constant
is not negative, τ ≥ 0.
SOURCE-FREE RL CIRCUIT
Source – Free RL Circuit
• Consider the circuit,
The energy stored at t = 0,
1
2
wL (0) = LI 0
2
KVL,
di
L + Ri = 0
dt
or
di R
+ i=0
dt L
Rearranging the terms and simplifying,
di
R
=− i
dt
L
di
R
= − dt
i
L
Source – Free RL Circuit
• From the circuit,
Taking the indefinite integral,
di
R
∫ i = − L ∫ dt
or
R
ln i = − t + K
L
ln i (0) = ln I 0 = K
Or ,
Finally,
i (t ) = I 0 e
− Rt
L
i (t ) = I 0 e
−t
τ
i
R
ln i − ln I 0 = ln = − t
I0
L
i
− Rt
=e L
I0
Source – Free RL Circuit
• Time Constant: τ = L / R
H / Ω = (V ⋅ s / A)(V / A) = s
• Units:
• Current Response:
Source – Free RL Circuit
• Instantaneous Power: p(t ) = Ri (t ) = RI 0 e
• Energy absorbed by resistor: w(∞) = 1 LI 0 2
2 −2 Rt / L
2
2
RESPONSE TO A CONSTANT
FORCING FUNCTION
Response to a constant forcing
function
• Consider the driven RC network:
Response to a constant forcing
function
• For t > 0:
Nodal Equation:
dv v
C + = I0
dt R
I0
1
dv
+
v=
dt RC
C
Separating the variables
v − RI 0
dv
=−
dt
RC
dv
1
∫ v − RI 0 = − RC ∫ dt
t
ln(v − RI 0 ) = −
+K
RC
Response to a constant forcing
function
• For t > 0:
K is the constant of integration:
t
ln(v − RI 0 ) = −
+K
RC
Rewriting the equation:
v − RI 0 = e
−
t
+K
RC
Solving for v:
v = Ae
Exponential function
−
t
RC
+ RI 0
Constant function
Response to a constant forcing
function
From the equation:
v = Ae
Exponential function
vn = Ae
−
t
RC
Natural response
−
t
RC
+ RI 0
Constant function
v f = RI 0
Forced response
Response to a constant forcing
function
Natural and Forced response:
v = Ae
−
t
RC
+ RI 0
v f = RI 0
vn = Ae
−
t
RC
Forced response
Natural response
Response to a constant forcing
function
Complete response:
v = Ae
−
t
RC
+ RI 0
Response to a constant forcing
function
– Evaluating the constant, A:
v = Ae
−
t
RC
+ RI 0
• At t=0+, observe that the constant A is now
determined not only by the initial voltage (or
energy) on the capacitor but also by the forcing
function Io
0
−
v(0) = V0 = Ae
V0 = A + RI 0
• Therefore,
RC
+ RI 0
or A = V0 − RI 0
v(t ) = RI 0 + (V0 − RI 0 )e
−
t
RC
THE GENERAL CASE
The General Case
• All equations of the previous sections are special
cases of a general expression given by,
dy
+ Py = Q
dt
Where:
y = unknown variable such as v or i
P and Q are constants
• Source-free RC circuit
dv
1
+
v=0
dt RC
• Forced RC circuit
I0
1
dv
+
v=
dt RC
C
• Source-free RL circuit
di R
+ i=0
dt L
The General Case
• Solving for y in
• We have,
dy
+ Py = Q
dt
y = e − Pt ∫ Qe Pt dt + Ae − Pt
Where:
A = constant of integration
Q = function of time or a constant
• In the important DC case, where Q is a
constant:
Q
− Pt
y = Ae +
P
y = yn + y f
Response of a First Order Circuit to a
Constant Input
• Thevenin and Norton equivalent circuits
simplify the analysis of first-order circuits by
showing that all first-order circuits are
equivalent to one of two simple first-order
circuits.
A plan for analyzing first-order circuits.
(a) First, separate the energy storage element
from the rest of the circuit.
(b) Next, replace the circuit connected to a
capacitor by its Thevenin equivalent circuit, or
replace the circuit connected to an inductor by
its Norton equivalent circuit.
Thevenin
Equivalent
Circuit
Capacitor
Norton
Equivalent
Circuit
Inductor
Response of a First Order Circuit to a
Constant Input (RC)
• Given this RC circuit,
• At t=0+, the
Thevenin equivalent
is in Fig. (b)
where,
R3
Voc =
VS
R2 + R3
R2 R3
Rt =
R2 + R3
d
i (t ) = C v(t )
dt
Response of a First Order Circuit to a
Constant Input (RC)
• KVL around the loop,
Voc = Rt i (t ) + v(t )

 d
= Rt  C v(t )  + v(t )

 dt
d
v(t ) Voc
=
v(t ) +
dt
Rt C Rt C
• The First-order D.E.
• Thus, the total response is:
v(t ) = Voc + (v(0) − Voc )e − t /( Rt C )
Response of a First Order Circuit to a
Constant Input (RL)
• Given this RL circuit,
• At t=0+, the Norton
equivalent is shown
in Fig. (b)
where,
VS
I sc =
R2
R2 R3
Rt =
R2 + R3
d
v(t ) = L i (t )
dt
Response of a First Order Circuit to a
Constant Input (RL)
• KCL at the top node,
v(t )
I sc =
+ i (t )
Rt
d
L i (t )
= dt
+ i (t )
Rt
Rt
Rt
d
i (t ) + i (t ) = I sc
dt
L
L
• The First-order D.E.
• Thus, the total response is:
i (t ) = I sc + (i (0) − I sc )e − ( Rt / L )t
SINGULARITY FUNCTIONS
Singularity functions
• Definition:
– Forcing functions whose values change abruptly
(e.g. voltage or current sources suddenly inserted
in switched circuits)
• There are many singularity functions that are
useful in circuit analysis:
– Step function
– Pulse
The Step Response
• the response of a circuit having only one input
which is a unit step function.
• can be a voltage or current
• initial energies are zero - ∞ < t < 0
The Unit Step Function
• The unit step function is the function equal to
zero for all negative values of its argument and
equal to 1 for all positive values of its
argument.
• Denoted as u(t),
u (t ) = 0,
t<0
= 1,
• Graphically,
t>0
Voltage step sources
• The unit step function may be used to
represent voltages or current with finite
discontinuities
• Voltage step source of V volts is represented
by the product Vu(t)
Voltage step source of V volts
Equivalent circuit
Current step sources
• Current step source of V volts is represented
by the product Iu(t)
Current step source of V volts
Equivalent circuit
Voltage and current step generators
Network with V applied at t=0
Network with I applied at t=0
The Unit Step Function delayed
• Generally,
u (t − t0 ) = 0,
t < t0
= 1,
t > t0
• Graphically,
Graph of step function u(t-to)
• The step response of a circuit is the response of the
circuit to the sudden application of a constant source
v= Vou(t-to) when all the initial conditions of the circuit
are equal to zero.
Pulse Signal
• A pulse signal has a constant nonzero value for
a time duration of Δt = t – to
• Consider the pulse source:
t < t0
0

v(t ) = V0 t0 < t < t1
0
t1 < t

Pulse Signal
• The pulse can be obtained using two-step voltage sources, Vo occurring at t
= to and –Vo occurring at t = t1
v1 (t ) = V0u (t − t0 )
v2 (t ) = −V0u (t − t1 )
Pulse Signal
• For t = to,
v1 (t ) = V0u (t − t0 )
• For t = t1,
v2 (t ) = −V0u (t − t1 )
v1 (t ) = V0u (t − t0 )
Thus,
v(t ) = v1 (t ) + v2 (t )
v(t ) = V0 [u (t − t0 ) − u (t − t1 )]
Checking:
For t < to,
v(t ) = V0 [0 − 0] = 0 V
v2 (t ) = −V0u (t − t1 )
For t0 < t < t1,
v(t ) = V0 [1 − 0] = V0
For t > t1,
v(t ) = V0 [1 − 1] = 0 V
t < t0
0

v(t ) = V0 t0 < t < t1
0
t1 < t

Pulse Signal: Equivalent Source
• The pulse can be obtained using two-step voltage sources, Vo occurring at t
= to and –Vo occurring at t = t1
v(t ) = V0u (t − t0 ) − V0u (t − t1 )
Exercise:
• Using unit step functions, write an expression
for the current i(t), that satisfies
i (t ) = 0,
= −2 A,
= 4 A,
= 0,
t < 10 ms
10 < t < 20 ms
20 < t < 40 ms
t > 40 ms
Exercise:
i2 (t ) = 6u (t − 0.02)
i1 (t ) = −2u (t − 0.01)
i3 (t ) = −4u (t − 0.04)
Exercise:
• Using unit step functions, write an expression
for the current i(t), that satisfies
i (t ) = 0,
= −2 A,
= 4 A,
= 0,
t < 10 ms
10 < t < 20 ms
20 < t < 40 ms
t > 40 ms
• Answer:
− 2u (t − 0.01) + 6u (t − 0.02) − 4u (t − 0.04) A
Exercise:
• Square wave
First Pulse
Second Pulse
Third Pulse
The Response of a First-Order Circuit
to a Nonconstant Source
• The differential equation of an RL or RC circuit:
dx(t )
+ ax(t ) = y (t )
dt
• Where y(t) is a constant if the source is
constant (DC) and α=1/τ the reciprocal of the
time constant.
• Using the integrating factor method, which
consists of multiplying the equation by a
factor that makes the left-hand side a perfect
derivative, and then integrating both sides.
The Response of a First-Order Circuit
to a Nonconstant Source
• The forced response was found to be dictated
by the form of y(t). dx(t )
+ ax(t ) = y (t )
dt
x f = e −αt ∫ e bt eαt dt
1
xf =
e −αt eα +b
α +b
x f = e −αt ∫ e (α +b ) t dt
e − bt
xf =
α +b
• The forced response of an RL or RC circuit is of
the same form as the forcing function itself.
The Response of a First-Order Circuit
to a Nonconstant Source
• Sources can be either:
 Aperiodic (nonperiodic) – non-repeating
e.g. exponential
 Periodic – repeats itself exactly after a fixed
length of time.
f (t + Τ) = f (t )
 Smallest positive number T that satisfies the equation is called
the period.
 If T has no value, the source is aperiodic.
The Response of a First-Order Circuit
to a Nonconstant Source
Special Case: Damped Exponential
Input : y (t ) = e − bt
e − bt
Response : x f =
a −b
For the special case where a=b, we have a-b=0, the form
of the response is indeterminate.
When the natural response contains a term of the same
form as the forcing function, we need to multiply the
assumed form of the forced response by t.
SAMPLE PROBLEMS
Ex. 1. Determining the Natural
Response of an RL Circuit
The switch in the circuit shown has been closed
for a long time before it is opened at t = 0. Find
a)
b)
c)
d)
iL(t) for t ≥ 0,
i0(t) for t ≥ 0+,
v0(t) for t ≥ 0+,
the percentage of the total energy stored in
the 2 H inductor that is dissipated in the 10 Ω
resistor.
Ex. 1. Determining the Natural
Response of an RL Circuit
Solution:
a) Find iL(t) for
t ≥ 0,
t
−
iL (t ) = I 0e τ
The switch has been closed for a long time prior
to t = 0, so we know the voltage across the
inductor must be zero at t = 0 . Therefore the
initial current in the inductor is 20 A at t = 0-.
Hence iL(0+) also is 20 A , because an
instantaneous change in the current cannot
occur in an inductor. Thus, I0 = 20 A.
Ex. 1. Determining the Natural
Response of an RL Circuit
Solution:
a) Find iL(t) for t ≥ 0+,
We replace the resistive circuit connected to the
terminals of the inductor with a single resistor of
10 Ω.
I0 = 20 A.
Req = 2 + (40 || 10) = 10 Ω
Time constant, τ = L/ Req = 0.2 s; 1/τ = 5
iL(t)=20e-5t A
Ex. 1. Determining the Natural
Response of an RL Circuit
Solution:
b) Find i0(t) for t ≥ 0+,
Using current division,
10
i0 = −iL
10 + 40
10
−5t
i0 = 20e
10 + 40
i0 = 4e −5t A
1
i0 = 20e  
5
−5t
Ex. 1. Determining the Natural
Response of an RL Circuit
Solution:
c) Find v0(t) for t ≥ 0+,
Using Ohm’s Law,
v0 = 40i0
(
v0 = 40 4e −5t
−5t
)
v0 = 160e V
Ex. 1. Determining the Natural
Response of an RL Circuit
Solution:
d) the percentage of
the total energy
stored in the 2 H inductor that is dissipated in
the 10 Ω resistor.
2
(
)
−5t 2
v 0 160e
p10 Ω (t ) =
=
10
10
p10 Ω (t ) = 2560e −10tW
∞
w10 Ω (t ) = ∫ 2560e −10t dt = 256 J
1 0 2 1
w(0) = Li = (2)(400)
2
2
w(0) = 400 J
256
×100 = 64%
400
Ex. 2. Determining the Natural
Response of an RC Circuit
The switch in the circuit shown in has been in
position x for a long time. At t = 0, the switch
moves instantaneously to position y.
Find
a) vc(t) for t ≥ 0,
b) v0(t) for t ≥ 0+,
c) i0(t) for t ≥ 0+,
d) the total energy dissipated in the 60 kΩ
resistor.
Ex. 2. Determining the Natural
Response of an RC Circuit
Solution:
a) Find vc(t) for t ≥ 0,
−t
vc (t ) = V0e τ
Because the switch has been in position x for a
long time, the 0.5 μF capacitor will charge to 100
V and be positive at the upper terminal. We can
replace the resistive network connected to the
capacitor at t = 0+ with an equivalent resistance of
80 kΩ. Hence the time constant of the circuit is
(0.5 x 10-6)(80 x 103) or 40 ms.
Thus,
−25t
vc (t ) = 100e
V ,t ≥ 0
Ex. 2. Determining the Natural
Response of an RC Circuit
Solution:
b) Find v0(t) for t ≥ 0+,
Using voltage division,
48
v0 = vc (t )
80
v0 = 60e
−25t
V ,t ≥ 0 +
Ex. 2. Determining the Natural
Response of an RC Circuit
Solution:
c) Find v0(t) for t ≥ 0+,
Using Ohm’s Law,
v0 (t )
i0 (t ) =
3
60 ×10
i0 (t ) = e −25t mA, t ≥ 0
Ex. 2. Determining the Natural
Response of an RC Circuit
Solution:
d) total energy
dissipated in the
60 kΩ resistor.
Power dissipated:
Total Energy dissipated:
∞
p60 kΩ (t ) = i 0 (t )(60 × 10 )
2
3
p60 kΩ (t ) = 60e −50t mW , t ≥ 0 +
w60 kΩ (t ) = ∫ i 2 0 (t )(60 ×103 )dt
0
w60 kΩ (t ) = 1.2mJ
Ex. 3. Determining the Step Response
of an RC Circuit
The switch in the circuit shown has been in
position a for a long time. At t = 0, the switch
moved to position b.
Find
a) vc(t) for t ≤ 0,
b) Final value of vc(t).
c) Time constant in position b.
d) Expression for vc(t) for t ≥ 0+.
e) Expression for i(t) for t ≥ 0+.
f) How long after the switch in position b does
the capacitor voltage equal zero.
g) Plot vc(t) and i(t) versus t.
Ex. 3. Determining the Step Response
of an RC Circuit
Solution:
a) Initial value of vc(t)
The switch has been in position a for a long t ime,
so the capacitor look s like an open circuit.
Therefore the voltage across the capacitor is the
voltage across the 60 Ω resistor. From the voltage
divider rule, the voltage across the 60 Ω resistor
is 40 x [60/ (60 + 20)] or 30 V. As the reference
for vc is positive at the upper terminal o f
the capacitor , we have vc(0) = -30 V.
Ex. 3. Determining the Step Response
of an RC Circuit
Solution:
b) Final value of vc(t)
After the switch has been in position b for a long
time, the capacitor will look like an open circuit
in terms o f the 90 V source. Thus the final value
of the capacitor voltage is +90 V.
Ex. 3. Determining the Step Response
of an RC Circuit
Solution:
c) Time constant
in position b.
τ = RC
τ = (400 ×10 )(0.5 ×10 )
3
τ = 0.2s
−6
Ex. 3. Determining the Step Response
of an RC Circuit
Solution:
d) Expression for vc(t) for t ≥ 0+.
v(t ) = Voc + (v(0) − Voc )e − t /( Rt C )
Voc = 90V
v(0) = −30V
τ = 0.2s
−5t
vc (t ) = 90 − 120e V , t ≥ 0
vc (t ) = 90 + (− 30 − 90 )e −5t