K. A. Saaifan, Jacobs University, Bremen
8. Basic RL and RC Circuits
This chapter deals with the solutions of the responses of RL and RC circuits
The analysis of RC and RL circuits leads to a linear differential equation
This chapter covers the following topics
The Source-Free RL Circuit
The Source-Free RC Circuit
The Unit-Step Function
Driven RL Circuit
Driven RC Circuit
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K. A. Saaifan, Jacobs University, Bremen
First Order Linear Differential Equations
A first order homogeneous linear differential equation is one of the form
dy
py=0
dt
where "Linear'' indicates that both dy/dt and y occur to the first power and
"homogeneous'' refers to the zero on the right hand side of the equation
In circuit analysis, y can either be the voltage v or the current i of the circuit
The solution of a homogeneous linear differential equation is called a
complementary function
●
In circuit analysis, we refer to the solution of the circuit as a natural response
or a transient response
K. A. Saaifan, Jacobs University, Bremen
A Direct Approach
Since the variables can be separated, the differential equation can be rewritten as
dy
=−p dt
y
We integrate both sides as
dy
∫ y =−p∫ d t
Then, we have
ln yt=−p tA
The constant of integration must be selected to satisfy the initial condition y(0)=Y0
ln Y 0 =A
Thus, we obtain
ln yt=−p tln Y 0
−p t
y t=Y 0 e
for t0
ln yt−ln Y 0 =−p t
ln
y t
=−p t
Y0
K. A. Saaifan, Jacobs University, Bremen
A General Solution Approach
We assume a general solution of y(t) in exponential form
y t=Ae
s1 t
where A and s1 are constant to be determined
Substitute the exponential solution into the differential equation
dy
py=0
dt
s1 t
s1 t
As1 e p Ae =0
where (s1 +p) is the characteristic equation
Determine the value of s1
 s1p =0
s1 =−p
s t
 s1p  Ae =0
1
The characteristic equation
dy
py=0
dt
sypy=sp y=0
Invoke the initial condition to determine the remaining constant A
y 0=Y 0 =Ae
s1 0
A=Y 0
The final form is
y t=Y 0 e−p t
for t0
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K. A. Saaifan, Jacobs University, Bremen
8.1 The Source-Free RL Circuit
We assume a series RL circuit for which i(t) to be determined
Apply KVL
R ivL =0
di R
 i=0
dt L
-
We will solve the natural response
it=Ae
+
s1 t
The characteristic equation
s
R
i=0
L
s1 =−
R
L
Using the initial condition i(0)=I0, we have
s1 0
i0=I 0 =Ae
A=I 0
The natural response is given as
it=I 0 e
−R
t
L
for t0
K. A. Saaifan, Jacobs University, Bremen
Example: RL with a switch
We have two different circuits: one with the switch
closed and one with the switch open
We are asked to find v(0.2) for the circuit shown
in Figure (c)
From Figure (b), we compute the current iL
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iL = =2.4 A
10
this current is used as initial conditions when the
switch is opened
Write the differential equation of the circuit (c)
di
40 iL 10 iL5 L =0
dt
diL
10 iL =0
dt
The general solution of i(t) in exponential form
st
it=Ae
Since i(0)=2.4 A, the solution is
1
it=2.4e−10 t
for t0
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K. A. Saaifan, Jacobs University, Bremen
The Exponential Response
The expression for the current in an RL series
circuit describes the natural response of the
inductor
−
it=I 0 e
R
t
L
+
The current decreases exponentially with time
The L/R term in the above equation is known
commonly as the time constant, τ, of the RL
series circuit
−
it=I 0 e
t

Power in The RL Series Circuit
The power being dissipated in the resistor is
2
−2
2
0
p R=vR it=R i t=RI e
t

The total energy is found as
∞
wR=∫0 pR dt=RI
1
= LI 20
2
2
0
∞
−2
∫0 e
t

dt
The time constant is
τ=L/R
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K. A. Saaifan, Jacobs University, Bremen
8.2 The Source-Free RC Circuit
We assume a series RL circuit for which i(t) to be determined
Apply KVL
iC
v
iC =−
R
dv
v

=0
dt RC
We will solve the natural response
vt=Ae
s1 t
The characteristic equation
1
1
 v=0
s1 =−
RC
RC
Using the initial condition v(0)=V0, we have
s
s1 0
v0=V 0 =Ae
A=V 0
The natural response is given as
vt=V 0 e
−1
t
RC
for t0
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K. A. Saaifan, Jacobs University, Bremen
The Exponential Response
The expression for the current in an RC series
circuit describes the natural response of the
inductor
−
vt=V 0 e
1
t
RC
The voltage decreases exponentially with time
The 1/RC term in the above equation is known
commonly as the time constant, τ, of the RC
series circuit
−
vt=V 0 e
t

Power in The RC Series Circuit
The power being dissipated in the resistor is
2
t
v2 t V 0 −2 RC
p R=viR t=
=
e
R
R
The total energy is found as
t
∞
V 20 ∞ −2 RC
wR=∫0 pR dt= ∫0 e
dt
R
1
= CV 20
2
The time constant is
τ=RC
K. A. Saaifan, Jacobs University, Bremen
Determine the inductor voltage v in the circuit for t > 0.
For t<0, l0 V appears across the 4 Ω
resistor, so a dc current of iL= l0/4 = 2.5 A
flows through the inductor (which acts as a
short circuit)
For t>0, the battery is removed so we write
the simple KVL equation:
diL
diL
6 iL4 iL5
=0
2 iL =0
dt
dt
Thus, we can represent the circuit with the
equation
it=iL 0e−2 t
=2.5 e−2 t A
Finally, the voltage is
vt=L
d it
=52.5−2e−2 t =−25 e−2 t V
dt
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K. A. Saaifan, Jacobs University, Bremen
Noting carefully how the circuit changes once the switch in the circuit is
thrown, determine v(t) at t = 0 and at t = 160 μs
Before the switch is thrown, the 80Ω resistor is connected only by one of its
terminals and therefore may be ignored (i=0)
With no current flow permitted through the capacitor, we know v(0)=50 V
since the capacitor voltage cannot change in zero time
After the switch is thrown, the only remaining circuit is a simple source-free
RC circuit. With τ = RC =160 μs
vt=v 0e
−t

=50e−6250 t V
Finally, the voltage is
v160 s=50 e−1 =18.39 V
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K. A. Saaifan, Jacobs University, Bremen
8.3 A More General Perspective
General RL Circuits
The time constant of a single-inductor circuit will be τ=L/Req where Req is the
resistance seen by the inductor
Example: Req=R3+R4+R1R2/ (R1+R2)
General RC Circuits
The time constant of a single-capacitor circuit will be τ=ReqC, where Req is the
resistance seen by the capacitor
Example: Req=R2+R1R3/ (R1+R3)
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K. A. Saaifan, Jacobs University, Bremen
At t=0.15 s in the circuit, find the value of (a) iL; (b) i1; (c) i2
For t< 0, (the switch is open)
−
i2 0 =0
2
=0.4 A
28
8
iL 0−=2
=1.6 A
28
For t>0, 100% of the 2-A source contributes to i2, The 8-Ω
resistor is shorted out so i1=0
i1 0−=2
Thus
i2 t=2−iL t
where iL t=iL 0e
−t

Finally, the currents are
,
=
L 0.4
−

= =0.2 s , and iL 0=iL 0 =iL 0 =1.6 A
R eq 2
−0.15
0.2
iL 0.15=1.6 e
=755.6 mA
i2 0.15=2−iL 0.15=1.244 A
K. A. Saaifan, Jacobs University, Bremen
1st Order Response Observations
The voltage on a capacitor or the current through an inductor is the same
prior to and after a switch at t=0
Resistor current (or voltage) prior to the switch i(0-) can be different from
the voltage after the switch i(0+)
All voltages and currents in an RC or RL circuit follow the same natural
response e-t/τ
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K. A. Saaifan, Jacobs University, Bremen
8.4 The Unit-Step Function
So far, we have been studying the natural response of RL and RC circuits (when no
sources or forcing functions were present)
In other words, we have been solving problems in which energy sources are
suddenly removed from the circuit
We shall consider that type of response which results when energy sources are
suddenly applied to a circuit
The unit-step function u(t) is a convenient notation to represent change:
{
ut= 0 t0
1 t0
{
ut−t0 = 0
1
tt0
tt0
K. A. Saaifan, Jacobs University, Bremen
Switches and Steps
In order to obtain an exact equivalent for the voltage-step forcing function, we
may provide a single-pole double-throw switch
The exact equivalent for the current-step forcing function, we may replace this
circuit by a dc source in series with a switch
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K. A. Saaifan, Jacobs University, Bremen
Modeling Pulses Using u(t)
By manipulating the unit-step forcing function, we can generate many useful
forcing functions
A rectangular voltage pulse
{
0
vt= V 0
0
tt0
t0 tt1
tt1
The two unit steps u(t−t0) and −u(t−t1) are needed to obtain the rectangular
voltage pulse
A pulsed sinewave
vt=V m ut−t0 −u t−t1 sin 2 f t
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K. A. Saaifan, Jacobs University, Bremen
8.5 Driven RL Circuit
Now, we consider the behavior of a simple RL network to the sudden
application of a dc source
The shown circuits represent an RL circuit subjects to a voltage-step forcing
function V0u(t)
Applying Kirchhoff’s voltage law
di
iRL =V 0 ut
dt
Since i(t)=0 for t<0, we study the solution for t>0
iRL
di
=V 0
dt
Then, we have
di
L
=dt
V 0 −iR
t0
t0
Integrating both sides
−L
ln  V 0−Ri =tK
R
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K. A. Saaifan, Jacobs University, Bremen
invoke the initial condition to find K (i(t)=0 for t<0)
−L
ln V 0 =K
R
and hence
L
−  ln V 0 −Ri−ln V 0 =t
R
Rearranging
−R
t
L
Which can be rewritten as
ln V 0 −Ri−ln V 0 =
or
t
V 0 V 0 −R
L
it= −
e
R R
−R
t
V0
it=
1−e L u t
R
ln
V 0 −Ri −R
=
t
V0
L
t0
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K. A. Saaifan, Jacobs University, Bremen
The expression for the voltage in an RL series circuit describes the energizing
characteristics of the inductor
Forced response
it=
V

0
R
Natural response
−V


e
0
−R
t
L
R
The Natural response: The exponential term has the functional form of the
natural response of the RL circuit; it is a negative exponential, it approaches
zero as time increases, and it is characterized by the time constant L/R
The Forced response: It is the response that is present a long time after the
switch is closed
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K. A. Saaifan, Jacobs University, Bremen
A General Solution Approach
The solution of any linear differential equation may be expressed as the sum of
two parts: the complementary solution (natural response) and the particular
solution (forced response)
The inhomogeneous linear differential equation has the form of
V
dy
di R
p y=f t
 i= 0 ut
dt
dt L
L
or
dyp y dt=f tdt
Forced response
The solution is given as
particular solution

y=e
∫ f te dt
− pt
pt
complementary solution

Ae
−pt
it=
V

0
R
Natural response
−V


e
0
R
We note that, when f(t) is zero (a source-free circuit), the solution is the
natural response
−pt
y n t=Ae
Since f(t)=F, the particular solution leads to the following forced response
F
y f t=e−pt∫ F e pt dt =
p
−R
t
L
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K. A. Saaifan, Jacobs University, Bremen
Example: RL with a forcing function
Applying KVL yields
V0
di R
 i=
ut
dt L
L
First, we evaluate the natural response by solving
di R
 i=0
dt L
s t
We assume in t=Ae , where A and s1 are constant
to be determined
1
The characteristic equation
R
R
s i=0
s1 =−
L
L
−R
Then, we have
in t=Ae
L t
Next, we determine the forced response if t=
Finally, the complete solution is i(t)=in(t)+if(t)
−R
V0
t
it= Ae L
R
where
V0
V0
i0=0= A
A=−
R
R
V0
R
K. A. Saaifan, Jacobs University, Bremen
Determine i(t) for all values of time in the circuit
We note that the circuit contains a dc voltage source as well as a step-voltage
source
Using superposition, we solve the circuit for each source alone
We compute the current due to only a dc voltage source
idc t=
50
=25 A
2
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K. A. Saaifan, Jacobs University, Bremen
We compute the current due to only a a step-voltage source
istep voltage t=if tin t
−2 t
=25−25 e
The complete response
−2 t
it=idc tistep voltage t=50−25e
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K. A. Saaifan, Jacobs University, Bremen
Find the current response in a simple series RL circuit when
the forcing function is a rectangular voltage pulse of
amplitude V0 and duration t0
We represent the forcing function as the sum of two stepvoltage sources V0u(t) and -V0u(t-t0)
Using superposition, assume i1(t) designate the part of i(t)
due to V0u(t) acting alone, and i2(t) represents that part
due to -V0u(t-t0) acting alone
it=i1 ti2 t
We solve the response i1(t) as
−R
V0
L t
i1 t=
1−e 
R
We solve the response i2(t) as
t0
−R
−V 0
t−t 
i2 t=
1−e L

t 00
R
We now add the two solutions, but do so carefully, since
each is valid over a different interval of time
0
K. A. Saaifan, Jacobs University, Bremen
{
it=
i1 t
i1 ti2 t
0tt0
tt0
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K. A. Saaifan, Jacobs University, Bremen
8.6 Driven RC Circuit
Similar to RL circuits, the complete response of an RC circuit consists of the
natural and the forced response
Since the procedure is virtually identical to what we have already discussed in
detail for RL circuits, we consider a relevant example for a driven RC circuit
Find the capacitor voltage vC(t) and the current i(t) in the 200 Ω resistor for
all time
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K. A. Saaifan, Jacobs University, Bremen
For t<0, the circuit has two separate loops
The capacitor acts as open circuits
50
−
vC 0 =120
=100 V
1050
The current in the 200 Ω resistor
1
i0− =50
=192 mA
60200
For t>0
The complete response of the capacitor is
vC t=vCf tvCn t
The natural response vCn(t)
vCn t=Ae
−t
R eq C
where Req =50∣∣200∣∣60=24 
The forced response is
50∣∣200
vCf t=50
=20 V
6050∣∣200
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K. A. Saaifan, Jacobs University, Bremen
The complete response is
vC t=20Ae
−t
1.2
We use vC 0 =vC 0=vC 0 =100 V , then
−

100=20A
A=80
Thus, we have
vC t=2080 e
−t
1.2
t0
and
vC t=100
t0
The current in the 200 Ω resistor
−t
vC t
it=
=100400 e 1.2 mA
200
it=192 mA
t0
t0
K. A. Saaifan, Jacobs University, Bremen
Homework Assignment 6
P9.7, P9.10, P9.20, P9.22, P9.24, P9.27, P9.29, P9.35, P9.38, P9.40, P9.44,
P9.46, P9.48, P9.52, P9.54, P9.57, P9.61 and 9.65
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