Chapter 7
EGR 260 – Circuit Analysis
Reading Assignment: Chapter 7 in Electric Circuits, 9th Ed. by Nilsson
Chapter 7 – Circuit Analysis with Capacitors and Inductors
KVL and KCL involving circuits with capacitors and inductors result in
differential equations (D.E.) rather than algebraic equations.
The order of a differential equation is equal to highest derivative.
dx
1st-order DE:
+ a 0 x(t) = f(t)
dt
d2x
dx
2nd-order DE:
+ a1
+ a 0 x(t) = f(t)
2
dt
dt


dn x
d n-1x
dx
nth-order DE:
+ a n-1 n-1 +    + a1
+ a 0 x(t) = f(t)
n
dt
dt
dt
1
Chapter 7
EGR 260 – Circuit Analysis
2
Circuit Order
Order
of a
circuit
=
Order of the
differential equation
(DE) required to
describe the circuit
=
The number of
independent* energy
storage elements
(C’s and L’s)
* C’s and L’s are independent if they cannot be combined with other C’s and
L’s (in series or parallel, for example)
Examples: Draw several circuits with inductors and capacitors and determine
the order of each circuit.
Chapter 7
EGR 260 – Circuit Analysis
3
First-Order Circuits
A first-order circuit can only contain one energy storage element (a capacitor or
an inductor). The circuit will also contain resistance (discuss). So there are
two types of first-order circuits:
• RC circuit
• RL circuit
Source-Free Circuits
A source-free circuit is one where all independent sources have been
disconnected from the circuit after some switch action. The voltages and
currents in the circuit typically will have some transient response due to initial
conditions (initial capacitor voltages and initial inductor currents). We will
begin by analyzing source-free circuits as they are the simplest type. Later we
will analyze circuits that also contain sources after the initial switch action.
Chapter 7
EGR 260 – Circuit Analysis
4
Source-free RC circuit
Consider the RC circuit shown below. Note that it is source-free because no
sources are connected to the circuit for t > 0. Use KCL to find the differential
equation:
dv
1

v(t)  0 for t  0
t=0
+
dt
RC
VX
+
_
R
C
v (t)
_
and solve the differential
equation to show that:
v(t)  VX e
-t
RC
for t  0
Chapter 7
EGR 260 – Circuit Analysis
Checks on the solution
• Verify that the initial condition is satisfied.
• Find dv(t)/dt and regenerate the D.E. from the solution.
• Show that the energy dissipated over all time by the resistor equals the initial
energy stored in the capacitor.
5
Chapter 7
EGR 260 – Circuit Analysis
General form of the D.E. and the response for a 1st-order source-free circuit
In general, a first-order D.E. has the form:
dx
1
 x(t)  0 for t  0
dt

Solving this differential equation (as we did with the RC circuit) yields:
x(t)  x(0)e
-t

for t  0
where  = (Greek letter “Tau”) = time constant (in seconds)
Notes concerning  :
dv
1
1) for the previous RC circuit the DE was:

v(t)  0 for t  0
dt
RC
so
 = RC
(for an RC circuit)
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EGR 260 – Circuit Analysis
Chapter 7
7
2)  is related to the rate of exponential decay in a circuit as shown below
x(t)  x(0)e
-t

Smaller  (faster decay)
x(0)
Larger  (slower decay)
t
0
0
3) It is typically easier to sketch a response in terms of multiples of  than to be
concerning with scaling of the graph (otherwise choosing an appropriate
scale can be difficult). This is illustrated on the following page.
Chapter 7
EGR 260 – Circuit Analysis
8
Graphing functions in terms of  :
Illustration:
A) Calculate values for x(t) = x(0)e-t/ for t = , 2, 3, 4, and 5.
x(t) = x(0)e-t/
t
B) Graph x(t) versus t for t = 0, , 2, 3, 4, and 5.
x(t)
0

2
3
4
5
t
Chapter 7
EGR 260 – Circuit Analysis
9
Time for a circuit to completely decay
From the last page note that x(5) = x(0)e-5/ = x(0)e-5 = 0.007x(0)
This means that magnitude of x(t) is only 0.7% of its initial value by time 5 (or
the function has lost 99.3% of its original value). Technically a decaying
exponential function never reaches zero, but we see that by time t = 5 it is very
close. So we generally use the approximation that:
5 = time for a circuit to decay
Example: Some circuits connect a “bleeder resistor” in parallel with a
capacitor when the circuit is turned off in order to safely discharge the capacitor
(which might otherwise have a significant voltage across it for a long time).
For the circuit shown below, what value of Rbleeder should be used in order to
discharge the capacitor in 10 seconds (the circuit is turned off at time t = tx)?
t = tx
Circuit
t = tx
470uF
Rbleeder
Chapter 7
EGR 260 – Circuit Analysis
Example: The switch in the circuit
shown had been closed for a long time
and then opened at time t = 0.
50V
A) Determine an expression for v(t).
B) Graph v(t) versus t.
10
t=0
+
_
15k
+
10 uF
v (t)
_
Chapter 7
EGR 260 – Circuit Analysis
Example: (continued)
C) How long will it take for the capacitor to completely discharge?
D) Determine the capacitor voltage at time t = 100 ms.
E) Determine the time at which the capacitor voltage is 10V.
11
Chapter 7
EGR 260 – Circuit Analysis
12
Equivalent Resistance seen by a Capacitor
For the RC circuit in the previous example, it was determined that  = RC. But
what value of R should be used in circuits with multiple resistors?
In general, a first-order RC circuit has the following time constant:
 = REQ C
where Req is the Thevenin resistance seen by the capacitor.
More specifically,
R EQ = R seen from the terminals of the capacitor for t > 0
with independent sources killed
Circuit
C
Circuit
t>0
independent
sources killed
REQ
Chapter 7
EGR 260 – Circuit Analysis
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Example: Determine an expression for v(t). Graph v(t) versus t.
50V
+
_
20
t=0
30
90
+
40
10 mF
v (t)
_
Chapter 7
EGR 260 – Circuit Analysis
14
Source-free RL circuit
Consider the RL circuit shown below. Use KVL to find the differential
equation:
t=0
di
R

i(t)  0 for t  0
i(t)
dt
L
IX
R
L
and use the general form of the
solution to a first-order D.E. to
show that:
 = L/R
i(t)  IX e
-tR
L
for t  0
Chapter 7
EGR 260 – Circuit Analysis
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Equivalent Resistance seen by an Inductor
For the RL circuit in the previous example, it was determined that  = L/R. As
with the RC circuit, the value of R should actually be the equivalent (or
Thevenin) resistance seen by the inductor.
In general, a first-order RL circuit has the following time constant:
 =
L
R EQ
where
R EQ = R seen from the terminals of the inductor for t > 0
with independent sources killed
Circuit
L
Circuit
t>0
independent
sources killed
REQ
Chapter 7
EGR 260 – Circuit Analysis
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Example: Determine an expression for i(t). Sketch i(t) versus t.
t=0
100 V
+
_
50
150
i(t)
10 mH
75
Chapter 7
EGR 260 – Circuit Analysis
17
First-order circuits with DC forcing functions:
In the last class we consider source-free circuits (circuits with no independent
sources for t > ). Now we will consider circuits having DC forcing functions
for t > 0 (i.e., circuits that do have independent DC sources for t > 0).
The general solution to a differential equation has two parts:
x(t) = xh + xp = homogeneous solution + particular solution
or
x(t) = xn + xf = natural solution + forced solution
where xh or xn is due to the initial conditions in the circuit
and xp or xf is due to the forcing functions (independent voltage and current
sources for t > 0).
xp or xf in general take on the “form” of the forcing functions, so DC sources
imply that the forced response function will be a constant (DC), sinusoidal
sources imply that the forced response will be sinusoidal, etc.
Since we are only considering DC forcing functions in this chapter, we assume
that
xf = B (a constant)
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EGR 260 – Circuit Analysis
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Recall that a 1st-order source-free circuit had the form Ae-t/ . Note that there
was a natural response only since there were no forcing functions (sources) for t
> 0. So the natural response was
xn = Ae-t/
The complete response for 1st-order circuit with DC forcing functions therefore
will have the form x(t) = xf + xn or
x(t) = B + Ae-t/
The “Shortcut Method”
An easy way to find the constants B and A is to evaluate x(t) at 2 points. Two
convenient points at t = 0- and t =  since the circuit is in steady-state at these
two points. This approach is sometimes called the “shortcut method.”
So, x(0) = B + Ae0 = B + A
And x() = B + Ae- = B
Show how this yields the following expression found in the text:
x(t) = x() +[x(0) - x()]e-t/
Chapter 7
EGR 260 – Circuit Analysis
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“Shortcut Method” - Procedure
The shortcut method will be the key method used in this chapter to analyze 1storder circuit with DC forcing functions:
1)
2)
3)
4)
Analyze the circuit at t = 0-: Find x(0-) = x(0+), where x = vC or iL .
Analyze the circuit at t = : Find x().
Find  = REQC or  = L/REQ .
Assume that x(t) has the form x(t) = B + Ae-t/ and solve for B and
A using x(0) and x().
Notes:
• The “shortcut method” also works for source-free circuits, but x() = B = 0
since the circuit is dead at t = .
• If variables other than vC or iL are needed, it is generally easiest to solve for
vC or iL first and then use the result to find the desired variable.
Chapter 7
EGR 260 – Circuit Analysis
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Example: Find v(t) and i(t) for t > 0.
t=0
15
i(t)
5
240V
_+
+
60
20
10 uF
v(t)
_
Chapter 7
EGR 260 – Circuit Analysis
21
Example: Find v(t) and i(t) for t > 0.
20
t=0
40 V
+
_
300
i(t)
+
+
_
100 V
75
3 mH
v(t)
_
EGR 260 – Circuit Analysis
Chapter 7
22
Charging and discharging capacitors – the forms of these two simple circuits
are commonly referred to in later courses.
Example: Capacitor charging circuit.
Example: Capacitor discharging
circuit. Find v(t) for t > 0.
Find v(t) for t > 0.
t=0
VX
+
_
R
t=0
+
C
v (t)
_
VX
+
_
+
R
C
v (t)
_
Chapter 7
EGR 260 – Circuit Analysis
23
1st-order Circuits with Dependent Sources
Dependent sources affect the resistance seen by the inductor or capacitor and
therefore affect the value of  for the circuit.
Two approaches can be used to find :
1) When REQ seen by an inductor or a capacitor, remove the L or C, kill any
independent sources, and place any value independent source at the
terminals. Then
terminal voltage
R EQ 
terminal current
2) Write a DE for the circuit (any variable) and  can be easily determined from
the DE since it has the form:
dx
1
 x(t)  f(t)
dt

Chapter 7
EGR 260 – Circuit Analysis
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Example: Find v(t) for t > 0 if v(0) = 2 V.
R EQ 
A) Use method 1: Find  = REQC using
+ 20
18 V
+
_
3v(t)
40
0.1 F
+
v (t)
_
terminal voltage
terminal current
Chapter 7
EGR 260 – Circuit Analysis
25
Example: (continued) Find v(t) for t > 0 if v(0) = 2 V.
B) Use method 2: Write a DE for v(t).
+ 20
18 V
+
_
3v(t)
40
0.1 F
+
v (t)
_
EGR 260 – Circuit Analysis
Chapter 7
26
Unit Step Functions
Unit step functions have several important uses in electrical engineering,
including:
• representing piecewise-continuous signals
• representing switches
• defining functions for use with one-sided Laplace transforms (in EGR 261)
1
where u(t)  
0
Definition: u(t) = unit step function
and u(t) is represented by the graph shown below.
u(t)
1
t
0
for t  0
for t  0
Chapter 7 EGR 260 – Circuit Analysis
A good way to think of a unit step function is as follows:
• u(argument) = 1 for argument > 0
• u(argument) = 0 for argument < 0
• the transition in u(argument) occurs where argument = 0
Example: Graph 4u(t - 2)
27
The function can also be described as follows:
u(t)
4
4u(t - 2)  
0
4
for t  2
for t  2
t
0
u(argument) = 0 for argument < 0.
Note that for t < 2,
u(t - 2) has a negative argument.
For example, when t = 1,
u(t - 2) = u(1 - 2) = u(-1) = 0.
2
u(argument) = 1 for argument > 0.
Note that for t > 2,
u(t - 2) has a positive argument.
For example, when t = 3,
u(t - 2) = u(3 - 2) = u(+1) = 1.
The transition occurs when the argument = 0.
Note that when t = 2, u(t - 2) = u(2 - 2) = u(0).
Chapter 7
EGR 260 – Circuit Analysis
Example: Graph the following functions:
1) -2u(t - 10)
2) 3u(t + 2)
3) 4u(-t)
4) 4u(2-t)
28
Chapter 7
EGR 260 – Circuit Analysis
Example: Graph the following functions (continued):
5) 6u(-4-t)
6) u(-t)
7) Show that 1 - u(t) = u(-t)
8) sin(t) and sin(t)u(t)
29
Chapter 7
EGR 260 – Circuit Analysis
Example: Graph the following functions (continued):
9) u(t) - u(t - 2)
10) f(t) = 2t
11) 2t[u(t) - u(t - 2)] - discuss the concept of a “window”
30
Chapter 7
EGR 260 – Circuit Analysis
Example: Graph the following functions (continued):
12) f(t)[u(t - 2) - u(t - 4)] for any f(t)
13) (2t + 6)[u(t + 2) - u(t - 2)]
31
Chapter 7
EGR 260 – Circuit Analysis
There are two common types of problems in representing functions using unit
step functions:
1) Determining the function that represents a given graph
Approach: Represent each unique portion of the function using unit step
“windows”
2) Graphing a function specified by unit steps
Approach: As each unit step function “turns on”, graph the cumulative
function.
32
Chapter 7
EGR 260 – Circuit Analysis
Determining the function that represents a given graph:
Approach: Represent each unique portion of the function using unit step
“windows”
Example:
1) Represent f(t) shown below using unit step functions.
f(t)
12
6
0
-6
2
4
t
33
Chapter 7
EGR 260 – Circuit Analysis
34
Example:
2) Represent f(t) shown below using unit step functions.
f(t)
60
20
0
8
16
24
t
Chapter 7
EGR 260 – Circuit Analysis
35
Graphing a function specified using unit steps functions:
Approach: As each unit step function “turns on”, graph the cumulative
function (i.e., add each part as it turns on).
Example:
1) Graph f(t) = 2u(t) + 4u(t – 2) – 8u(t – 4)
Solution: f(t) = 2u(t) + 4u(t – 2) – 8u(t – 4)
turns ON turns ON
at t = 0
at t = 2
turns ON
at t = 4
f(t)
6
 0
 2

so f(t)  
2  4  6
6 - 8  -2
t0
0t2
2t4
2
4t
2
-2
4
t
Chapter 7
EGR 260 – Circuit Analysis
2) Graph f(t) = (t + 2)u(t + 1) + (3 – t)u(t – 2) - (2t-5)u(t – 4)
3) Graph f(t) = 4sin(4t)[u(t) – u(t – 1)]
36
EGR 260 – Circuit Analysis
Chapter 7
37
Using unit step functions to replace switches in circuits
Unit step functions are commonly used to represent switches in circuits.
Consider the following examples.
Example: A unit step function is used below to replace a switch connecting a
voltage source.
• Discuss the value of VX.
• Discuss the forms of the solution.
t=0
+
10 V
+
_
VX
R
C
_
v(t) =
R
+
10(1 - e -t/RC ) V, t > 0
0,
t<0
v (t)
10u(t) V
+
_
+
C
v (t)
_
_
v(t) = 10(1 - e-t/RC )u(t) V
Chapter 7
EGR 260 – Circuit Analysis
38
Example: A unit step function can be used to replace a switch disconnecting a
current source.
• Discuss the concept of a “make before break” switch
• Discuss the value of IX.
• Draw two possible circuits using unit step functions that are equivalent to the
circuit shown below.
t=0
IX
+
2A
R
C
v (t)
_
Chapter 7
EGR 260 – Circuit Analysis
39
Example: Determine an expression for v(t) in the circuit below. Use the
“shortcut method.”
_
+
10
40
15
2u(t) A
10 V
20
+
5mF
+
_
v (t)
_
20u(-t) V
Chapter 7
EGR 260 – Circuit Analysis
40
Unit step response to a circuit
Suppose that you wished to compare the outputs of two circuits. It would be
misleading to compare them if the circuits had different inputs or different
initial conditions. A common way to compare them is to use a unit step input
[1u(t)V or 1u(t)A] and zero initial conditions (or zero initial stored energy).
Unit step response – the output of a circuit where the input is a unit step and
there are zero initial conditions.
Circuit with zero
initial conditions
Input is
1u(t) V
or
1u(t) A
Output (a voltage or
a current) is the
“unit step response”
Chapter 7
EGR 260 – Circuit Analysis
Example: Find the unit step response for VC(t) in the circuit shown below.
1k
v(t)
_+
+
1mF
VC(t)
_
41