ECS 203 - Part C - For CPE2
Asst. Prof. Dr.Prapun Suksompong
April 22, 2015
CHAPTER 10
First-Order Circuits
We have been assuming that the circuits under our consideration have
reached their steady-states. This assumption has helped us analyze circuits
under dc conditions in Chapter 6 (see Examples 6.2.11, 6.4.13, and 6.4.14)
and perform ac analysis in Chapters 7, 8, and 9.
In this chapter and the next chapter, we relax the steady-state assumption. This means we will need to return to solving differential equations.
We will start with circuits that contain only one (equivalent) capacitor or
inductor. This is enough to see how the voltage or current behaves during
the charging/discharging of these storage elements.
10.1. Introduction and a Mathematical Fact
10.1.1. In this chapter, we will examine two types of simple circuits
with a storage element:
(a) A circuit with a resistor and one capacitor (called an RC circuit);
and
(b) A circuit with a resistor and an inductor (called an RL circuit).
These circuits may look simple but they find applications in electronics,
communications, and control systems.
10.1.2. Applying Kirchhoff’s laws to the RC and RL circuits produce
first order differential equations. Hence, the circuits are collectively
known as first-order circuits.
10.1.3. There are two ways to excite the circuits.
(a) By initial conditions of the storage elements in the circuit.
• Also known as source-free circuits
• Assume that energy is initially stored in the capacitive or inductive element.
• This is the discharging process.
(b) By using independent sources
• This is the charging process
• For this chapter, we will consider independent dc sources.
125
126
10. FIRST-ORDER CIRCUITS
Before we start our circuit analysis, it is helpful to consider one mathematical fact which we will use throughout this chapter:
10.1.4. The solution of the first-order differential equation
d
x(t) = ax + b
dt
is given by
b
b
− .
(10.10)
x(t) = ea(t−t0 ) x(t0 ) +
a
a
This result takes a bit of effort to proof. However, if you have MATLAB, you
can get 10.10 using one line of code:
dsolve(’Dx = a*x+b’,’x(t0) = x0’, ’t’)
In addition, if a < 0, the solution is given by
(10.11)
x(t) = ea(t−t0 ) (x(t0 ) − x(∞)) + x(∞).
where x(∞) = limt→∞ x(t).
When a < 0, from (10.10), x(∞) = −b/a. We can then get (10.11) by
substituting b/a in (10.10) with x(∞).
10.1.5. It turns out that you only have two kinds of plots to worry
about when dealing with (10.11):
10.2. SOURCE-FREE RC CIRCUITS
127
10.2. Source-Free RC Circuits
Definition 10.2.1. A source-free RC circuit occurs when its dc source
is suddenly disconnected. The energy already stored in the capacitor is
released to the resistor.
iC
C
+
v
iR
R
–
10.2.2. Consider a series combination of a resistor an initially charged
capacitor. We assume that at time t = 0, the initial voltage is v(0) = Vo .
(Hence, the initial stored energy is w(0) = 12 CVo2 .)
Applying KCL, we get
i C + iR = 0
dv
v
C +
= 0
dt R
dv
1
v
= −
dt
RC
128
10. FIRST-ORDER CIRCUITS
Hence,
(10.12)
t
t
v(t) = Vo e− RC = Vo e− τ ,
where τ = RC.
This shows that the voltage response of RC circuit is an exponential
decay of the initial voltage.
• Since the response is due to the initial energy stored and the physical
characteristics of the circuit and not due to some external voltage
or current source, it is called the natural response of the circuit.
10.2.3. Remarks:
(a) τ = RC is the time constant which is the time required for the
response to decay to 36.8 percent of its initial value. (e ≈ 2.718 and
hence 1/e ≈ 0.368.)
(b) The smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response.
10.2. SOURCE-FREE RC CIRCUITS
129
(c) Energy Consideration:
(i) Capacitor:
−2t
−2t
1 2
1 −t 2 1
wC (t) = Cv (t) = C V0 e τ
= CV02 e τ = wC (0)e τ
2
2
2
(ii) Resistor:
iR (t) =
v(t) Vo −t
= eτ
R
R
Vo2 −2t
pR (t) = v(t) × iR (t) =
eτ
R
Z t
−2t
1
wR (t) =
p(µ)dµ = CVo2 (1 − e τ ) = wC (0) − wC (t)
2
0
• Notice that as t → ∞, wR → 12 CV02 , which is the same as
wC (0), the energy initially stored in the capacitor.
• The energy that was initially stored in the capacitor is eventually dissipated in the resistor.
(d) The voltage v(t) is less than 1 percent of V0 after 5τ (five time
constants). Thus, it is customary to assume that the capacitor is
fully discharged (or charged) after five time constants. In other
words, it takes 5τ for the circuit to reach its final state or steady
state when no changes take place with time.
10.2.4. In summary: The key in working with a source-free RC circuit
is to determine the voltage across the capacitor which can be found from :
1. the initial voltage v(0) = V0 .
2. the time constant τ (= RC)
3. Equation (10.12)
130
10. FIRST-ORDER CIRCUITS
10.2.5. There are basically three extensions of what we found:
(a) The initial condition v(t0 ) can be given at non-zero t0 . This has
already been taken care of by (10.11).
(b) There can be more than one resistors. In which case, find the equivalent resistance at the capacitor terminals. See Example 10.2.6.
(c) The initial value of the voltage v(t0 ) might not be given but condition(s) of the circuit before time t0 is given instead. See Example 10.2.8. In which case, we may have to consider the long-term
(steady-state) behavior of the circuit before t0 . In such situation,
recall two properties of capacitors studied in 6.2.7:
(i) For DC, capacitor becomes open circuit.
(ii) Voltage across the capacitor cannot change instantaneously.
Example 10.2.6. Let vC (0) = 15 V. Determine vC (t), vx (t), and ix (t)
for t ≥ 0.
8Ω
ix
5Ω
0.1 F
+
vC
–
12 Ω
+
vx
–
10.2. SOURCE-FREE RC CIRCUITS
131
Example 10.2.7. Circuit with switch(es):
Example 10.2.8. The switch in the circuit below has been closed for a
long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the
initial energy stored in the capacitor.
3Ω
20 V
t=0
9Ω
1Ω
+
v
–
20 mF
132
10. FIRST-ORDER CIRCUITS
10.3. Source-Free RL Circuits
10.3.1. Consider the series connection of a resistor and an inductor.
i
L
–
vL
+
+
vR
–
R
Assume that the inductor has an initial current Io or i(0) = Io . (Hence,
the energy stored in the inductor is w(0) = 12 LIo2 .)
Applying KVL, we get
vL + vR = 0
di
L + Ri = 0
dt
di R
+ i = 0
dt L
From (10.10),
(10.13)
i(t) = Io e
−Rt
L
−t
= Io e τ .
Note that:
(a) τ = RL is the time constant which is the time required for the
response to decay to 36.8 percent of its initial value.
(b) The energy dissipated at the resistor can be found by
−t
vR (t) = i(t)R = Io Re τ
−2t
pR (t) = vR (t) × i(t) = Io2 Re τ
Z t
−2t
1
wR (t) =
p(µ)dµ = LIo2 (1 − e τ ) = wL (0) − wL (t).
2
0
10.3. SOURCE-FREE RL CIRCUITS
133
10.3.2. In summary: The key in working with a source free RL circuit
is to determine the current through the inductor which can be found from:
(a) the initial current i(0) = Io .
(b) the time constant τ (= RL ).
(c) Equation (10.13)
10.3.3. Similar extensions of the results to the three cases discussed in
(10.2.5) are also possible.
Example 10.3.4. The switch in the circuit below has been closed for a
long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.
2Ω
t=0
4Ω
i(t)
40 V
12 Ω
16 Ω
2H
Example 10.3.5. Determine i, io and vo for all t in the circuit. Assume
that the switch was closed for a long time.
3Ω
t=0
i
1H
i0
18 A
4Ω
2Ω
+
v0
–
134
10. FIRST-ORDER CIRCUITS
10.4. Unit step function
10.4.1. Mathematically, we use the step function(s) to model abrupt
change(s) in voltage or current.
Definition 10.4.2. The unit step function u(t) is 0 for negative values
of t and 1 for positive values of t. In mathematical terms,
0, t < 0
u(t) =
1, t > 0
In [Alexander and Sadiku], the unit step function is undefined at t = 0.
u(t)
1
0
t
Example 10.4.3. A voltage source of V0 u(t)
t=0
a
V0u(t)
a
V0
(a)
b
(b)
b
10.4. UNIT STEP FUNCTION
135
Example 10.4.4. A current source of I0 u(t)
i
t=0
a
a
I0u(t)
I0
b
(a)
b
(b)
10.4.5. We can also use the step functions to represent the gate function
(pulse) which may be regarded as a step function that switches on at one
value of t and switches off at another value of t.
Example 10.4.6. The gate function below switches on at t = 2 s and
switches off at t = 5 s.
v(t)
10
0
1
2
3
4
5
t
It consists of the sum of two unit step functions:
v(t) = 10u(t − 2) − 10u(t − 5).
10u(t – 2)
–10u(t – 5)
10
10
0
1
2
t
0
1
2
3
4
5
t
–10
Definition 10.4.7. When the dc source of an RC circuit is suddenly
applied (i.e., this happens when the capacitor is being charged), the voltage
or current source can be modeled as a step function, and the response is
known as a step response.
136
10. FIRST-ORDER CIRCUITS
10.5. Step Response of an RC Circuit
In the previous sections, we have been considering RC and RL circuits
with no source. Here, we start to look at the cases with dc sources.
10.5.1. Consider an RC circuit with voltage step input below:
R
t=0
+
VS
C
–
v
The voltage across the capacitor is v. We assume an initial voltage1
v(0) = V0 on the capacitor.
For t ≥ 0, applying KCL, we have
C
dv v − Vs
+
=0
dt
R
dv
v
Vs
+
=
dt RC RC
dv
1
Vs
= −
v+
dt
RC
RC
From (10.10), the solution is
−t
v(t) = Vs + (Vo − Vs )e τ ,
t ≥ 0.
1Since the voltage of a capacitor cannot change instantaneously v(0− ) = v(0+ ) = V , where v(0− ) is
0
the voltage across the capacitor just before switching and v(0+) is its voltage immediately after switching.
10.5. STEP RESPONSE OF AN RC CIRCUIT
137
v(t)
VS
V0
0
t
10.5.2. In general, the step response of an RC circuit whose switch
changes position at time t = 0 can be written as
(10.14)
−t
v(t) = v(∞) + (v(0) − v(∞))e τ ,
t > 0.
where v(0) is the initial voltage and v(∞) is the final or steady-state value.
We obtain
(a) v(0) from the given circuit for t < 0 and
(b) v(∞) and τ from the circuit for t > 0.
10.5.3. If the switch changes position at time t = to instead of at t = 0,
the response becomes
(10.15)
v(t) = v(∞) + (v(to ) − v(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial value at t = to .
10.5.4. The key in finding the step response of an RC circuit is to
determine the voltage across the capacitor which can be found from:
1. the initial capacitor voltage v(0) or v(t0 ),
2. the final capacitor voltage v(∞),
3. the time constant τ ,
4. Equations (10.14) or (10.15).
10.5.5. Extension:
138
10. FIRST-ORDER CIRCUITS
Example 10.5.6. The switch in the circuit below has been in position
A for a long time. At t = 0, the switch moves to B. Determine v(t) for
t > 0 and calculate its value at t = 1 s and 4 s.
3 kΩ
A
B 4 kΩ
t=0
24 V
+
v
–
5 kΩ
0.5 mF
30 V
Example 10.5.7. Find v(t) for t > 0 in the circuit below. Assume the
switch has been open for a long time and is closed at t = 0. Numerically
evaluate v(t) at t = 0.5.
t=0
2Ω
6Ω
10 V
+
v
–
1
-F
3
5V
3. [Alexander and Sadiku, 2009, Q7.7] Assuming that the switch in Figure 3 has been in
position A for a long time and is moved to position B at t = 0, find vo(t) for t  0.
Figure 3
4. [F2010] Consider the circuit in Figure 4 below. Assume the switch has been at position 1 for
a long time and moves to position 2 at t = 0 sec.
R1 Ω
R2 Ω
1
2
Vs1
t=0
Vs2
+
v(t)
-
C
Figure 4
Let
Vs1 = 5 V, Vs2 = 0 V, R1 = 6 , R2 = 3 , and C = 10 F.
(a) (3 pt) Find v(0-). Do not forget to justify your answer.
(b) (1 pt) Find v(0). Do not forget to justify your answer.
(c) (4 pt) Find v(t) for t > 0.
5. [F2010] Consider the circuit in Figure 5 below. Assume the switch has been at position 1 for
a long time and moves to position 2 at t = 5 sec.
R1 Ω
1
2
Vs1
t=5
R2 Ω
Vs2
C
+
v(t)
-
Figure 5
Let
Vs1 = 16 V, Vs2 = 8 V, R1 = 3 , R2 = 5 , and C = 8 F.
(a)
(b)
(c)
(d)
(3 pt) Find v(0).
(2 pt) Find v(5).
(4 pt) Find v(t).
(1 pt) Evaluate v(t) at t = 7.
10.5. STEP RESPONSE OF AN RC CIRCUIT
139
Example 10.5.8. An electronic flash unit provides a common example
of an RC circuit. This application exploits the ability of the capacitor
to oppose any abrupt change in voltage. A simplified circuit is shown
below. It consists essentially of a high-voltage dc supply, a current-limiting
large resistor R1 , and a capacitor C in parallel with the flashlamp of low
resistance R2 .
R1
1
i
High
voltage
dc supply
2
vS
R2
C
+
–
v
• When the switch is in position 1, the capacitor charges slowly due
to the large time constant (τ = R1 C). The capacitor voltage rises
gradually from zero to Vs , while its current decreases gradually from
I1 = Vs /R1 to zero.
• With the switch in position 2, the capacitor voltage is discharged.
The low resistance R2 of the photolamp permits a high discharge
current with peak I2 = Vs /R2 in a short duration.
This simple RC circuit provides a short-duration, high current pulse.
Such a circuit also finds applications in electric spot welding and the radar
transmitter tube.
9. Consider the circuit in Figure 9 below. Let
Vs = 10 V, R1 = 30 k, R2 = 10 k, and C = 4 F.
Figure 9
Assume that the switch has been in position 1 during time t < 0. Then, during time t  0 the
switch changes its position five times: at t1 = 0 ms, t2 = 25 ms, t3 = 50 ms, t4 = 75 ms, t5 = 100
ms.
(At time t1, the switch changes to position 2. At time t2, the switch changes back to position 1.
At time t3, the switch changes again to position 2….)
Plot the voltage v(t) for time t > 0.
Hint: You should have v(t5)  4.59 V.
140
10. FIRST-ORDER CIRCUITS
Example 10.5.9. An RC circuit can be used to provide various time
delays. Consider the circuit below. It basically consists of an RC circuit
with the capacitor connected in parallel with a neon lamp. The voltage
source can provide enough voltage to fire the lamp.
R1
+
110 V
–
S
R2
C
0.1 µF
70 V
Neon
lamp
• When the switch is closed, the capacitor voltage increases gradually
toward 110 V at a rate determined by the circuit’s time constant,
(R1 + R2 )C. The lamp will act as an open circuit and not emit light
until the voltage across it exceeds a particular level, say 70 V.
• When the voltage level is reached, the lamp fires (goes on), and
the capacitor discharges through it. Due to the low resistance of
the lamp when on, the capacitor voltage drops fast and the lamp
turns off. The lamp acts again as an open circuit and the capacitor
recharges.
• By adjusting R2 , we can introduce either short or long time delays
into the circuit and make the lamp fire, recharge, and fire repeatedly
every time constant τ = (R1 + R2 )C, because it takes a time period
τ to get the capacitor voltage high enough to fire or low enough to
turn off.
• The warning blinkers commonly found on road construction sites
are one example of the usefulness of such an RC delay circuit.
10.6. STEP RESPONSE OF AN RL CIRCUIT
141
10.6. Step Response of an RL Circuit
Finding step response of an RL circuit follows almost the same procedure as finding step response of an RC circuit. However, here, we use the
inductor current i as the circuit response.
10.6.1. In general, the step response of an RL circuit whose switch
changes position at time t = 0 can be written as
−t
i(t) = i(∞) + (i(0) − i(∞))e τ ,
t > 0.
where i(0) is the initial current and i(∞) is the final or steady-state value.
If the switch changes position at time t = to instead of at t = 0, the
response becomes
i(t) = i(∞) + (i(to ) − i(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial inductor current value at time t = to .
Example 10.6.2. Find i(t) in the circuit below for t > 0. Assume that
the switch has been closed for a long time.
t=0
2Ω
3Ω
i
10 V
1
-H
3
142
10. FIRST-ORDER CIRCUITS
Example 10.6.3. The switch in the circuit below has been closed for a
long time. It opens at t = 0. Find i(t) for t > 0.
i
5Ω
1.5 H
10 Ω
t=0
9A
Example 10.6.4. At t = 0, switch 1 in the circuit below is closed, and
switch 2 is closed 4 s later. Find i(t) for t > 0.
4Ω
S1 t = 0
6Ω
P
i
S2
t=4
40 V
2Ω
10 V
5H