CHAPTER 8: SECOND ORDER CIRCUITS
Second Oder-Circuits
 A second-order circuit is characterized by a second order differential equation. It consists
of resistors and the equivalent of two energy storage elements.
Finding Initial and Final Values
 There two key points to keep in mind in determining the initial conditions.
 First – is always in circuit analysis - we must be carefully handle the polarity of voltage
v(t) across the capacitor and the direction of current i(t) through the inductor. Keep in
mind that v and i are defined strictly according to the passive sign convention.
 Second – Keep in mind that the capacitor voltage is always continuous so that
Where t = 0- denotes the time just before a switching event and
t = 0+ is the time just after the switching event, assuming that the
switching event takes place at t = 0
EXAMPLE 8.1
The switch in the circuit below has been closed for a long time. It is open at
t = 0. Find: (a) i(0+), v(0+), (b) di(0+)/dt , dv(0+)/dt , (c) i(∞), v(∞).
8. 1
Solution:
(a) If the switch is closed a long time before t = 0, it means that the circuit has reached dc
steady state at t = 0. At dc steady state, the inductor acts like a short circuit, while the capacitor
acts like an open circuit, so we have the circuit in Fig.(a) at t = 0−. Thus,
(b) At t = 0+, the switch is open; the equivalent circuit is as shown in Fig. (b). The same current
flows through both the inductor and capacitor. Hence,
8. 2
(c) Fort > 0, the circuit undergoes transience. But as t →∞, the circuit reaches steady state
again. The inductor acts like a short circuit and the capacitor like an open circuit, so that the
circuit becomes that shown in Fig.(c), from which we have
EXAMPLE 8.2
In the circuit shown, calculate: (a) iL(0+), vC(0+), vR(0+),(b) diL(0+)dt , dvC(0+)/dt ,
dvR(0+)/dt , (c) iL(∞), vC(∞), vR(∞).
8. 3
Solution:
(a) For t < 0, 3u(t) = 0. At t = 0−, since the circuit has reached steady state, the inductor can be
replaced by a short circuit, while the capacitor is replaced by an open circuit as shown in Fig.(a).
From this figure we obtain
 For t > 0, 3u(t) = 3, so that the circuit is now equivalent to that in Fig.(b). Since the
inductor current and capacitor voltage cannot change abruptly,
 Although the voltage across the 4- resistor is not required, we will use it to apply KVL
and KCL; let it be called vo. Applying KCL at node a in Fig.(b) gives
8. 4
(Eq. 1)
Applying KVL to the middle mesh in Fig.(b) yields
Since vC(0+) = −20 V, it implies that
(Eq. 2)
Thus; eq. 1 substitute eq. 2
(b) Since L diL/dt = vL,
But applying KVL to the right mesh in Fig.(b) gives
Hence,
Similarly, since C dvC/dt = iC, then dvC /dt = iC /C. We apply KCL at node b in
Fig.(b) to get iC:
8. 5
Since vo(0+) = 4 and iL(0+) = 0, iC(0+) = 4/4 = 1 A. Then
To get dvR(0+)/dt , we apply KCL to node a and obtain
Taking the derivative of each term and setting t = 0+ gives
(Eq. 3)
We also apply KVL to the middle mesh in Fig.(b) and obtain
Again, taking the derivative of each term and setting t = 0+ yields
Substituting for dvC(0+)/dt = 2 gives
(Eq. 4)
From Eqs. 3 and 4, we get
8. 6
We can find diR(0+)/dt although it is not required. Since vR = 5iR,
(c) As t →∞, the circuit reaches steady state. We have the equivalent circuit in Fig.(a) except
that the 3-A current source is now operative. By current division principle,
The Source-Free Series RLC Circuit

8. 7
Consider the series RLC circuit shown. The circuit is being excited by the energy initially
stored in the capacitor and inductor. The energy is represented by the initial capacitor
voltage V0 and initial inductor current I0. Thus, at t = 0,

Applying KVL around the loop,

To eliminate the integral, we differentiate with respect to t and rearrange terms. We get

This is a second-order differential equation and is the reason for calling the RLC circuits
second-order circuits.
To solve such a second-order differential equation requires that we have two initial
conditions, such as the initial value of i and its first derivative or initial values of some i
and v. The initial value of i is given in Eq. 2. We get the initial value of the derivative of i
from Eqs. 2 and 3; that is,

Or
8. 8

Our experience in the preceding chapter on first-order circuits suggests that the solution
is of exponential form. So we let

Where A and s are constants to be determined. Substituting Eq.6 into Eq.4and carrying
out the necessary differentiations, we obtain
8. 9

Since i = Aest is the assumed solution we are trying to find, only the expression in
parentheses can be zero:

This quadratic equation is known as the characteristic equation of the differential Eq.4.
The two roots of Eq.8 are

A more compact way of expressing the roots is

Where

The roots s1 and s2 are called natural frequencies, measured in nepers per second
(Np/s), because they are associated with the natural response of the circuit; ω0 is known
as the resonant frequency or strictly as the undamped natural frequency, expressed in
radians per second (rad/s); and α is the neper frequency or the damping factor,
expressed in nepers per second. In terms of α and ω0, Eq.8 can be written as

The two values of s indicate that there are two possible solutions for i, each of which is
of the form of the assumed; that is,

Thus, the natural response of the series RLC circuit is

From Eq. 11, we can infer that there are three types of solutions:
1. If α > ω0, we have the overdamped case.
2. If α = ω0, we have the critically damped case.
3. If α < ω0, we have the underdamped case.
Overdamped Case (α > ω0)

When C > 4L/R2. When this happens, both roots s1 and s2 are negative and real. The
response is
Critically Damped Case (α = ω0)
 When α = ω0, C = 4L/R2 and
Underdamped Case (α < ω0)
 For α < ω0,C < 4L/R2. The roots may be written as
8. 10
8. 11
EXAMPLE 8.3
In the figure shown, R = 40Ω , L = 4 H, and C = 1/4 F. Calculate the characteristic roots
of the circuit. Is the natural response overdamped, underdamped, or critically damped?
Solution:
Figure 8.8
EXAMPLE 8.4
Find i(t) in the circuit shown. Assume that the circuit has reached steady state at
t = 0−.
8. 12
Solution:
8. 13
The Source-Free Parallel RLC Circuit
8. 14
 Consider the parallel RLC circuit shown above. Assume initial inductor current I0 and
initial capacitor voltage V0,
 Applying KCL at the top node gives
 Taking the derivative with respect to t and dividing by C results in
 We obtain the characteristic equation by replacing the first derivative by s and the second
derivative by s2. The characteristic equation is obtained as
 The roots of the characteristic equation are
8. 15
1. Overdamped Case (α > ω0)

When L > 4R2C. The roots of the characteristic equation are real and negative.
The response is
2. Critically Damped Case (α = ω0)

When α < ω0 , L < 4R2C. In this case the roots are complex and may be
expressed as
3. Underdamped Case (α < ω0)

8. 16
When α < ω0 , L < 4R2C. In this case the roots are complex and may be
expressed as
EXAMPLE 8.5
In the parallel circuit shown, find v(t) for t > 0, assuming v(0) = 5 V, i(0) =
0A, L = 1 H, and C = 10 mF. Consider these cases: R = 1.923Ω , R = 5Ω ,
and RΩ = 6.25Ω .
Solution:
8. 17
8. 18
8. 19
EXAMPLE 8.6
Find v(t) for t > 0 in the RLC circuit of shown.
8. 20
8. 21
Step Response of a Series RLC Circuit
 Step response is obtained by the sudden application of a dc source. Consider
the series RLC circuit shown above. Applying KVL around the loop for t >
0,
8. 22
 The natural response is the solution when we set Vs = 0 in Eq. (8.40). The natural
response vn for the overdamped, underdamped, and critically damped cases are:
 The forced response is the steady state or final value of v(t). In the circuit shown, the final
value of the capacitor voltage is the same as the source voltage Vs . Hence,
 Thus, the complete solutions for the overdamped, underdamped, and critically damped
cases are:
8. 23
EXAMPLE 8.7
For the circuit shown in Figure 8.19, find v(t) and i(t) for t > 0. Consider these cases:
R = 5Ω ,R = 4Ω , and R = 1Ω .
Solution:
8. 24
8. 25
8. 26
8. 27
8. 28
Step Response of a Parallel RLC Circuit
 Consider the parallel RLC circuit shown above. We want to find i due to a sudden
application of a dc current. Applying KCL at the top node for t > 0,
 The forced response is the steady state or final value of i. In the circuit shown, the final
value of the current through the inductor is the same as the source current Is . Thus,
8. 29
EXAMPLE 8.8
In the circuit shown, find i(t) and iR(t) for t > 0.
Solution:
8. 30
8. 31
REVIEW QUESTIONS
1. When a step input is applied to a second-order circuit, the final values of the circuit
variables are found by:
A. Replacing capacitors with closed circuits and inductors with open circuits.
B. Replacing capacitors with open circuits and inductors with closed circuits.
C. Doing neither of the above.
2. If the roots of the characteristic equation of an RLC circuit are -2 and -3 , the response is:
A. (A cos 2t + B sin 2t)e-3t
B. (A + 2Bt)e-3t
C. Ae-2t + Bte-3t
D. Ae-2t + Be-3t
3. In a series RLC circuit, setting R = 0 will produce:
A. An overdamped response
B. A critically damped response
C. An underdamped response
D. An undamped response
E. None of the above
4. A parallel RLC circuit has L = 2 H and C = .25 F. The value of R that will produce unity
damping factor is:
A. 0.5 Ω
B. 1 Ω
C. 2 Ω
D. 4 Ω
5. In an electric circuit, the dual of resistance is:
A. Conductance
B. Capacitance
C. Inductance
D. Open Circuit
E. Short Circuit
PROBLEM SOLVING
1. Refer to the circuit in Fig. 8.66. Determine:
(a) i(0+) and v(0+)
(b) di(0+)/dt and dv(0+)/dt
(c) i(∞) and v(∞)
8. 32
2. If R = 50 Ω, L = 1.5 H, what value of C will make an RLC series circuit:
(a) overdamped,
(b) critically damped,
(c) underdamped
3. The switch in Fig. 8.77 moves from position A to position B at t = 0 (please note that the
switch must connect to point B before it breaks the connection at A, a make-before-break
switch). Determine i(t) for t > 0.
4. Obtain v(t) and i(t) for t > 0 in the circuit of Fig. 8.84.
5. For the circuit in Fig. 8.101, find v(t) for t > 0. Assume that v(0+) = 4 V and i(0+) = 2 A.
8. 33