Assist .Prof. Aysun Taşyapı ÇELEBİ
Electronics and Communication Engineering
University of Kocaeli
Course Book :
• Fundamentals of Electric Circuits, by Charles K. Alexander
and Matthew N. O. Sadiku, McGraw Hill; 5rd edition
(2007)
Reference Books:
1) Electric Circuits, by James W. Nilsson and Susan Riedel,
Prentice Hall; 8th edition (2007)
2) Schaum's Outline of Electric Circuits, by Mahmood
Nahvi and Joseph Edminister, McGraw-Hill; 4th edition
(2002)
3) Introduction to Electric Circuits, by Richard C. Dorf and
James A. Svoboda, Wiley, 7th edition (2006)
4) Schaum's Outline of Basic Circuit Analysis, by John
O'Malley and John O'Malley, McGraw-Hill; 2nd edition
(1992)
Course Outline
1) Second Order DC Circuits (Fund. of Electric Circuits, CH 8)
2) Sinusoids and Phasors (Fund. of Electric Circuits, CH 9)
3) Sinusoidal Steady-State Analysis (Fund. of Electric Circuits,
CH 10)
4) AC Power Analysis (Fund. of Electric Circuits, CH 11)
5) Frequency Response (Fund. of Electric Circuits, CH 14)
6) Laplace Transform (Fund. of Electric Circuits, CH 15
Overview
• The previous chapter introduced the concept of first order
circuits.
• This chapter will expand on that with second order circuits:
those that need a second order differential equation.
• RLC series and parallel circuits will be discussed in this
context.
• The step response of these circuits will be covered as well.
• Finally the concept of duality will be discussed.
Second Order Circuits
• The previous chapter considered circuits which only
required first order differential equations to solve.
• However, when more than one “storage element”, i.e.
capacitor or inductor is present, the equations require
second order differential equations
• The analysis is similar to what was done with first
order circuits
• This time, though we will only consider DC
independent sources
1.1. Introduction
1.1. Introduction
Keep in mind;
 Capacitor voltage always continuous…
 Inductor current always continuous…
EXAMPLE 1.1.
• The switch in Figure has been closed for a long time. İt is
open at t=0.
• Find;
a) 𝑖(0+ ), 𝑣(0+ )
b) 𝑑𝑖((0+ )/dt, 𝑑𝑣((0+ )/dt,
c) 𝑖(∞) , 𝑣(∞)
EXAMPLE 1.1. /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.
8.(a) at t = 0−. Thus,
As the inductor current and the capacitor
voltage cannot change abruptly,
EXAMPLE 1.1. /Solution:
• (b) At t = 0+, the switch is open; the equivalent circuit is as
shown in Fig. 8 (b). The same current flows through both the
inductor and capacitor.
We now obtain 𝑣𝐿 by applying KVL to the loop
EXAMPLE 1.1. /Solution:
(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. 8(c), from which we have
Homework
EXAMPLE 1.2.
• In figure calculate;
a) 𝑖𝑙 (0+ ), 𝑣𝑐 (0+ ), 𝑣𝑅 (0+ )
b) 𝑑𝑖𝐿 ((0+ )/dt, 𝑑𝑣𝑐 ((0+ )/dt
c) 𝑖𝐿 ∞ , 𝑣𝑐 ∞ , 𝑣𝑅 (∞)
EXAMPLE 1.2. /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
EXAMPLE 1.2. /Solution:
• 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,
Applying KCL at node a in Fig. (b) gives
Applying KVL to the middle mesh in Fig.(b) yields
EXAMPLE 1.2. /Solution:
But applying KVL to the right mesh in Fig. (b) gives
EXAMPLE 1.2. /Solution:
• (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,
Homework
1.3. The Source-Free Series
RLC Circuits
(1.a)
(1.b)
1.3. The Source-Free Series
RLC Circuits
 Applying KVL around the loop;
(2)
 To eliminate the integral, we differentiate with respect to t and
rearrange terms:….
(3)
1.3. The Source-Free Series
RLC Circuits
 with initial values equation (2)…
(4)
or
Based on the first order solutions, we can expect that the solution will be in
exponential form.
 equation (3) becomes…
or
i
0
1.3. The Source-Free Series
RLC Circuits
(5)
Known as characteristic equation
İt’s roots;
Where;
1.3. The Source-Free Series
RLC Circuits
𝜔0 : resonant frequency or undamped natural frequency (rad/s)
∝: neper frequency or damping factor (Np/s)
İn terms of 𝜔0 𝑎𝑛𝑑 ∝ equation (5) gets…
(6)
The two values of s in Eq. (5) indicate that there are two possible solutions in
Eq. (6); that is,
Natural response of series RLC;
1.3. The Source-Free Series
RLC
Circuits
There are three types of solutions;
Overdamped Case (∝> 𝝎𝟎 )
İn this situation;
𝑠1 𝑎𝑛𝑑 𝑠2 negative and real
From this, we should not expect to see an
oscillation
1.3. The Source-Free Series
RLC Circuits
Critically Damped Case (∝= 𝝎𝟎 )
1.3. The Source-Free Series
RLC Circuits
Under Damped Case (∝< 𝝎𝟎 )
• ω0 is often called the undamped natural frequency
• ωd is called the damped natural frequency
Properties of RLC network
• Behavior of RLC network is described as damping, which is a
gradual loss of the initial stored energy
• The resistor R causes the loss
• a determined the rate of the damping response
• If R = 0, the circuit is loss-less and energy is shifted back and forth
between the inductor and capacitor forever at the natural frequency.
• Oscillatory response of a lossy RLC network is possible because
the energy in the inductor and capacitor can be transferred from
one component to the other.
• Underdamped response is a damped oscillation, which is called
ringing.
Properties of RLC network
• Critically damped circuits reach the final steady state in the
shortest amount of time as compared to overdamped and
underdamped circuits.
• However, the initial change of an overdamped or underdamped
circuit may be greater than that obtained using a critically
damped circuit.
EXAMPLE 1.3.
R=40 Ω, L=4H and C=1/4 F. Calculate the characteristic
roots of the circuit. Is the natural response overdamped,
underdamped, or critically damped?
Solution:
Since α > ω0, we conclude that the response is overdamped. This is also
evident from the fact that the roots are real and negative.
EXAMPLE 1.4.
Find i(t) in the circuit. Assume that the circuit
has reached steady state at t=0− .
Solution:
For t < 0, the switch is closed. The capacitor acts like an open circuit while the inductor
acts like a shunted circuit. The equivalent circuit is shown in Fig. (a). Thus, at t = 0,
EXAMPLE 1.4./ Solution
• For t > 0, the switch is opened and the voltage source is
disconnected. The equivalent circuit is shown in Fig. 8.11(b),
which is a source-free series RLC circuit.
Hence, the response is underdamped (α < ω); that is,
EXAMPLE 1.4./ Solution
• We now obtain 𝐴1 and 𝐴2 using the initial conditions. At t = 0,
Homework