Assist .Prof. Sibel ÇİMEN
Electronics and Communication Engineering
University of Kocaeli
Course Book :
• Fundamentals of Electric Circuits, by Charles K. Alexander
and Matthew N. O. Sadiku, McGraw Hill; 3rd 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
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
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,
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
Bobinin uçlarından akan akımın exp. Karakteristiği olduğunu biliyoruz…
 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
1.3. The Source-Free Series
RLC Circuits
Critically Damped Case (∝= 𝝎𝟎 )
1.3. The Source-Free Series
RLC Circuits
Under Damped Case (∝< 𝝎𝟎 )
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,