Advanced Electrical Circuits
Analysis ( CCE 102 )
Prof. Dr. Fahmy El-khouly
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•
•
•
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Weighting of assessments:
-Quizzes
20 (Degrees)
20 %
Activities
20 (degrees)
20 %
-Mid-Term Exam 20 (Degrees) 20 %
- Final-Term Exam
40
(Degrees) 40 %
Total
100 (Degrees) 100 %
• 6- List of References :
• “ Electric Circuits “ James Nilsson and Susan Riedel, Ninth
Edition
COURSE OUTLINE
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Chapter1 : Response of First-Order RL and
RC Circuit
Chapter 2 : Sinusoidal Steady-State Analysis
Chapter 3 : Active Filter Circuits
Chapter 4: Two-Port Circuits
‫‪Title‬‬
‫‪date‬‬
‫‪15-2-2016‬‬
‫‪22-2-2016‬‬
‫‪29-2-2016‬‬
‫‪7-3-2016‬‬
‫‪14-3-2016‬‬
‫‪21-3-2016‬‬
‫‪28-3-2016‬‬
‫أسبوع امتحانات منتصف الفصل‬
‫‪4-4-2016‬‬
‫‪11-4-2016‬‬
‫‪18-4-2016‬‬
‫عطلة تحرير سيناء‬
‫‪25-4-2016‬‬
‫عطلة شم النسيم‬
‫‪2-5-2016‬‬
‫‪9-5-2016‬‬
‫أسبوع امتحانات نهاية الفصل‬
‫‪16-5-2016‬‬
Intended Learning Outcomes (ILOs):
A : Knowledge and understanding:
On completing this course, students will be able to:
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a1-Explain transients in reactive circuits (inductive and capacitive).
a2- Describe the Response of RL and RC Circuits
a3- Identify the meaning of active, reactive, and apparent power
a4- Define the different theories that can be applied to AC electrical circuits
B-Intellectual
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Skills: ‫مهارات ذهنية‬
b1- Analyze of AC circuits using Vectors.
b2- Solve AC circuits and calculate active, reactive, and apparent power.
b3- Solve AC circuits using circuits' theorems, Mesh and Nodal analysis
b4- Calculate the response of first order RL-RC circuits.
C-Professional and Practical Skills ‫مهارات مهنية وعملية‬
c1- Develop skills and practice in the design and fabrication of practical AC
circuits used in electrical systems.
c2- Design resonance circuits.
D- General and Transferable Skills ‫مهارات عامة‬
d1- Use IT effectively.
d2- Work in a team or individually
d3- Communicate effectively using written, oral,
graphical, and presentational skills
Chapter 1
Response of First-Order RL and RC Circuit
1-The Natural Response of an RL Circuit
the natural response is the currents and voltages that arise when stored
energy in an inductor or capacitor is suddenly released to a resistive
network. This happens when the inductor or capacitor is abruptly
disconnected from its dc source.
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The switch is closed
for long time
Then Is reach its final
value Is=constant
Then VL=0 because
di/dt=0
Then VRo=VR=0
Then IRo=IR=0
Then iL=Is
When switch is opened
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If we use 0- to denote the time just prior to
switching, and 0+ for the time immediately
following switching, then
Natural response of an RL circuit is:
We derive the power dissipated in the resistor from
any of the following expressions:
The energy delivered to the resistor during any interval of
time after the switch has been opened is
The Significance of the Time
Constant
Using the time-constant concept, we write the expressions for current,
voltage, power, and energy as
a long time: implies that five
or more time constants have
elapsed. also means the time
it takes the circuit to reach its
steady-state value.
the time constant is
that it gives the time required for the
current to reach its final value if the
current continues to change at its initial
rate
−𝑡
𝑑𝑖
1
= − 𝐼𝑜 𝑒 τ
𝑑𝑡
τ
When i=0
t= τ
The above Equation
indicates that i would
reach its final value of
zero in τ seconds
Example 1
Determining the Natural Response of an RL Circuit
The switch in the circuit shown in Fig. 7.7 has
been closed for a long time before it is opened at
t = 0. Find
a) iL(t)
for
t > 0,
b) io(t)
for
t > 0+,
c) v0(t)
for
t > 0+,
d) the percentage of the total energy stored in the
2 H inductor that is dissipated in the 10 Ω resistor.
Solution
(a) the voltage across the inductor must be zero at t =
0-. Therefore the initial current in the inductor is 20 A
at t = 0-. Hence, IL(0+) also is 20 A
= 0.2
(b)
(c)
(d)
The initial energy stored in the 2 H inductor is
Therefore the percentage of energy dissipated in the 10 Ω resistor
is
Example 2
In the circuit shown in Fig. 7.8, the initial currents in inductors L1 and L2
have been established by sources not shown. The switch is opened at t = 0.
a) Find i1 , i2, and i3 for t ≥ 0.
b) Calculate the initial energy stored in the parallel inductors.
c) Determine how much energy is stored in the inductors as t —> ∞.
d) Show that the total energy delivered to the resistive network equals the
difference between the results obtained in (b) and (c).
10∗15
+
10+15
10∗40
𝑅𝑒𝑞 =
50
𝑅𝑥 =
4 = 10
=8
Solution
𝐿𝑒𝑞 =
20∗5
20+5
=4
( a ) the initial value of i(t) is 12 A and the time constant is 4/8, or 0.5 s.
Therefore
(b)
(d)
2- The Natural Response
of an RC Circuit
Example 3
The switch in the circuit shown in Fig. 7.13 has been
in position x for a long time. At t = 0, the switch
moves instantaneously to position y. Find
a) vc(t) for t > 0,
b) Vo(t) for t > 0+,
c) i0(t) for t > 0+, and
d) the total energy dissipated in the 60 kΩ resistor.
Solution
a)
capacitor will charge to 100 V and be positive at the upper
terminal. We can replace the resistive network connected to
the capacitor at t = 0+ with an equivalent resistance of 80 kΩ.
Hence the time constant of the circuit is (0.5 X 10-6)(80 X 103)
or 40 ms. Then,
b)
c)
d)
3- The Step Response of RL
and RC Circuits
when either dc voltage or current sources are suddenly applied. The
response of a circuit to the sudden application of a constant voltage or
current source is referred to as the step response of the circuit.
3-1 The Step Response of an RL Circuit
If the current were to continue to increase at its initial rate,
it would reach its final value at t= τ that is, because
If the current were to continue to increase at this rate
at t= τ
The voltage across an inductor is Ldi/dt,
When the initial inductor current is zero
Example
The switch in the circuit shown in Fig. 7.19 has been in position a for a long
time. At t = 0, the switch moves from position a to position b. The switch is a
make-before-break type; that is, the connection at position b is established
before the connection at position a is broken, so there is no interruption of
current through the inductor.
a) Find the expression for i(t) for t ≥ 0.
b) What is the initial voltage across the inductor just after the switch has been
moved to position b?
c) How many milliseconds after the switch has been moved does the inductor
voltage equal 24 V?
d) Does this initial voltage make sense in terms of circuit behavior?
e) Plot both i(t) and v(t) versus t.
solution :
a) Io
is -8 A
When the switch is in position b, the final value of i will be 24/2, or
12 A. The time constant of the circuit is 200/2, or 100 ms.
.
c) Yes; in the instant after the switch has been moved to
position b, the inductor sustains a current of 8 A
counterclockwise around the newly
formed closed path. This current causes a 16 V drop across the
2 Ω resistor. This voltage drop adds to the drop across the
source, producing a 40 V drop across the inductor.
d) We find the time at which the inductor voltage equals 24 V by
solving the expression
Differential equation form
Solution form
Differential equation form
Solution form
The Step Response of an RC Circuit
For current
A General Solution for Step and Natural Responses
where the value of the constant K can be zero
Because the sources in the circuit are constant voltages and/or currents, the
final value of x will be constant
when x reaches its final value, the derivative dx/dt must be zero. Hence
where xf represents the final value of the variable.
When computing the step and natural responses of circuits, it
may help to follow these steps:
1. Identify the variable of interest for the circuit. For RC circuits,
it is most convenient to choose the capacitive voltage; for RL
circuits, it is best to choose the inductive current.
2. Determine the initial value of the variable, which is its value at
t o.
Note that if you choose capacitive voltage or inductive current
as your variable of interest, it is not necessary to
distinguish between t = to- and t = to+. This is because they
both are continuous variables. If you choose another
variable, you need to remember that its initial value is
defined at t = to+.
3. Calculate the final value of the variable, which is its value as t
—> ∞.
4. Calculate the time constant for the circuit.
Example
The switch in the circuit shown in Fig. 7.22 has been in position 1
for a long time. At t = 0, the switch moves to position 2. Find
a) v0(t) for t > 0 and
b) io{t) for t > 0+.
Solution
a) The switch has been in position 1 for a long time,
so the initial value of v0 is 40(60/80), or 30 V
we find the Norton equivalent with respect to the terminals of the capacitor
for t > 0. To do this, we begin by computing the open-circuit voltage, which is
given by the -75 V source divided across the 40 kΩ and 160 k Ω resistors:
The value of the Norton current
source is the ratio of the opencircuit voltage to the Thevenin
resistance, or -60/(40 X 103) = 1.5 mA.