Advanced Electrical Circuits Analysis ( CCE 102 )

Prof. Dr. Fahmy El-khouly

• 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

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

date Title

15-2-2016

22-2-2016

29-2-2016

7-3-2016

14-3-2016

21-3-2016

28-3-2016

4-4-2016 امتحانات منتصف الفصلأسبوع

11-4-2016

18-4-2016

25-4-2016 عطلة تحرير سيناء

2-5-2016 عطلة شم النسيم

9-5-2016

16-5-2016 امتحانات نهاية الفصلأسبوع

Intended Learning Outcomes (ILOs):

A : Knowledge and understanding:

On completing this course, students will be able to:

– 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 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 individuallyd3- Communicate effectively using written, oral,

graphical, and presentational skills

Chapter 1

Response of First-Order RL and RC 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.

1-The Natural Response of an RL Circuit

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

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

The above Equation

indicates that i would

reach its final value of

zero in τ seconds

𝑑𝑖

𝑑𝑡= −

1

τ𝐼𝑜𝑒

−𝑡τ

When i=0

t= τ

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).

𝑅𝑥 =10∗15

10+15+ 4 = 10

𝑅𝑒𝑞 =10∗40

50= 8

Solution

( a ) the initial value of i(t) is 12 A and the time constant is 4/8, or 0.5 s.

Therefore

𝐿𝑒𝑞 =20∗5

20+5= 4

(b)

(d)

2- The Natural Response

of an RC Circuit

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+, andd) the total energy dissipated in the 60 kΩ resistor.

Example 3

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

3-1 The Step Response of an RL Circuit

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.

If the current were to continue to increase at its initial rate,

it would reach its final value at t= τ that is, because

at t= τ

If the current were to continue to increase at this rate

The voltage across an inductor is Ldi/dt,

When the initial inductor current is zero

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.

Example

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

to.

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.

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) iot) for t > 0+.

Example

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 open-

circuit voltage to the Thevenin

resistance, or -60/(40 X 103) = -

1.5 mA.